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Copyright 2012 Cengage Learning
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Copyright 2012 Cengage Learning
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Tenth Edition

Copyright 2012 Cengage Learning
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May not be copied, scanned, or duplicated, in whole or in part
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Copyright 2012 Cengage Learning
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Tenth Edition

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States

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A First Course in Differential
Equations with Modeling
Applications, Tenth Edition

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Contents

Preface

ix

Projects

P-1

1
1
...
2

Initial-Value Problems

1
...
1

Solution Curves Without a Solution
2
...
1

Direction Fields

2
...
2

Autonomous First-Order DEs

36

36

2
...
3

Linear Equations

2
...
5

Solutions by Substitutions

2
...
1

Linear Models

3
...
3

Modeling with Systems of First-Order DEs

84
95

Chapter 3 in Review

106

113

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●

4
4
...
1
...
1
...
1
...
2

Reduction of Order

4
...
4

Undetermined Coefficients—Superposition Approach

4
...
6

Variation of Parameters

156

4
...
8

Green’s Functions

129
139
149

169

4
...
1

Initial-Value Problems

4
...
2
4
...
10 Nonlinear Differential Equations
Chapter 4 in Review

180

185

190

5
5
...
1
...
1
...
1
...
1
...
2

Linear Models: Boundary-Value Problems

5
...
1

Review of Power Series

232

6
...
3

Solutions About Singular Points

247

6
...
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1

Definition of the Laplace Transform

7
...
2
...
2
...
3

Inverse Transforms

274

Transforms of Derivatives

Operational Properties I

281
284

289

7
...
1

Translation on the -Axis

290

7
...
2
7
...
4
...
4
...
4
...
5

The Dirac Delta Function

7
...
1

Preliminary Theory—Linear Systems

8
...
2
...
2
...
2
...
3

Complex Eigenvalues

342

334

Nonhomogeneous Linear Systems
8
...
1

Variation of Parameters

348

Undetermined Coefficient

8
...
2
8
...
1

Euler Methods and Error Analysis

9
...
3

Multistep Methods

9
...
5

Second-Order Boundary-Value Problems

363

368
373

Chapter 9 in Review

375
380

384

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●

I

Gamma Function

II

Matrices

III

Laplace Transforms

APP-1

APP-3
APP-21

Answers for Selected Odd-Numbered Problems
Index

ANS-1

I-1

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Preface

Authors of books live with the hope that someone actually
them
...
True, the topics covered in the text are
chosen to appeal to instructors because they make the decision on whether to use it
in their classes, but everything written in it is aimed directly at you, the student
...
Think of it as a
book
...
Before attempting any of the exercises, work
the examples in a section; the examples are constructed to illustrate what I consider the most important aspects of the section, and
therefore, reflect the procedures necessary to work most of the problems in the exercise sets
...
Try working it, then compare your
results against the solution given, and, if necessary, resolve any differences
...
This may not be easy, but that is part of
the learning process
...

Specifically for you, a
(
) is available as an
optional supplement
...
Bear in mind you do
not have to purchase the
; by following my pointers given at the beginning of
most sections, you can review the appropriate mathematics from your old precalculus
or calculus texts
...
I hope you enjoy the text and
the course you are about to embark on—as an undergraduate math major it was one
of my favorites because I liked mathematics with a connection to the physical
world
...
taylor@cengage
...
The
longer version of the textbook,
, can be used for either a one-semester course, or a two-semester
course covering ordinary and partial differential equations
...
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For a one semester course, I assume that the students have successfully completed at
least two semesters of calculus
...
You will not
find a “suggested syllabus” in this preface; I will not pretend to be so wise as to tell
other teachers what to teach
...
The textbook strikes a reasonable balance between the analytical, qualitative, and quantitative approaches to the study of differential equations
...

For those who are familiar with the previous editions, I would like to mention a
few of the improvements made in this edition
...
Each project includes
a related problem set, and a correlation of the project material with a section
in the text
...
In
like manner, some exercise sets have been improved by sending some problems into retirement
...

• Several instructors took the time to e-mail me expressing their concerns
about my approach to linear first-order differential equations
...
3, Linear Equations, has been rewritten with the intent to simplify
the discussion
...
Section 4
...

• Section 5
...

• At the request of users of the previous editions, a new section on the review
of power series has been added to Chapter 6
...
In particular, the discussion of the
modified Bessel functions and the spherical Bessel functions in Section 6
...


•

(
), prepared by Warren S
...
Wright (ISBN 9781133491927 accompanies
and
ISBN 9781133491958 accompanies
), provides important review material from
algebra and calculus, the solution of every third problem in each exercise
set (with the exception of the Discussion Problems and Computer Lab
Assignments), relevant command syntax for the computer algebra systems
and
, lists of important concepts, as well as helpful
hints on how to start certain problems
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Wright and
Carol D
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is an online instructor database that offers complete,
worked-out solutions for all exercises in the text, allowing you to create
customized, secure solutions printouts (in PDF format) matched exactly to
the problems you assign in class
...
cengage
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Included is a test bank with hundreds of questions customized directly to the text, with all questions also provided in PDF and Microsoft
Word formats for instructors who opt not to use the software component
...
Available for this title, Enhanced WebAssign allows you to assign,
collect, grade, and record assignments via the Web
...
Enhanced WebAssign is more than a homework system—it is
a complete learning system for students
...
Many thanks to Molly
Taylor (senior sponsoring editor), Shaylin Walsh Hogan (assistant editor), and Alex
Gontar (editorial assistant) for orchestrating the development of this edition and its
component materials
...
Ed Dionne (project manager, MPS) worked tirelessly to provide top-notch
publishing services
...
Once again an especially heartfelt thank you to Leslie Lahr, developmental
editor, for her support, sympathetic ear, willingness to communicate, suggestions,
and for obtaining and organizing the excellent projects that appear at the front of
the text
...
J
...
Lewis,
Michael Olinick,
Finally, over the years these textbooks have been improved in a countless number of ways through the suggestions and criticisms of the reviewers
...


William Atherton,
Philip Bacon,
Bruce Bayly,
William H
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G
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Brown,
David Buchthal,
Nguyen P
...
Chow,
Dominic P
...
Crooke,
Bruce E
...
Davis,
Richard A
...
Edmondson,
John H
...
Fennell,
W
...
Fitzgibbon,
Harvey J
...
Gormley,
Layachi Hadji,
Ruben Hayrapetyan,
Terry Herdman,
Zdzislaw Jackiewicz,
S
...
Jain,
Anthony J
...
Johnson,
Harry L
...
Johnson,
Joseph Kazimir,
J
...
Khlief,
C
...
Knickerbocker,
Carlon A
...
Kudzma,
Alexandra Kurepa,
G
...
Latta,
Cecelia Laurie,
James R
...
Meek,
Gary H
...
Merrill,
Vivien Miller,
Gerald Mueller,
Philip S
...
J
...
Newton,
Brian M
...
K
...
O’Dell,
A
...
Perryman,
Joseph H
...
Poxon,
Robert Pruitt,

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●

K
...
B
...
Rosenthal,
Barbara Shabell,
Seenith Sivasundaram,
Don E
...
W
...
B
...
Zill
Los Angeles

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Copyright 2012 Cengage Learning
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Tenth Edition

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
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Project for Section 3
...


Is AIDS an Invariably
Fatal Disease?

Cell infected with HIV

This essay will address and answer the question: Is the acquired immunodeficienc
syndrome (AIDS), which is the end stage of the human immunodeficiency virus
(HIV) infection, an invariably fatal disease?
Like other viruses, HIV has no metabolism and cannot reproduce itself outside of
a living cell
...
To reproduce, HIV must use the reproductive apparatus of the cell it invades
and infects to produce exact copies of the viral RNA
...
The double-stranded viral DNA migrates into the nucleus of the invaded cell and
is inserted into the cell’s genome with the aid of another viral enzyme (integrase)
...

When the infected cell is stimulated to reproduce, the proviral DNA is transcribed into
viral DNA, and new viral particles are synthesized
...

What makes HIV infection so dangerous is the fact that it fatally weakens a
host’s immune system by binding to the CD4 molecule on the surface of cells vital
for defense against disease, including T-helper cells and a subpopulation of natural
killer cells
...

Modeling suggests that HIV infection of natural killer cells
[ ]
...
g
...
Indeed, about 1% of Caucasians lack coreceptor
molecules, and, therefore, are completely
to becoming HIV infected
...
This is clearly a war of attrition, one in which the
immune system invariably loses
...
Because HIV rapidly mutates, its
ability to infect T4 cells on contact (its infectivity) eventually increases and the
rate T4 cells become infected increases
...
There comes a point, however, when the production rate of T4
cells reaches its maximum possible limit and any further increase in HIV’s infectivity must necessarily cause a drop in the T4 density leading to AIDS
...
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In over 95% of hosts, the immune system gradually loses its long battle with the
virus
...
The host reaches the AIDS stage of the infection
when one
of the more than twenty opportunistic infections characteristic of AIDS develops
(clinical AIDS) when the T4 cell density falls below 250 cells/mm3 (an additional
definition of AIDS promulgated by the CDC in 1987)
...

In order to model survivability with AIDS, the time at which a host develops
AIDS will be denoted by ϭ 0
...
For the remaining part of the
cohort, the probability of dying per unit time at time will be assumed to be a constant , where, of course, must be positive
...


(1)

Using the integrating-factor method discussed in Section 2
...


(2)

Instead of the parameter appearing in (2), two new parameters can be defined for
Ϫ1
a host for whom AIDS is fatal: the
and
aver given by aver ϭ
the
given by 1͞2 ϭ ln(2)͞
...
See Problem 8 in Exercise 3
...
In terms of these parameters
the entire time-dependence in (2) can be written as
Ϫ

ϭ

Ϫ>

aver

ϭ 2Ϫ >

1>2

(3)

Using a least-squares program to fit the survival fraction function in (2) to the
actual survival data for the 159 Marylanders who developed AIDS in 1985 produces
an immortal fraction value of ϭ 0
...
666 year, with the average survival time being aver ϭ 0
...
See Figure 1
...
The 1985 Maryland AIDS survival curve is virtually identical to those of 1983 and 1984
...
Since zidovudine was not
known to have an impact on the HIV infection before 1985 and was not common
1
...
8
0
...
4
0
...

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●

therapy before 1987, it is reasonable to conclude that the survival of the 1985
Maryland AIDS patients was not significantly influenced by zidovudine therap
...
Residents with AIDS who
changed their name and then died or who died abroad would still be counted as alive
by the Maryland Department of Health and Mental Hygiene
...
0665 (6
...

Detailed data on the survivability of 1,415 zidovudine-treated HIV-infected
hosts whose T4 cell densities dropped below normal values were published by
Easterbrook et al
...
As their T4 cell densities drop towards zero, these people develop clinical AIDS and begin to die
...
If the time ϭ 0 is redefined to
mean the moment the T4 cell density of a host falls below 10 cells/mm3, then the
survivability of such hosts was determined by Easterbrook to be 0
...
316, and
0
...
5 years, and 2 years, respectively
...
878 year
[ ]; equivalently, the average survival time is aver ϭ 1
...
These results clearly
show that zidovudine is not effective in halting replication in all strains of HIV,
since those who receive this drug eventually die at nearly the same rate as those who
do not
...
5 months between the survival half-life
for 1993 hosts with T4 cell densities below 10 cells/mm3 on zidovudine therapy
( 1͞2 ϭ 0
...
666 year) may be entirely due to improved hospitalization and improvements in the treatment of the opportunistic infections associated with AIDS over the
years
...
Zidovudine
therapy has been estimated to extend the survivability of an HIV-infected patient by
perhaps 5 or 6 months on the average [ ]
...
0665
and the average survival time falls within the range 0
...
27 years
...
65%
and may be zero
...
7 months [
...


Related Problems
Suppose the fraction of a cohort of AIDS patients that survives a time after
AIDS diagnosis is given by ( ) ϭ exp(Ϫ )
...

The fraction of a cohort of AIDS patients that survives a time after AIDS
diagnosis is given by ( ) ϭ exp(Ϫ )
...
4
months
...
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...
The time it takes for ( ) to reach the value of
0
...

Show that ( ) can be written in the form ( ) ϭ 2Ϫ > 1>2
...
Thus, it is always true that 1͞2 Ͻ aver
...
e
...
Only 14%
of lung cancer patients survive 5 years after diagnosis
...
Find an expression for the fraction ( ) of lung cancer patients that
survive a time after being diagnosed with the disease
...
What fraction of lung cancer patients
survives two years with the disease?

References
Kramer, Ivan
...
8, No
...

Kramer, Ivan
...

15, no
...

Easterbrook, Philippa J
...

Progressive CD4 cell depletion and death in zidovudine-treated patients
...
6, 1993, No
...

Kramer, Ivan
...

,
Vol
...
3, Feb
...

Stehr-Green, J
...
, Holman, R
...
, Mahoney, M
...
Survival analysis of
hemophilia-associated AIDS cases in the US
...
1989, 79
(7): 832–835
...
, Tan, Wai-Yuan, Pee, David, Goedert, James J
...

, Aug
...
15, No
...


Courtesy of Ivan Kramer

ABOUT THE AUTHOR
earned a BS in Physics and Mathematics from The City College of
New York in 1961 and a PhD from the University of California at Berkeley in theoretical particle physics in 1967
...
Dr
...

In addition to his many published articles on HIV infection and AIDS, his current
research interests include mutation models of cancers, Alzheimers disease, and
schizophrenia
...
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2

Courtesy of Jo Gasoigne

The Allee Effect

Dr Jo with Queenie; Queenie is on the left

The top five most famous Belgians apparently include a cyclist, a punk singer, the inventor of the saxophone, the creator of Tintin, and Audrey Hepburn
...
He had a fairly short life, dying at
the age of 45, but did manage to include some excitement—he was deported from
Rome for trying to persuade the Pope that the Papal States needed a written constitution
...

Aside from this episode,
(1804–1849) was a mathematician who
concerned himself, among other things, with the dynamics of natural populations—
fish, rabbits, buttercups, bacteria, or whatever
...
) Theorizing on the growth of natural populations had up to this point been relatively limited, although scientists had reached the
obvious conclusion that the growth rate of a population ( ͞ , where ( ) is the
population size at time ) depended on ( ) the birth rate and ( ) the mortality rate ,
both of which would vary in direct proportion to the size of the population :
ϭ


...


(2)

This model of population growth has a problem, which should be clear to you—if
not, plot ͞ for increasing values of
...
Clearly, something
eventually has to step in and slow down ͞
...


(3)

Equation (3) is called the
, and it forms to this day the basis of much
of the modern science of population dynamics
...
Thus the size
of the population cannot exceed the carrying capacity of the environment
...
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...
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...
To get
growth rate, we just divide each side of equation (3) by :
1

ϭ

΂1 Ϫ ΃ ϭ


...

1
Er, hang on a minute
...
zero sharks? Here is clearly a flaw in the
logistic model
...
When we use this equation
to try and analyze how populations might work, it becomes a model
...
This assumption to some extent underlies all our ideas about sustainable management of natural
resources—a fish population cannot be fished indefinitely unless we assume that when
a population is reduced in size, it has the ability to grow back to where it was before
...
But for very depleted or endangered populations, the idea that individuals keep
doing better as the population gets smaller is a risky one
...
1% of , has been protected since the early 1990s, and has yet to show convincing signs of recovery
...
Allee showed
that, in fact, individuals in a population can do worse when the population becomes
very small or very sparse
...
As a result of Allee’s work, a population where the
growth rate declines at low population size is said to show an

...
On the other hand, there is nothing that an adult cod likes more than a snack of
baby cod—they are not fish with very picky eating habits—so these arguments may
not stack up
...

Allee effects can be modelled in many ways
...


(4)

where is called the

...


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●

a value of somewhere between ϭ 0 and ϭ , that is, 0 Ͻ Ͻ , depending on
the species (but for most species a good bit closer to 0 than , luckily)
...
If you work through Problems 2
and 3, you will see that the consequences of equation (4) can be disastrous for endangered populations
...
Do this, and plot ( ) as a function of for 0 Յ Յ 10
...
01 (fish per
cubic metre of seawater, say)
...

The value of can tell us a lot about the ecology of a species—sardines,
where females mature in less than one year and have millions of eggs, have
a high , while sharks, where females bear a few live young each year, have
a low
...

:
If a marine protected area is put in place to stop overfishing, which species
will recover quickest—sardines or sharks?
Find the population equilibria for the model in (4)
...
You should find three values of for which the population is at equilibrium
...
If, when a population deviates a
bit from the equilibrium value (as populations inevitably do), it tends to return to
it, this is a stable equilibrium; if, however, when the population deviates from
the equilibrium it tends to diverge from it ever further, this is an unstable equilibrium
...
Unstable equilibria are a feature of Allee effect models such as
(4)
...
[
:
See Section 2
...
]
Discuss the consequences of the result above for a population ( ) fluctuatin
close to the Allee threshold
...
, Berec L
...

Oxford University Press
...
1997
...


Springer-Verlag,

Courtesy of Jo Gascoigne

ABOUT THE AUTHOR
After a degree in Zoology,
thought her first job, on conservation in
East Africa, would be about lions and elephants—but it turned out to be about fish
Despite the initial crushing disappointment, she ended up loving them—so much, in
fact, that she went on to complete a PhD in marine conservation biology at the
College of William and Mary, in Williamsburg, Virginia, where she studied lobster and
Caribbean conch, and also spent 10 days living underwater in the Aquarius habitat in
Florida
...
She now works to promote environmentally sustainable fisheries
...
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...
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...
3

Damien Richard/Shutterstock
...
During
this 70-year absence, significant changes were recorded in the populations of other
predator and prey animals residing in the park
...
With the
reintroduction of the wolf in 1995, we anticipated changes in both the predator and
prey animal populations in the Yellowstone Park ecosystem as the success of the
wolf population is dependent upon how it influences and is influenced by the other
species in the ecosystem
...
Recent studies have shown that the elk population
has been negatively impacted by the reintroduction of the wolves
...
This article
asks the question of whether the wolves could have such an effect and, if so, could
the elk population disappear?
Let’s begin with a more detailed look at the changes in the elk population independent of the wolves
...
Using the simplest differential equation model for population dynamics, we can determine the growth rate for elks (represented by the variable ) prior to the reintroduction of the wolves
...
0, (10) ϭ 18
...
The solution, which is left as an exercise for the reader, finds the
combined birth/death growth rate to be approximately 0
...
0

0
...
In
2007, biologists estimated the number of wolves to be approximately 171
...
0325 Ϫ 0
...
6
(0) ϭ 18
...
05

(2)

(0) ϭ 0
...
All populations are
measured in thousands of animals
...
So, from the initial conditions, we have 18,000 elk and 21 wolves in the
year 1995
...
0325
...
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...
The first equation
shows that the growth rate of the elk ( > ) is positively impacted by the size of
the herd (0
...
This can be interpreted as the probability of breeding increases with the number of elk
...
8 ) has
a negative impact on the growth rate of the elk since it measures the interaction
between predator and prey
...
6 ϩ 0
...
But, the interaction between the elk and wolves (0
...

Since an analytical solution cannot be found to the initial-value problem (2), we
need to rely on technology to find approximate solutions
...


The graphs in Figures 1 and 2 show the populations for both species between 1995
and 2009
...
In this model, we see the population decline from 18,000 in 1995 to approximately 7,000 in 2009
...


180

16000

160

14000

140

Wolf population

200

18000

Elk population

20000

12000
10000
8000
6000

120
100
80
60

4000

40

2000

20

0
1995 1997 1999 2001 2003 2005 2007 2009

Elk population

0
1995 1997 1999 2001 2003 2005 2007 2009

Wolf population

The alert reader will note that the model also shows a decline in the wolf population after 2004
...

Figure 3 below shows the long-term behavior of both populations
...

Information on the reintroduction of wolves into Yellowstone Park and central
Idaho can be found on the Internet
...
S
...


Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

20000

200
Elk

18000

180

Wolf

160

14000

140

12000

120

10000

100

8000

80

6000

60

4000

40

2000

Elk and Wolf Populations

16000

20

0
1990

2000

2010

2020

2030

2040

2050

2060

2070

0
2080

Long-term behavior of the populations

Related Problems
Solve the pre-wolf initial-value problem (1) by first solving the differential
equation and applying the initial condition
...
What other factors might
account for the decrease in the elk population?
Consider the long-term changes in the elk and wolf populations
...

What does the initial-value problem (1) tell us about the growth of the elk population without the influence of the wolves? Find a similar model for the introduction of rabbits into Australia in 1859 and the impact of introducing a prey
population into an environment without a natural predator population
...
J
...
J
...
Lawrence University
Principal Research Engineer
Sensis Corporation
C
...
Knickerbocker received his PhD in mathematics from Clarkson University in
1984
...
Lawrence University, where he authored numerous articles in a variety of topics,
including nonlinear partial differential equations, graph theory, applied physics, and
psychology
...
Currently, Dr
...


Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


Project for Section 5
...
com

Bungee Jumping

Bungee jumping from a bridge

Bridge

Suppose that you have no sense
...
Suppose that you plan to jump off that bridge
...
Instead, you plan to attach a bungee cord to your feet, to dive gracefully
into the void, and to be pulled back gently by the cord before you hit the river that is
174 feet below
...
You
need to choose the stiffness and length of the cord so as to avoid the unpleasantness
associated with an unexpected water landing
...
Call the position at the bottom of the cord 0, and
measure the position of your feet below that “natural length” as ( ), where increases
as you go down and is a function of time
...
Then, at the time you
jump, (0) = -100, while if your six-foot frame hits the water head first, at that time
( ) = 174 - 100 - 6 = 68
...
Note also
that you plan to dive so your head will be six feet below the end of the chord when
it stops you
...
You know that when you leap from the
bridge, air resistance will increase proportionally to your speed, providing a force in
the opposite direction to your motion of about b , where b is a constant and is your
velocity
...
Thus, you know that the force of the cord pulling you back
from destruction may be expressed as
( )ϭ

100

Bungee
100 ft

174 ft
0

Water

The bungee setup

Յ0
Ͼ0

The number is called the
, and it is where the stiffness of the cord
you use influences the equation
...
This could lead to discomfort, injury, or even a
Darwin award
...
Consequently, you are interested in finding the distance you fall below the natural length of the cord as a function of the spring constant
...


Here
is your weight, 160 lb
...
e
...
The constant b for air resistance depends on a number of things, including whether you wear your skin-tight
pink spandex or your skater shorts and XXL T-shirt, but you know that the value
today is about 1 0
...
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...
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...
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...
We will work with such equations more extensively
in later chapters, but we already know how to solve such equations from our past
experience
...

In Problem 1 you find an expression for your position seconds after you step off
the bridge, before the bungee cord starts to pull you back
...
When you pass the natural length of the bungee cord, it does start
to pull back, so the differential equation changes
...
We can thus describe the motion
for ( ) Ͼ 0 using the problem Љ = - - b Ј ( 1) = 0 Ј( 1) = 1 An illustration
of a solution to this problem in phase space can be seen in Figure 2
...
All
we have to do is to find out the time 2 when you stop going down
...
e
...

As you can see, knowing a little bit of math is a dangerous thing
...
By the time you swoop past the natural length of the cord, that approximation is only wishful thinking, so your actual mileage may vary
...
Do not trust your life to an approximation made by a man who has been
dead for 200 years
...


Related Problems
Solve the equation Љ + b Ј =
for ( ) given that you step off the bridge—no
jumping, no diving! Stepping off means (0) = -100 Ј(0) = 0 You may use
= 160 b = 1 and = 32
Use the solution from Problem 1 to compute the length of time 1 that you freefall
(the time it takes to go the natural length of the cord: 100 feet)
...
Call the result 1 You have found your downward speed when you pass the point where the cord starts to pull
...


For now, you may use the value = 14 but eventually you will need to replace
that with the actual values for the cords you brought
...

Compute the derivative of the expression you found in Problem 4 and solve for
the value of where it is zero
...

To do so, you suspend a weight of 10 lb
...
2 feet
...

Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

ABOUT THE AUTHOR

Courtesy of Kevin Cooper

PhD, Colorado State University, is the Computing Coordinator for
Mathematics at Washington State University, Pullman, Washington
...
Dr
...


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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


Project for Section 5
...
com

Collapse of the Tacoma Narrows Bridge

The rebuilt Tacoma Narrows bridge (1950)
and new parallel bridge (2009)

In the summer of 1940, the Tacoma Narrows Suspension Bridge in the State of
Washington was completed and opened to traffic
...
The bridge became a tourist attraction as people came
to watch, and perhaps ride, the undulating bridge
...
Soon, the vertical oscillations became rotational, as observed by looking down the roadway
...
Figure 1 shows a picture of the bridge
during the collapse
...
Or, do an Internet search with the key words
“Tacoma Bridge Disaster” in order to find and view some interesting videos of the
collapse of the bridge
...
He and his coauthors [ ] claimed that the wind blowing perpendicularly across
the roadway separated into vortices (wind swirls) alternately above and below the
roadbed, thereby setting up a periodic, vertical force acting on the bridge
...
Others further hypothesized that the frequency of
this forcing function exactly matched the natural frequency of the bridge, thus leading to resonance, large oscillations, and destruction
...

As we can see from equation (31) in Section 5
...
3, resonance is a linear phenomenon
...

Furthermore, there must be absolutely no damping in the system
...

If resonance did not cause the collapse of the bridge, what did? Recent research
provides an alternative explanation for the collapse of the Tacoma Narrows Bridge
...
The theory involves partial differential equations
...

The development of the model below is not exactly the same as that of Lazer and
McKenna, but it results in a similar differential equation
...

Consider a single vertical cable of the suspension bridge
...
When stretched, the cable acts like a spring with Hooke’s constant, ,
while, when compressed, it acts like a spring with a different Hooke’s constant,
...
Let the vertical deflectio
(positive direction downward) of the slice of the roadbed attached to this cable be

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

denoted by ( ), where represents time, and ϭ 0 represents the equilibrium position of the road
...
This
change in the Hooke’s Law constant at ϭ 0 provides the nonlinearity to the differential equation
...
Note that
the differential equation is linear on any interval on which does not change sign
...
We will
assume that ϭ 1 kg, ϭ 4 N/m, a ϭ 1N/m, and ( ) ϭ sin(4 ) N
...
We also
assign the following initial values to : (0) ϭ 0, Ј(0) ϭ 0
...

Because of the downward initial velocity and the positive applied force, ( ) will
initially increase and become positive
...
01
...
1
...
It is easy to see
1
that ( ) ϭ 1cos(2 ) ϩ 2sin(2 ) (equation (9), Section 4
...
4
...
4)
...

12

(2)

The initial conditions give
(0) ϭ 0 ϭ

1,

1
Ϫ ,
3
1
so that 2 ϭ (0
...
Therefore, (2) becomes
Ј(0) ϭ 0
...
01 ϩ sin(2 ) Ϫ sin(4 )
2
3
12
1
1
΄1΂0
...

2

ϭ sin(2 )

(3)

We note that the first positive value of for which ( ) is again equal to zero is ϭ p
...
01 ϩ 2)
...

2
3
After ϭ p, becomes negative, so we must now solve the new problem
2
Љϩ

ϭ sin(4 ),

2
΂p΃ ϭ 0, Ј΂p΃ ϭ Ϫ΂0
...

2
2

(4)

Proceeding as above, the solution of (4) is

΂

( ) ϭ 0
...
01 ϩ 5΃ Ϫ 15 sin cos(2 )΅
...
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...
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...
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...
01 ϩ 15, so that equation (5) holds on [p>2, 3p>2]
...

2
During this cycle, the section of the roadway started at the equilibrium with positive
velocity, became positive, came back to the equilibrium position with negative velocity, became negative, and finally returned to the equilibrium position with positive
velocity
...

2
The solution for the next cycle is
Ј(3p)
2

΄ ΂

΃

΅

1
7
1
( ) ϭ sin(2 ) Ϫ 0
...
01 ϩ

΃

on

΅

8
4
Ϫ
cos cos(2 )
15
15

[3p>2, 2p],
(6)

on

[2p, 3p]
...
01 ϩ 15 ), while at the beginning of the third cycle it is (0
...
In fact, the
2
velocity at the beginning of each cycle is 15 greater than at the beginning of the previous cycle
...
See Figure 2 for a
graph of the
on the interval [0, 3p]
...

It must be remembered that the model presented here is a very simplified onedimensional model that cannot take into account all of the intricate interactions of
real bridges
...
More recently, McKenna [ ] has refined that model to provide
a different viewpoint of the torsional oscillations observed in the Tacoma Bridge
...
It is likely that
the models will be refined over time, and new insights will be gained from the
research
...


0
...
0

2

4

6

8

0
...
4
0
...
Note that resonance occurs in the first problem but not in the second
Љ ϩ ϭ Ϫcos , (0) ϭ 0, Ј(0) ϭ 0
...


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...
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...
Editorial review has
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...


●

Solve the initial-value problem Љ ϩ ( ) ϭ sin(4 ), (0) ϭ 0, Ј(0) ϭ 1, where
( )ϭ

Ά

·

if
0
,
if Ͻ 0

and
ϭ 1, ϭ 4, (Compare your answer with the example in this project
...

Note that, in part (a), the condition Ͼ of the text is not satisfied
...
What happens in each case as increases? What would happen in each case
if the second initial condition were replaced with Ј(0) ϭ 0
...
01
ϭ 0
...
5

References
Lewis, G
...
, “Tacoma Narrows Suspension Bridge Collapse” in
, Dennis G
...
Boston: PWS-Kent, 1993
...
,
, 167–169
...

Amman, O
...
, T
...
B
...
Washington D
...
: Federal Works Agency, 1941
...
C
...
J
...
Large amplitude periodic oscillations in suspension bridges: Some new connections with nonlinear analysis
...

Peterson, I
...

137 (1991): 344–346
...
J
...

106 (1999):1–18
...
Lewis

ABOUT THE AUTHOR
is professor emeritus at Michigan Technological University,
where he has taught and done research in Applied Math and Differential Equations
for 34 years
...
His hobbies include
travel, food and wine, fishing, and birding, activities that he intends to continue in
retirement
...
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...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...
3

© Ronald C
...
The amber glow of streetlights mixed with the violent
red flash of police cruisers begins to fade with the rising of a furnace orange sun
...
Taking a seat on the
bumper of her tan LTD, Detective Marlow begins to review the evidence
...
m
...
Wood was found in the walk in refrigerator in
the diner’s basement
...
m
...
Thirty minutes later the coroner
again measured the core body temperature
...
The thermostat inside the refrigerator reads 50 degrees Fahrenheit
...
She knows that Newton’s Law of
Cooling says that the rate at which an object cools is proportional to the difference
between the temperature of the body at time and the temperature
of the environment surrounding the body
...
Because Daphne wants to investigate the past using
positive values of time, she decides to correspond ϭ 0 with 6:00 a
...
, and so, for
example, ϭ 4 is 2:00 a
...
After a few scratches on her yellow pad, Daphne realizes
that with this time convention the constant in (1) will turn out to be

...
m
...
It reads 70 degrees Fahrenheit
...
She decides to leave this question unanswered for now, simply letting denote the number of hours the body has
been in the refrigerator prior to 6:00 a
...
For example, if ϭ 6, then the body was
moved at midnight
...
As the rapidly cooling
coffee begins to do its work, she realizes that the way to model the environmental temperature change caused by the move is with the unit step function ᐁ( )
...


(3)

Daphne’s mustard-stained polyester blouse begins to drip sweat under the blaze
of a midmorning sun
...
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...
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...
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...
She settles into the faux leather booth
...
The intense chill serves as a gruesome reminder of the tragedy
that occurred earlier at the Mayfair
...
She then carefully constructs a table that relates refrigeration
time to time of death while eating her scrapple and eggs
...
“Any suspects?” Daphne asks
...
The first is the late Mr
...
She was seen in the Mayfair between 5 and 6 p
...

in a shouting match with Wood
...
The second suspect is a South
Philly bookie who goes by the name of Slim
...
Nobody overheard the conversation, but
witnesses say there was a lot of hand gesturing, like Slim was upset or something
...
He left quietly around 11
...
”
“The cook?”
“Yep, the cook
...
The cashier says he heard Joe and
Shorty arguing over the proper way to present a plate of veal scaloppine
...
m
...
m
...
”
“Great work, partner
...
”

Related Problems
Solve equation (1), which models the scenario in which Joe Wood is killed in the
refrigerator
...
6 degrees Fahrenheit)
...
Your solution ( )
will depend on both and (Use the value of found in Problem 1
...
In particular, explain why large values of give
the same time of death
time body moved
12

time of death

6:00 p
...


11
10
9
8
7
6
5
4
3
2
Who does Daphne want to question and why?
(

The process of temperature change in a dead body is known as
is the process of body stiffening), and although it is not

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

perfectly described by Newton’s Law of Cooling, this topic is covered in most
forensic medicine texts
...
In particular, chemical processes in the body continue for several hours after death
...

A linear equation, known as the
, is sometimes used to give
a preliminary estimate of the time since death
...
4 Ϫ
1
...
Although we do not have all of the tools to
derive this equation exactly (the 1
...

Use equation (1) with an initial condition of (0) ϭ 0 to compute the equation of the tangent line to the solution through the point (0, 0) Do not use the
values of
or found in Problem 1
...
Next, let
ϭ 98 4 and solve for to get
ϭ

98
...


(5)

Courtesy of Tom LoFaro

ABOUT THE AUTHOR
is a professor and chair of the Mathematics and Computer Science
Department at Gustavus Adolphus College in St
...
He has been involved
in developing differential modeling projects for over 10 years, including being a principal investigator of the NSF-funded IDEA project (http://www
...
wsu
...
Dr
...
His oldest daughter
(age 12) aspires to be a forensic anthropologist much like Detective Daphne Marlow
...
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...
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...
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...
2

s76/ZUMA Press/Newscom

Earthquake Shaking of
Multistory Buildings

Collapsed apartment building in San
Francisco, October 18, 1989, the day after
the massive Loma Prieta earthquake

Large earthquakes typically have a devastating effect on buildings
...
More recently, that area was hit by the Loma Prieta earthquake that many people in the
United States and elsewhere experienced second-hand while watching on television the Major League Baseball World Series game that was taking place in San
Francisco in 1989
...
Let represent the horizontal
displacement of the th floor from equlibrium
...
During an earthquake, the ground moves
horizontally so that each floor is considered to be displaced relative to the ground
...

Typically, the structural elements in large buildings are made of steel, a highly
elastic material
...
We assume that Hooke’s Law holds, with proportionality constant between the th and the ( ϩ 1)st floors
...

We also assume a similar reaction between the first floor and the ground, with proportionality constant 0
...

Ϫ1
Ϫ1
Ӈ

Ϫ2
Ӈ

2

1

1

0

ground

Floors of building

ϩ1
Ϫ1( Ϫ

(

ϩ1

Ϫ )

Ϫl)
Ϫ1

Forces on th floo

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

We can apply Newton’s second law of motion (Section 5
...

2
1
2
2

1
2

ϭϪ

2
2

ϭ Ϫ 1(

o

1( 2

ϩ

0 1

1)

Ϫ

2

1)

Ϫ

2( 3

ϩ

Ϫ

2)

o
2
2

ϭϪ

Ϫ1(

Ϫ1)
...

Then the differential equations are
2

2

1
2

ϭ Ϫ4

2
2

ϭ2

1

1

ϩ2

2

Ϫ 2 2
...
2 is
1(

)ϭ2

2(

) ϭ ΂4 Ϫ v2΃
1

1

cos v1 ϩ 2

2

sin v1 ϩ 2

3

cos v2 ϩ 2

4

sin v2 ,

where v1 ϭ 23 ϩ 15 ϭ 2
...
874
...
2, 2(0) ϭ 0,
Ј(0)
Ј(0) ϭ 0
...
2 m/s
...
1>[΂v2 Ϫ v2΃v1] ϭ 0
...
See Figures 3 and 4 for
2
1
2
graphs of 1( ) and 2( )
...
It continues to the right, eventually pulling 1 along until
the two-second mark
...
2 seconds and continues moving left,
draging 1 along with it
...
There is no damping
in the system, so that the oscillatory behavior continues forever
...
10
0
...
05

1

2

3

4

5

0
...
05
1
0
...
1

Graph of

1(

)

Graph of

2(

)

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

If a horizontal oscillatory force of frequency v1 or v2 is applied, we have a situation analogous to resonance discussed in Section 5
...
3
...

Let’s define the following matrices and vector

ϭ

΂

0
o
1

0

΂

ϩ

0


...


0

2

Ϫ(

ϭ

0
0

0

1


...

0
...


0
0

0
...


0
0
0

0
0
0

Ϫ2

Ϫ(

Ϫ2

0

ϩ
Ϫ1

0
0
0
o
Ϫ1)

Ϫ1

Ϫ

Ϫ1

΃

΂΃
1(

()ϭ

)
()
2
o
()

Then the system of differential equations can be written in matrix form
2
2

or

ϭ


...
Matrix has an inverse given by

Ϫ1

ϭ

΂

Ϫ1
1

0

0
o
0

Ϫ1
2

0
0


...


0

0


...


We can therefore represent the matrix differential equation by
Љϭ(

Ϫ1

)

or

Љϭ


...

The eigenvalues of the matrix reveal the stability of the building during an earthquake
...
In the first example, the eigenvalues are Ϫ3 ϩ 15 ϭ Ϫ0
...
236
...
If l is the th eigenvalue, then v ϭ 1Ϫl is the th frequency, for ϭ 1, 2,
...
If this is oscillatory in
nature, say of the form ( ) ϭ cosg , then large displacements may develop in the
building, especially if the frequency g of the forcing term is close to one of the natural
frequencies of the building
...
1
...

Where

ϭ

Ϫ1

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

As another example, suppose we have a 10-story building, where each floor has a
mass 10000 kg, and each value is 5000 kg/s2
...
5
0
0
0
0
0
0
0
0

0
...
5
0
0
0
0
0
0
0

0
0
...
5
0
0
0
0
0
0

0
0
0
...
5
0
0
0
0
0

0
0
0
0
...
5
0
0
0
0

0
0
0
0
0
...
5
0
0
0

0
0
0
0
0
0
...
5
0
0

0
0
0
0
0
0
0
...
5
0

0
0
0
0
0
0
0
0
...
5

0
0
0
0
0
0
0
0
0
...
5

΃

The eigenvalues of are found easily using
or another similar computer
package
...
956, Ϫ1
...
623, Ϫ1
...
075, Ϫ0
...
5, Ϫ0
...
099, and Ϫ0
...
399, 1
...
274, 1
...
037, 0
...
707, 0
...
315, and 0
...
491, 4
...
932, 5
...
059, 7
...
887, 12
...
947, and 59
...

During a typical earthquake whose period might be in the range of 2 to 3 seconds, this
building does not seem to be in any danger of developing resonance
...
253 seconds, while the fifth through seventh are all on the order of
2–3 seconds
...


Related Problems
Consider a three-story building with the same and values as in the first example
...
What are the
matrices , , and ? Find the eigenvalues for
...
Write down the corresponding system of differential equations
...
What range of frequencies of an earthquake would place the building in danger of destruction?
Consider the tallest building on your campus
...
If you
have trouble coming up with such values, use the ones in the example problems
...
Is your building safe from a modest-sized period2 earthquake? What if you multiplied the matrix
by 10 (that is, made the
building stiffer)? What would you have to multiply the matrix by in order to
put your building in the danger zone?
Solve the earthquake problem for the three-story building of Problem 1:
Љϭ

ϩ ( ),

where ( ) = cosg , =
, = [1 0 0] , = 10,000 lbs is the amplitude
of the earthquake force acting at ground level, and g = 3 is the frequency of the
earthquake (a typical earthquake frequency)
...
3 for the method of
solving nonhomogeneous matrix differential equations
...


Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


Project for Section 8
...
The outbreak of World War I climaxed a rapid buildup of armaments
among rival European powers
...
The United States and the Soviet Union engaged in a costly nuclear arms race during the forty years of the Cold War
...

British meteorologist and educator Lewis F
...
Arms race models generally assume that each nation adjusts its accumulation of weapons in some manner dependent on the size of its own stockpile and the
armament levels of the other nations
...
Richardson’s model takes into account
internal constraints within a nation that slow down arms buildups: The more a
nation is spending on arms, the harder it is to make greater increases, because it
becomes increasingly difficult to divert society’s resources from basic needs such
as food and housing to weapons
...

The mathematical structure of this model is a linked system of two first-orde
linear differential equations
...
The constants and measure mutual fear; the constants and
represent proportionality factors for the “internal brakes” to further arms increases
...
Negative values for and
indicate a contribution based on goodwill
...
Although the model
is a relatively simple one, it allows us to consider several different long-term outcomes
...
A vicious cycle of unbounded
increases in and is another possible scenario
...
In other cases, the eventual outcome depends on the
starting point
...
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...
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...

Although “real world” arms races seldom match exactly with Richardson’s model,
his pioneering work has led to many fruitful applications of differential equation models
to problems in international relations and political science
...
”
Arms races are not limited to the interaction of nation states
...
Arms phenomena have also been observed between rival urban gangs
and between law enforcement agencies and organized crime
...
Colleges have engaged in “amenities
arms races,” often spending millions of dollars on more luxurious dormitories,
state- of-the-art athletic facilities, epicurean dining options, and the like, to be more
competitive in attracting student applications
...
Most generally, the assumptions represented in a Richardson-type
model also characterize many competitions in which each side perceives a need to
stay ahead of the other in some mutually important measure
...
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...

The qualitative long-term behavior of a Richardson arms race model can, in
some cases, depend on the initial conditions
...
The assumption here is that each country sets a desired
level of arms expenditures for itself and then changes its weapons stock proportionally to the gap between its current level and the desired one
...

How will and evolve over time under such a model?
Generalize the stock adjustment model of (a) to a more realistic one where
the desired level for each country depends on the levels of both countries
...
Show that, under these assumptions, the stock adjustment model is equivalent to a Richardson model
...
How might the equations change if two of the nations
are close allies not threatened by the arms buildup of each other, but fearful of the
armaments of the third
...

In the real world, an unbounded runaway arms race is impossible since there is
an absolute limit to the amount any country can spend on weapons;
gross national product minus some amount for survival
...


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●

References
Richardson, Lewis F
...

Olinick, Michael,
Reading, MA: Addison-Wesley, 1978
...
, and Dagobert L
...
Midlarsky, ed
...
Boston: Unwin
Hyman, 1989
...
Dr
...
He is the author or
co-author of a number of books on single and multivariable calculus, mathematical
modeling, probability, topology, and principles and practice of mathematics
...


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Introduction to Differential Equations

1
...
2 Initial-Value Problems
1
...
Analogous to a course in algebra and
trigonometry, in which a good amount of time is spent solving equations such as
2
ϩ 5 ϩ 4 ϭ 0 for the unknown number , in this course
of our tasks will be
to solve differential equations such as Љ ϩ 2 Ј ϩ ϭ 0 for an unknown function
ϭ ␾( )
...
As the course unfolds, you will see that there is
more to the study of differential equations than just mastering methods that
mathematicians over past centuries devised to solve them
...
In order to read, stu , and be conversant in a specialized
subject, you have to master some of the terminology of that discipline
...
In the last section we briefly examin
the link between differential equations and the real world
...
And so the mathematical description—or
—of phenomena, experiments, observations, or theories may
be a differential equation
...
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...
The exponential function ϭ 0
...
2 0
...
If we replace 0
...
2
...


An equation containing the derivatives of one or more unknown functions (or
dependent variables), with respect to one or more independent variables, is said
to be a
To talk about them, we shall classify differential equations according to
and
If a differential equation contains only ordinary derivatives of one or more unknown functions with respect to a
independent
variable, it is said to be an
An equation involving partial derivatives of one or more unknown functions of two or more independent variables is called a
Our first example
illustrates several of each type of differential equation
...


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●

The following equations are partial differential equations:*
Ѩ2
Ѩ2
ϩ 2 ϭ 0,
Ѩ 2 Ѩ

Ѩ2
Ѩ2
Ѩ
ϭ 2 Ϫ2 ,
2
Ѩ
Ѩ
Ѩ

Ѩ
Ѩ
ϭϪ
...
This means and must be functions of
independent variables
...
or the
Ј, Љ,
ٞ,
...
Actually, the
prime notation is used to denote only the first three derivatives; the fourth derivative
is written (4) instead of ЉЉ
...
Although less convenient to write and to typeset, the Leibniz notation has an advantage over the prime notation in that it clearly displays both the dependent and
independent variables
...
Partial
derivatives are often denoted by a
indicating the independent variables
...

The
(either ODE
or PDE) is the order of the highest derivative in the equation
...
In Example 1, the first and third
equations in (2) are first-order ODEs, whereas in (3) the first two equations are
second-order PDEs
...
For example, if we assume that
denotes the dependent variable in ( Ϫ )
ϩ4
ϭ 0, then Ј ϭ ͞ , so by
dividing by the differential , we get the alternative form 4 Ј ϩ ϭ
...
,

( )

) ϭ 0,

(4)

where is a real-valued function of ϩ 2 variables: , , Ј,
...
In that text the word
and
the abbreviation DE refer only to ODEs
...


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...
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...


●

highest derivative
equation

( )

in terms of the remaining

ϩ 1 variables
...
,

),

(5)

where is a real-valued continuous function, is referred to as the
Thus when it suits our purposes, we shall use the normal forms

of (4)
...
For example,
the normal form of the first-order equation 4 Ј ϩ ϭ is Ј ϭ ( Ϫ )͞4 ; the normal
form of the second-order equation Љ Ϫ Ј ϩ 6 ϭ 0 is Љ ϭ Ј Ϫ 6
...
This means that an th-order ODE is
linear when (4) is ( ) ( ) ϩ Ϫ1( ) ( Ϫ1) ϩ и и и ϩ 1( ) Ј ϩ 0 ( ) Ϫ ( ) ϭ 0 or
Ϫ1

( )

ϩ

Ϫ1(

)

Ϫ1

1(

ϩи иϩ

)

0(

ϩ

) ϭ ( )
...
(7)

In the additive combination on the left-hand side of equation (6) we see that the characteristic two properties of a linear ODE are as follows:
• The dependent variable and all its derivatives Ј, Љ,
...

• The coefficients 0, 1,
...
, ( ) depend at most on the
independent variable
...
Nonlinear
functions of the dependent variable or its derivatives, such as sin or Ј , cannot
appear in a linear equation
...
We
have just demonstrated that the first equation is linear in the variable by writing it in
the alternative form 4 Ј ϩ ϭ
...


2

nonlinear term:
power not 1

and

4

–––– ϩ
4

2

ϭ0

first-, second-, and fourth-order ordinary differential equa-

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...
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...


●

As was stated before, one of the goals in this course is to solve, or
find solutions of, differential equations
...


Any function ␾, defined on an interval and possessing at least derivatives
that are continuous on , which when substituted into an th-order ordinary differential equation reduces the equation to an identity, is said to be a
of
the equation on the interval
...
, ␾ ( )( )) ϭ 0

for all in
...
In our introductory discussion we
2
saw that ϭ 0
...
2 on the interval (Ϫϱ, ϱ)
...

You cannot think
of an ordinary differential
equation without simultaneously thinking
The interval in Definition 1
...
2
is variously called the
the
the
or the
and can be an open interval ( , ), a closed
interval [ , ], an infinite interval , ϱ), and so on
...

>

1
ϭ 16

1/2

;

ϭ

4

ЉϪ2 Јϩ

ϭ 0;

ϭ

One way of verifying that the given function is a solution is to see,
after substituting, whether each side of the equation is the same for every in the
interval
...

4
From the derivatives Ј ϭ
ϩ and Љ ϭ
ϩ 2 we have, for every real
number ,
-

:
-

:

ЉϪ2 Јϩ
0
...
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...
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...
A solution of a differential equation that is identically
zero on an interval is said to be a
The graph of a solution ␾ of an ODE is called a
Since ␾ is a differentiable function, it is continuous on its interval of defini
tion
...
Put another way, the domain of the function ␾ need not be
the same as the interval of definition (or domain) of the solution ␾
...


The domain of ϭ 1͞ , considered simply as a
, is the set of all real
numbers except 0
...
The rational function ϭ 1͞ is discontinuous at 0, and its graph, in a neighborhood of
the origin, is given in Figure 1
...
1(a)
...

Now ϭ 1͞ is also a solution of the linear first-order differential equation
Ј ϩ ϭ 0
...
) But when we say that ϭ 1͞ is a
of this DE, we
mean that it is a function defined on an interval on which it is differentiable and
satisfies the equation
...
Because
2
the solution curves defined by ϭ 1͞ for Ϫ3 Ͻ Ͻ Ϫ1 and 1 Ͻ Ͻ 10 are sim2
ply segments, or pieces, of the solution curves defined by ϭ 1͞ for Ϫϱ Ͻ Ͻ 0
and 0 Ͻ Ͻ ϱ, respectively, it makes sense to take the interval to be as large as
possible
...
The solution curve on (0, ϱ)
is shown in Figure 1
...
1(b)
...
A solution in
which the dependent variable is expressed solely in terms of the independent
variable and constants is said to be an
For our purposes, let us
think of an explicit solution as an explicit formula ϭ ␾( ) that we can manipulate,
evaluate, and differentiate using the standard rules
...
Moreover, the trivial solution ϭ 0 is an explicit solution of all three equations
...
This is particularly true when we attempt to solve nonlinear first-orde
differential equations
...


A relation ( , ) ϭ 0 is said to be an
of an ordinary
differential equation (4) on an interval , provided that there exists at least one
function ␾ that satisfies the relation as well as the differential equation on

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...
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...


●

It is beyond the scope of this course to investigate the conditions under which a
relation ( , ) ϭ 0 defines a differentiable function ␾
...
See ( ) in the

5

5

ϩ

The relation

ϭ

2

ϩ

2

ϭ 25 is an implicit solution of the differential equation
(8)

ϭϪ

5

on the open interval (Ϫ5, 5)
...
Moreover, solving
2
ϩ 2 ϭ 25 for
in terms of
yields ϭ Ϯ 225 Ϫ 2
...
The solution curves given in Figures 1
...
2(b) and 1
...
2(c) are segments of
the graph of the implicit solution in Figure 1
...
2(a)
...


Any relation of the form 2 ϩ 2 Ϫ ϭ 0
satisfies (8) for any constant
However, it is understood that the relation should always make sense in the real number
system; thus, for example, if ϭ Ϫ25, we cannot say that 2 ϩ 2 ϩ 25 ϭ 0 is an
implicit solution of the equation
...
”

5

5

5

ϭ Ϫ͙

Ϫ

Ϫ Ͻ

The study of differential equations is similar to that of
integral calculus
...

Analogously, when solving a first-order differential equation ( , , Ј) ϭ 0, we
obtain a solution containing a single arbitrary constant or parameter A
solution containing an arbitrary constant represents a set ( , , ) ϭ 0 of solutions
called a
When solving an th-order differential
equation ( , , Ј,
...
, ) ϭ 0
...
A solution of a differential equation that is free of arbitrary
parameters is called a

Ͻ

An implicit solution
and two explicit solutions of (8) in
Example 5

y
c>0
c=0
c<0

x

The one-parameter family
first-order equation

ϭ
ЈϪ

Some solutions of DE
in part (a) of Example 6

Ϫ cos is an explicit solution of the linear
ϭ

2

sin

on the interval (Ϫϱ, ϱ)
...
) Figure 1
...
3 shows the graphs of some particular
solutions in this family for various choices of
...


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...
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...


●

y

The two-parameter family ϭ 1 ϩ 2
is an explicit solution of the linear
second-order equation
ЉϪ2 Јϩ ϭ0
x

Some solutions
of DE in part (b) of Example 6

in part (b) of Example 3
...
) In Figure 1
...
4 we have shown seven of the “double infinity” of solutions in the family
...

2 ϭ 0), and ϭ 5 Ϫ 2
Sometimes a differential equation possesses a solution that is not a member of a
family of solutions of the equation —that is, a solution that cannot be obtained by specializing
of the parameters in the family of solutions
...
In Section 2
...
When ϭ 0, the resulting particular
4
1
solution is ϭ 16 4
...

In all the preceding examples we used and to denote the independent and
dependent variables, respectively
...
For example, we could
denote the independent variable by and the dependent variable by

(
(

)
)

The functions ϭ 1 cos 4 and ϭ 2 sin 4 , where 1 and 2 are arbitrary constants
or parameters, are both solutions of the linear differential equation
Љ ϩ 16 ϭ 0
...

2

2

1

sin 4 , and so

sin 4 ) ϭ 0
...

The next example shows that a solution of a differential equation can be a
piecewise-defined function

1,
0

The one-parameter family of quartic monomial functions ϭ
tion of the linear first-order equatio

1,
0

4

is an explicit solu-

ЈϪ4 ϭ0
piecewise-defined solution

Some solutions of DE
in Example 8

on the interval (Ϫϱ, ϱ)
...
) The blue and red solution curves shown in
Figure 1
...
5(a) are the graphs of ϭ 4 and ϭ Ϫ 4 and correspond to the choices
ϭ 1 and ϭ Ϫ1, respectively
...
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...
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...
As seen in Figure 1
...
5(b) the solution is constructed
from the family by choosing ϭ Ϫ1 for Ͻ 0 and ϭ 1 for Ն 0
...
But
often in theory, as well as in many applications, we must deal with systems of
differential equations
...
For example, if and denote dependent variables and
denotes the independent variable, then a system of two first-orde differential
equations is given by
ϭ (, , )
(9)
ϭ ( , , )
...

( ) A few last words about implicit solutions of differential equations are in
order
...
But don’t read too much
into this one example
...
1
...
An implicit solution ( , ) ϭ 0 can define a perfectly good
differentiable function ␾ that is a solution of a DE, yet we might not be able to
solve ( , ) ϭ 0 using analytical methods such as algebra
...
See Problems 45
and 46 in Exercises 1
...
Also, read the discussion following Example 4 in
Section 2
...

( ) Although the concept of a solution has been emphasized in this section,
you should also be aware that a DE does not necessarily have to possess
a solution
...
1
...

( ) It might not be apparent whether a first-order ODE written in differential
form ( , ) ϩ ( , ) ϭ 0 is linear or nonlinear because there is nothing
in this form that tells us which symbol denotes the dependent variable
...
1
...
, ( ) ) ϭ 0 can
be solved for ( ), but one should be a little bit careful here
...
See
Problems 52 and 53 in Exercises 1
...

( ) You may run across the term
in DE texts or in
lectures in courses in differential equations
...
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...
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...

( ) If
solution of an th-order ODE ( , , Ј,
...
, ) ϭ 0 by
appropriate choices of the parameters , ϭ 1, 2,
...
In solving linear ODEs, we shall impose relatively simple restrictions on the coefficients of the equation; with these
restrictions one can be assured that not only does a solution exist on an interval
but also that a family of solutions yields all possible solutions
...
Furthermore, if we happen to
obtain a family of solutions for a nonlinear equation, it is not obvious whether
this family contains all solutions
...
Don’t be concerned about
this concept at this point, but store the words “general solution” in the back of
your mind —we will come back to this notion in Section 2
...


In Problems 1 – 8 state the order of the given ordinary differential equation
...

(1 Ϫ ) Љ Ϫ 4 Јϩ 5 ϭ cos
Ϫ

΂ ΃ϩ

Ϫ

3

3

5 (4)
2

ϩ

2

ϭ

2

ϭϪ

2

2

ϭ0

Ϫ /2

ϭ

ϩ 20 ϭ 24;

Љϩ6 ϭ0

1ϩ
B

2

ϩ

ϭ cos( ϩ )

΂ ΃

2

΃

Ј ϭ 25 ϩ

ϩ( ϩ

ϭ 0; in ; in

ϩ
Ϫ

)

Љ ϩ ϭ tan ;

( Ϫ ) Јϭ

In Problems 9 and 10 determine whether the given
first-order differential equation is linear in the indicated
dependent variable by matching it with the first differential
equation given in (7)
...
Proceed as in Example 2, by considering ␾ simply as a
give its domain
...
2

...
Assume
an appropriate interval of definition for each solution

ϭ 0; in ; in

2

Јϭ2
2 Јϭ

;

3

Ϫ

2

;

ϩ 8;

ϭ

ϩ 42 ϩ 2

ϭ 5 tan 5
ϭ 1͞(4 Ϫ

cos ;

2

)

ϭ (1 Ϫ sin )Ϫ1/2

In Problems 19 and 20 verify that the indicated expression is
an implicit solution of the given first-order differential equation
...


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...
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...


●

Use a graphing utility to obtain the graph of an explicit solution
...


΂2

΃

Ϫ1
ϭ
Ϫ1

ϭ ( Ϫ 1)(1 Ϫ 2 ); ln
2

ϩ(

2

Ϫ )

ϭ 0;

2

Ϫ2

2

ϩ

ϭ1

In Problems 33– 36 use the concept that ϭ , Ϫϱ Ͻ Ͻ ϱ,
is a constant function if and only if Ј ϭ 0 to determine
whether the given differential equation possesses constant
solutions
...
Assume
an appropriate interval of definition for each solution
ϭ (1 Ϫ );

ϩ2

1

ϭ

ϭ 1;

1ϩ
Ϫ

ϭ

͵

2

ϩ

0

Ϫ

1

( Ϫ 1) Ј ϭ 1
Љ ϩ 4 Ј ϩ 6 ϭ 10
In Problems 37 and 38 verify that the indicated pair of
functions is a solution of the given system of differential
equations on the interval (Ϫϱ, ϱ)
...
Explain why the piecewisedefined functio
is a solution of the differential equation
on (Ϫϱ, ϱ)
...

In Problems 27–30 find values of
so that the function
ϭ
is a solution of the given differential equation
...

Make up a differential equation that you feel confiden
possesses only the trivial solution ϭ 0
...

What function do you know from calculus is such that
its first derivative is itself? Its first derivative is a
constant multiple of itself? Write each answer in
the form of a first-order differential equation with a
solution
...


Јϩ2 ϭ0

5 Јϭ2

order differential equation

ЉϪ5 Јϩ6 ϭ0

2 Љϩ7 ЈϪ4 ϭ0

terval of definition [

In Problems 31 and 32 find values of so that the function
ϭ is a solution of the given differential equation
...
Find an in-

the interval (Ϫϱ, ϱ)
...
Then find specific constants
and so that ϭ sin ϩ cos is a particular solution of the DE
...
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...
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...
In each case the relation ( , ) ϭ 0
implicitly defines several solutions of the DE
...
Use different
colored pencils to mark off segments, or pieces, on each
graph that correspond to graphs of solutions
...
Use
the solution curve to estimate an interval of definition of
each solution ␾
...

The differential equation ( Ј) 2 Ϫ 4 Ј Ϫ 12 3 ϭ 0 has
the form given in (4)
...

The normal form (5) of an th-order differential equation is equivalent to (4) whenever both forms have
exactly the same solutions
...

Find a linear second-order differential equation
( , , Ј, Љ) ϭ 0 for which ϭ 1 ϩ 2 2 is a twoparameter family of solutions
...


1
1

Graph for Problem 45

Qualitative information about a solution ϭ ␾( ) of a
differential equation can often be obtained from the
equation itself
...


Consider the differential equation > ϭ Ϫ
...

>
and lim > ? What does
What are lim
2

1
1

: Ϫϱ

Graph for Problem 46

The graphs of members of the one-parameter family
3
ϩ 3ϭ3
are called
Verify
that this family is an implicit solution of the first-orde
differential equation
ϭ

( 3 Ϫ 2 3)

...
1
...
Discuss:
How can the DE in Problem 47 help in finding points
on the graph of 3 ϩ 3 ϭ 3 where the tangent line
is vertical? How does knowing where a tangent line is
vertical help in determining an interval of definitio
of a solution ␾ of the DE? Carry out your ideas,
and compare with your estimates of the intervals in
Problem 46
...
Why can’t the interval of
definition be the closed interval Ϫ5, 5]?
In Problem 21 a one-parameter family of solutions of
the DE Ј ϭ (1 Ϫ ) is given
...

Sketch the graph of a solution ϭ ␾( ) of the differential equation whose shape is suggested by
parts (a) – (c)
...

Using only the differential equation, find intervals on
the -axis on which a nonconstant solution ϭ ␾( )
is increasing
...

Consider the differential equation ͞ ϭ ( Ϫ ),
where and are positive constants
...

Using only the differential equation, find intervals on
the -axis on which a nonconstant solution ϭ ␾( )
is increasing
...

Using only the differential equation, explain why
ϭ ͞2 is the -coordinate of a point of inflectio
of the graph of a nonconstant solution ϭ ␾( )
...
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...
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...
These
constant solutions partition the -plane into three
regions
...


In Problems 59 and 60 use a CAS to compute all derivatives
and to carry out the simplifications needed to verify that the
indicated function is a particular solution of the given differential equation
...

Explain why there exist no constant solutions of
the DE
...
For
example, can a solution curve have any relative
extrema?
Explain why ϭ 0 is the -coordinate of a point of
inflection of a solution curve
Sketch the graph of a solution ϭ ␾( ) of the
differential equation whose shape is suggested by
parts (a) –(c)
...
On some interval
containing 0 the problem of solving an th-order differential equation subject to side conditions
specified at 0:
ϭ ΂ , , Ј,
...
,

(1)
( Ϫ1)

( 0) ϭ

Ϫ1,


...
,
( Ϫ1)
( 0) ϭ Ϫ1 are called
Solving an th-order initial-value problem such as (1) frequently entails first finding an
-parameter family of solutions of the given differential equation and then using the initialconditions at 0 to determine the constants in this family
...


where

0,

( 0) ϭ

΃

( Ϫ1)

1,

The cases ϭ 1 and ϭ 2 in (1),
:

ϭ ( , )
:

( 0) ϭ

0

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...
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...
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...


(2)

●

2

solutions of the DE

and

:

2

:
( 0,

0)

Solution curve of
first-orde IVP
solutions of the DE

1

( 0,

ϭ ( , , Ј)

( 0) ϭ

0,

Ј( 0) ϭ

(3)
1

are examples of
and
initial-value problems, respectively
...
For (2) we are seeking a solution
( ) of the differential equation Ј ϭ ( , ) on an interval containing 0 so that its
graph passes through the specified point ( 0, 0)
...
2
...
For (3) we want to find a solution ( ) of the differential equation
Љ ϭ ( , , Ј) on an interval containing 0 so that its graph not only passes through
( 0, 0) but the slope of the curve at this point is the number 1
...
2
...
The words
derive from physical systems where the independent variable is time and where ( 0) ϭ 0 and Ј( 0) ϭ 1 represent the position and velocity, respectively, of an object at some beginning, or initial,
time 0
...
1 you were asked to deduce that ϭ
is a oneparameter family of solutions of the simple first-order equation Ј ϭ All the
solutions in this family are defined on the interval (Ϫϱ, ϱ)
...
In this case ϭ
Ϫ1
Ϫ2
is a solution of the IVP
Јϭ ,

Solution curves of two

(0) ϭ 3
...


The two solution curves are shown in dark blue and dark red in Figure 1
...
3
...
In this
example notice how the interval of definition of the solution ( ) depends on the
initial condition ( 0) ϭ 0
...
2 you will be asked to show that a one-parameter family
of solutions of the first-order differential equation Ј ϩ 2 2 ϭ 0 is ϭ 1͞( 2 ϩ )
...
Thus ϭ 1͞( 2 Ϫ 1)
...
See Figure 1
...
4(a)
...
As can be seen in
Figure 1
...
4(a), the largest intervals on which ϭ 1͞( 2 Ϫ 1) is a solution
are (Ϫϱ,Ϫ1), (Ϫ1, 1), and (1, ϱ)
...
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...
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...
See
the red curve in Figure 1
...
4(b)
...
2 for a continuation of Example 2
...
1 we saw that ϭ 1 cos 4 ϩ 2 sin 4 is a two-parameter
family of solutions of Љ ϩ 16 ϭ 0
...

2
2

(4)

We first apply (␲͞2) ϭ Ϫ2 to the given family of solutions: 1 cos 2␲ ϩ
ϭ Ϫ2
...
We next apply
Ј(␲͞2) ϭ 1 to the one-parameter family ( ) ϭ Ϫ2 cos 4 ϩ 2 sin 4 Differentiating
and then setting ϭ ␲͞2 and Ј ϭ 1 gives 8 sin 2␲ ϩ 4 2 cos 2␲ ϭ 1, from which we
see that 2 ϭ 1
...

4
4
2 sin 2␲

1

1
(0, 1)

Two fundamental questions arise in considering
an initial-value problem:
?
?

solution defined on interval containing = 0

Graphs of function
and solution of IVP in Example 2

For the first-order initial-value problem (2) we ask

{
{

͞

ϭ (

)

?
(

(

0

0

0 )?

0 )?

Note that in Examples 1 and 3 the phrase “ solution” is used rather than “ solution” of the problem
...
At this point it has not been demonstrated
that there is a single solution of each problem
...


4

16

1
0

(0, 0)

Two solutions curves
of the same IVP in Example 4

1
Each of the functions ϭ 0 and ϭ 16 4 satisfies the differential equation
1/2
and the initial condition (0) ϭ 0, so the initial-value problem
͞ ϭ

ϭ

1/2

,

(0) ϭ 0

has at least two solutions
...
2
...

Within the safe confine of a formal course in differential equations one can be
fairly confiden that
differential equations will have solutions and that solutions of
initial-value problems will
be unique
...

Therefore it is desirable to know in advance of trying to solve an initial-value problem

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...
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...


●

whether a solution exists and, when it does, whether it is the only solution of the problem
...
We shall wait until Chapter 4 to address
the question of existence and uniqueness of a second-order initial-value problem
...
If ( , ) and Ѩ ͞Ѩ are continuous
on , then there exists some interval 0: ( 0 Ϫ , 0 ϩ ), Ͼ 0, contained in
[ , ], and a unique function ( ), defined on 0, that is a solution of the initialvalue problem (2)
...
The geometry of Theorem 1
...
1 is illustrated
in Figure 1
...
6
...
Inspection of the functions
( , )ϭ

1/2

and

Ѩ
ϭ
Ѩ
2

1/2

shows that they are continuous in the upper half-plane defined by Ͼ 0
...
2
...
Thus, for example, even without solving it, we know
that there exists some interval centered at 2 on which the initial-value problem
͞ ϭ 1/2, (2) ϭ 1 has a unique solution
...
2
...
This follows from the fact that ( , ) ϭ and
Ѩ ͞Ѩ ϭ 1 are continuous throughout the entire -plane
...

Suppose ( ) represents a solution of
the initial-value problem (2)
...
Example 2 of
is defined or exists, and the interval 0 of existence
Section 1
...
Now suppose ( 0, 0) is a point in the interior of the rectangular region in Theorem 1
...
1
...
The interval
depends on both ( , ) and the initial condition ( 0) ϭ 0
...
2
...
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...
However, Theorem 1
...
1
does not give any indication of the sizes of intervals and 0;
0

The number Ͼ 0 that defines the interval
: ( 0 Ϫ , 0 ϩ ) could be very small, so it is best to think that the solution ( )
0
is
— that is, a solution defined near the point ( 0, 0)
...
2
...
2
...
This means
that when ( , ) and Ѩ ͞Ѩ are continuous on a rectangular region , it must
always follow that a solution of (2) exists and is unique whenever ( 0, 0) is a
point interior to
However, if the conditions stated in the hypothesis of
Theorem 1
...
1 do not hold, then anything could happen: Problem (2)
still
have a solution and this solution
be unique, or (2) may have several solutions, or it may have no solution at all
...
2
...
On the other hand, the hypotheses of Theorem 1
...
1 do
not hold on the line ϭ 1 for the differential equation
͞ ϭ ͉ Ϫ 1͉
...
Can you guess this solution?
( ) You are encouraged to read, think about, work, and then keep in mind
Problem 49 in Exercises 1
...

( ) Initial conditions are prescribed at a
point 0
...
Conditions such as
(1) ϭ 0,

(5) ϭ 0

and called
ary conditions is called a

or

(p>2) ϭ 0,

Ј(p) ϭ 1


...
For example,

Љ ϩ l ϭ 0,

Ј(0) ϭ 0,

Ј(p) ϭ 0

is a boundary-value problem
...
2
...
The mathematical description of many problems in science and engineering involve
second-order IVPs or two-point BVPs
...


In Problems 1 and 2, ϭ 1͞(1 ϩ 1 Ϫ ) is a one-parameter
family of solutions of the first-order DE Ј ϭ Ϫ 2
...

(0) ϭ Ϫ 1
3

(Ϫ1) ϭ 2

In Problems 3 –6, ϭ 1͞( 2 ϩ ) is a one-parameter family
of solutions of the first-order DE Ј ϩ 2 2 ϭ 0
...
Give the largest
interval over which the solution is defined
(2) ϭ 1
3
(0) ϭ 1

(Ϫ2) ϭ 1
2

(1) ϭ Ϫ4
2

In Problems 7 –10, ϭ 1 cos ϩ 2 sin is a two-parameter
family of solutions of the second-order DE Љ ϩ ϭ 0
...
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...

(0) ϭ Ϫ1,

Ј(0) ϭ 8

(␲> 4) ϭ 12,
(␲> 6) ϭ 1,
2

(␲ ͞2) ϭ 0,

Ј(␲> 6) ϭ 0

Ј(␲> 4) ϭ 212

In Problems 11 –14, ϭ 1 ϩ 2 Ϫ is a two-parameter
family of solutions of the second-order DE Љ Ϫ ϭ 0
...

Ј(0) ϭ 2

(1) ϭ 0,

Ј(1) ϭ

(Ϫ1) ϭ 5,
(0) ϭ 0,

Ј(Ϫ1) ϭ Ϫ5
Ј(0) ϭ 0

In Problems 15 and 16 determine by inspection at least two
solutions of the given first-order IV
...

ϭ1

2/3

ϭ
ϭ

Ϫ

2

2

ϭ
3

2

In Problems 25 –28 determine whether Theorem 1
...
1 guarantees that the differential equation Ј ϭ 1 2 Ϫ 9 possesses a unique solution through the given point
...
Determine whether
this function is also a solution of the initial-value
problem in part (a)
...

Explain part (a) by determining a region in the
-plane for which the differential equation Ј ϭ
would have a unique solution through a point ( 0, 0)
in

Verify that ϭ tan ( ϩ ) is a one-parameter family
of solutions of the differential equation Ј ϭ 1 ϩ 2
...
2
...
Use the family of
solutions in part (a) to find an explicit solution of
the first-order initial-value problem Ј ϭ 1 ϩ 2,
(0) ϭ 0
...

Determine the largest interval of definition for the
solution of the initial-value problem in part (b)
...

Since ( , ) ϭ 2 and Ѩ ͞Ѩ ϭ 2 are continuous
everywhere, the region in Theorem 1
...
1 can be
taken to be the entire -plane
...
Then
find a solution from the family in part (a) that
satisfies (0) ϭ Ϫ1
...

Determine the largest interval of definition for the
solution of the first-order initial-value problem
Ј ϭ 2, (0) ϭ 0
...
]
Show that a solution from the family in part (a)
of Problem 31 that satisfies Ј ϭ 2, (1) ϭ 1, is
ϭ 1͞(2 Ϫ )
...

Are the solutions in parts (a) and (b) the same?
Verify that 3 2 Ϫ 2 ϭ is a one-parameter family of solutions of the differential equation
͞ ϭ3
By hand, sketch the graph of the implicit solution
3 2 Ϫ 2 ϭ 3
...
Give the
interval of definition of each explicit solution
The point (Ϫ2, 3) is on the graph of 3 2 Ϫ 2 ϭ 3,
but which of the explicit solutions in part (b) satisfies (Ϫ2) ϭ 3?
Use the family of solutions in part (a) of Problem 33
to find an implicit solution of the initial-value

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...
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...
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...


●

problem
͞ ϭ 3 , (2) ϭ Ϫ4
...
Match the solution curve with
at least one pair of the following initial conditions
...
If possible, find a solution of the differential
equation that satisfies the given side conditions
...

(0) ϭ 0, (p>4) ϭ 3

(0) ϭ 0, (p) ϭ 0

Ј(0) ϭ 0, Ј(p>6) ϭ 0

(0) ϭ 1, Ј(p) ϭ 5

(0) ϭ 0, (p) ϭ 2

Ј(p>2) ϭ 1, Ј(p) ϭ 0

5

In Problems 45 and 46 use Problem 51 in Exercises 1
...

5

Graph for Problem 35
5

Find a function ϭ ( ) whose graph at each point ( , )
has the slope given by 8 2 ϩ 6 and has the
-intercept (0, 9)
...


5

Consider the initial-value problem Ј ϭ Ϫ 2 ,
(0) ϭ 1
...
2
...

Explain your reasoning
...
Explain your reasoning
...
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...
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...
2
...
Explain why two different solution curves
cannot intersect or be tangent to each other at a point
( 0, 0) in
1
The functions ( ) ϭ 16

( )ϭ

Ά

4

, Ϫϱ Ͻ

0,
1
16

4

,

Ͻ ϱ and
ϭ 0
...
See
Figures 1
...
12(a) and 1
...
12(b), respectively
...

Resolve the apparent contradiction between this fact
and the last sentence in Example 5
...
In
this problem suppose that a model of the growing population of a small community is given by the initial-value
problem
(0) ϭ 100,

where is the number of individuals in the community
and time is measured in years
...
Once we
have studied some methods for solving DEs in Chapters 2 and 4, we return to, and solve, some of
these models in Chapters 3 and 5
...
The mathematical description of a system of phenomenon is
called a
and is constructed with certain goals in mind
...


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...


●

Construction of a mathematical model of a system starts with
()

identification of the variables that are responsible for changing the
system
...
In this step we are specifying the
of
the model
...
These assumptions will also include
any empirical laws that may be applicable to the system
...
For example, you may already be aware that in beginning
physics courses, the retarding force of air friction is sometimes ignored in modeling
the motion of a body falling near the surface of the Earth, but if you are a scientist
whose job it is to accurately predict the flight path of a long-range projectile,
you have to take into account air resistance and other factors such as the curvature
of the Earth
...

Once we have formulated a mathematical model that is either a differential
equation or a system of differential equations, we are faced with the not insignifican
problem of trying to solve it
...
But if the predictions produced by the solution are poor,
we can either increase the level of resolution of the model or make alternative assumptions about the mechanisms for change in the system
...
3
...

Express assumptions
in terms of DEs
If necessary,
alter assumptions
or increase resolution
of model

Solve the DEs

Display predictions
of model
(e
...
, graphically)

Steps in the modeling process with differential equations

Of course, by increasing the resolution, we add to the complexity of the mathematical model and increase the likelihood that we cannot obtain an explicit solution
...

One of the earliest attempts to model human
by means of mathematics was by the English clergyman and economist Thomas Malthus in 1798
...
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...
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...
In other words, the more
people there are at time , the more there are going to be in the future
...
This simple model, which fails to take into
account many factors that can influence human populations to either grow or decline
(immigration and emigration, for example), nevertheless turned out to be fairly accurate in predicting the population of the United States during the years 1790 – 1860
...

The nucleus of an atom consists of combinations of protons
and neutrons
...
Such nuclei are
said to be radioactive
...
To model the phenomenon of
it is assumed that the rate
͞ at which the nuclei of a substance decay is proportional to the amount (more precisely, the number of nuclei)
( ) of the substance remaining at time :
ϰ

or

ϭ


...
For growth, as we
expect in (1), Ͼ 0, and for decay, as in (2), Ͻ 0
...
The model (2) for decay also occurs in biological applications such as
determining the half-life of a drug —the time that it takes for 50% of a drug to be
eliminated from a body by excretion or metabolism
...
The
point is this:

Mathematical models are often accompanied by certain side conditions
...
If the initial point in time is
taken to be ϭ 0, then we know that (0) ϭ 0 and (0) ϭ 0
...
2, a boundary-value problem
...
If ( ) represents the temperature of a body at time ,
the temperature of the surrounding

*

If two quantities and are proportional, we write ϰ
multiple of the other: ϭ

This means that one quantity is a constant

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...
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...
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...


●

medium, and ͞ the rate at which the temperature of the body changes, then
Newton’s law of cooling/warming translates into the mathematical statement
ϰ

or

Ϫ

ϭ ( Ϫ

),

where is a constant of proportionality
...


(3)
is a

A contagious disease —for example, a flu virus —is
spread throughout a community by people coming into contact with other people
...
It seems reasonable to assume that
the rate ͞ at which the disease spreads is proportional to the number of encounters, or
between these two groups of people
...
Suppose a small community has a
fixed population of people
...
Using
this last equation to eliminate in (4) gives us the model
ϭ

( ϩ 1 Ϫ )
...

The disintegration of a radioactive substance, governed
by the differential equation (1), is said to be a
In chemistry
a few reactions follow this same empirical law: If the molecules of substance
decompose into smaller molecules, it is a natural assumption that the rate at which
this decomposition takes place is proportional to the amount of the first substance
that has not undergone conversion; that is, if ( ) is the amount of substance
remaining at any time, then
͞ ϭ , where is a negative constant since is
decreasing
...

Only the concentration of the -butyl chloride controls the rate of reaction
...
In this case the rate at which the reaction
proceeds is proportional to the product of the remaining concentrations of CH3Cl and
NaOH
...
Hence the rate of formation of is given by
ϭ (␣ Ϫ )(␤ Ϫ ),

(6)

where is a constant of proportionality
...
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...
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...
Let us suppose that a large mixing tank initially holds 300 gallons of brine (that
is, water in which a certain number of pounds of salt has been dissolved)
...
When the solution in
the tank is well stirred, it is pumped out at the same rate as the entering solution
...
3
...
If ( ) denotes the amount of salt (measured in pounds) in the tank at
time , then the rate at which ( ) changes is a net rate:
ϭ

output rate of brine
3 gal/min

΂

΃Ϫ΂

΃ϭ

Ϫ


...
Note that
is measured in pounds per
minute:
concentration
of salt
in inflo

Mixing tank

input rate
of brine

input rate
of salt

ϭ (2 lb/gal) и (3 gal/min) ϭ (6 lb/min)
...
Hence the concentration of the salt in the tank as well as in the outflow is
of salt is
( ) ϭ ( )͞300 lb/gal, so the output rate
concentration
of salt
in outflo

(

output rate
of brine

output rate
of salt

)

()
()
ϭ –––– lb/gal и (3 gal/min) ϭ –––– lb/min
...


(8)

denote general input and output rates of the brine solutions,* then
If and
there are three possibilities: ϭ
, Ͼ
, and Ͻ

...
In the latter two cases the number of gallons of brine in the tank is either increasing ( Ͼ
) or decreasing ( Ͻ
) at
the net rate Ϫ

...
3
...
This last expression comes from equating the kinetic energy 1 2 with the
2
potential energy
and solving for Suppose a tank filled with water is allowed to
drain through a hole under the influence of gravity
...
3
...
If
the area of the hole is
(in ft 2) and the speed of the water leaving the tank is
ϭ 12 (in ft/s), then the volume of water leaving the tank per second is 12
(in ft 3/s)
...
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...
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...
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...
Note here that we are ignoring
the possibility of friction at the hole that might cause a reduction of the rate of flo
there
...
3
...


-series circuit

ϭϪ
inductance : henries (h)
voltage drop across:

(10)

It is interesting to note that (10) remains valid even when
is not constant
...
See Problem 14 in Exercises 1
...

Consider the single-loop
-series circuit shown in Figure 1
...
4(a), containing an inductor, resistor, and capacitor
...
The letters , , and are known as inductance, resistance, and capacitance, respectively, and are generally constants
...
Figure 1
...
4(b) shows the symbols and the formulas for the
respective voltage drops across an inductor, a capacitor, and a resistor
...
Current ( ) and charge ( ) are
measured in amperes (A) and coulombs
(C), respectively

0

rock

0

and

()

building
ground

Position of rock
measured from ground level

ϩ

ϩ

1

ϭ ( )
...
1
...
Recall from elementary physics
that
states that a body either will remain at rest or will
continue to move with a constant velocity unless acted on by an external force
...

indicates that when the net
force acting on a body is not zero, then the net force is proportional to its acceleration or, more precisely, ϭ , where is the mass of the body
...
3
...
What is the position ( ) of the rock relative to the ground at time ?
The acceleration of the rock is the second derivative 2 ͞ 2
...


(12)

In other words, the net force is simply the weight ϭ 1 ϭ Ϫ of the rock near the
surface of the Earth
...
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...
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...
The minus sign in (12) is
used because the weight of the rock is a force directed downward, which is opposite
to the positive direction
...


(13)

Although we have not been stressing solutions of the equations we have constructed, note that (13) can be solved by integrating the constant Ϫ twice with
respect to The initial conditions determine the two constants of integration
...

2

positive
direction

air resistance

gravity

Falling body of mass

Before the famous experiment by the
Italian mathematician and physicist
(1564–1642) from the leaning
tower of Pisa, it was generally believed that heavier objects in free fall, such as a cannonball, fell with a greater acceleration than lighter objects, such as a feather
...
The difference in rates is due to air resistance
...
Under some circumstances a falling body of mass , such as a
feather with low density and irregular shape, encounters air resistance proportional
to its instantaneous velocity If we take, in this circumstance, the positive direction
to be oriented downward, then the net force acting on the mass is given by ϭ 1 ϩ
Ϫ , where the weight 1 ϭ
of the body is force acting in the positive
2ϭ
direction and air resistance 2 ϭ Ϫ is a force, called
acting in the
opposite or upward direction
...
3
...
Now since is related to acceleration
by ϭ ͞ , Newton’s second law becomes ϭ
ϭ
͞ By equating the
net force to this form of Newton’s second law, we obtain a first-order differential
equation for the velocity ( ) of the body at time ,
ϭ

Ϫ


...
If ( ) is the distance the body falls in
time from its initial point of release, then ϭ ͞ and ϭ ͞ ϭ 2 ͞ 2
...


(15)

Suppose a flexible cable, wire, or heavy rope is suspended between two vertical supports
...
3
...
3
...
Our
goal is to construct a mathematical model that describes the shape that such a cable
assumes
...
As drawn in blue in Figure 1
...
8, this
element of the cable is the curve in a rectangular coordinate system with -axis chosen to pass through the lowest point 1 on the curve and the -axis chosen units
below 1
...
Let 1 ϭ ͉ 1͉, 2 ϭ ͉ 2͉, and
ϭ ͉ ͉ denote the magnitudes of these vectors
...


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...
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...
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...


●

Because of static equilibrium we can write
1

ϭ

2

cos ␪

and

ϭ

By dividing the last equation by the first, we eliminate
because ͞ ϭ tan ␪, we arrive at
ϭ

2

sin ␪
...
But


...
We will come back to
equation (16) in Exercises 2
...
3
...
Over the centuries differential
equations would often spring from the efforts of a scientist or engineer to describe
some physical phenomenon or to translate an empirical or experimental law into
mathematical terms
...

With a solution in hand, the study of its properties then followed
...
Once they
realized that explicit solutions are at best difficul to obtain and at worst impossible
to obtain, mathematicians learned that a differential equation itself could be a font of
valuable information
...
In Chapter 2 we start with qualitative considerations of first
order ODEs, then examine analytical stratagems for solving some special first-orde
equations, and conclude with an introduction to an elementary numerical method
...
3
...


Different approaches to the study of differential equations

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...
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...


●

Each example in this section has described a dynamical system —a system that
changes or evolves with the flow of time Since the study of dynamical
systems is a branch of mathematics currently in vogue, we shall occasionally
relate the terminology of that field to the discussion at hand
In more precise terms, a
consists of a set of timedependent variables, called
together with a rule that enables
us to determine (without ambiguity) the state of the system (this may be a past,
present, or future state) in terms of a state prescribed at some time 0
...

In this course we shall be concerned only with continuous-time systems —
systems in which
variables are defined over a continuous range of time
...
The
at a time is the value of the state variables at that time; the specified state of
the system at a time 0 is simply the initial conditions that accompany the mathematical model
...
1)
...
In the case of the rock tossed from the roof
of a building, the response of the system — the solution of the differential
equation 2 ͞ 2 ϭ Ϫ , subject to the initial state (0) ϭ 0, Ј(0) ϭ 0 , is the
function ( ) ϭ Ϫ1 2 ϩ 0 ϩ 0, 0 Յ Յ , where
represents the time
2
when the rock hits the ground
...
The acceleration Љ( ) is
a state variable, since we have to know
only any initial position and initial velocity at a time 0 to uniquely determine
the rock’s position ( ) and velocity Ј( ) ϭ ( ) for any time in the interval
The acceleration Љ( ) ϭ ( ) is, of course, given by the differential
0Յ Յ
equation Љ( ) ϭ Ϫ , 0 Ͻ Ͻ
One last point: Not every system studied in this text is a dynamical system
...


Under the same assumptions that underlie the model in
(1), determine a differential equation for the population
( ) of a country when individuals are allowed to
immigrate into the country at a constant rate Ͼ 0
...

In another model of a changing population of a community it is assumed that the rate at which the population
changes is a
rate — that is, the difference between

the rate of births and the rate of deaths in the community
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

A cup of coffee cools according to Newton’s law of
cooling (3)
...
3
...


number of people ( ) who have adopted the innovation
at time if it is assumed that the rate at which the innovations spread through the community is jointly proportional to the number of people who have adopted it and
the number of people who have not adopted it
...
Pure water is pumped into the tank at a rate of
3 gal/min, and when the solution is well stirred, it is
then pumped out at the same rate
...
3
...
Devise a mathematical
model for the temperature ( ) of a body within this
environment
...
Another brine solution is pumped into the tank
at a rate of 3 gal/min, and when the solution is well
stirred, it is then pumped out at a
rate of 2 gal/min
...
5 gal/min?
Generalize the model given in equation (8) on page 24
by assuming that the large tank initially contains 0
number of gallons of brine, and
are the input and
output rates of the brine, respectively (measured in gallons per minute),
is the concentration of the salt in
the inflo , ( ) the concentration of the salt in the tank
as well as in the outflow at time (measured in pounds
of salt per gallon), and ( ) is the amount of salt in the
tank at time Ͼ 0

40
20
0

12

24

36

48

midnight noon midnight noon midnight

Ambient temperature in Problem 6

Suppose a student carrying a fl virus returns to an isolated college campus of 1000 students
...

At a time denoted as ϭ 0 a technological innovation is
introduced into a community that has a fixed population
of people
...
When water leaks through a
hole, friction and contraction of the stream near the hole
reduce the volume of water leaving the tank per second to
12 , where (0 Ͻ Ͻ 1) is an empirical constant
...
3
...
The
radius of the hole is 2 in
...


10 ft

circular
hole

Cubical tank in Problem 13

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...
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...
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...


●

The right-circular conical tank shown in Figure 1
...
13
loses water out of a circular hole at its bottom
...
, ϭ 32 ft/s 2,
and the friction/contraction factor introduced in Problem
13 is ϭ 0
...


2

SKYD IVING
MADE
EASY

8 ft

20 ft

Air resistance proportional to square of
velocity in Problem 17

circular hole

Conical tank in Problem 14

A series circuit contains a resistor and an inductor as
shown in Figure 1
...
14
...


A cylindrical barrel feet in diameter of weight lb
is floating in water as shown in Figure 1
...
17(a)
...
Using
Figure 1
...
17(b), determine a differential equation for
the vertical displacement ( ) if the origin is taken to be
on the vertical axis at the surface of the water when the
barrel is at rest
...
Assume that the
downward direction is positive, that the weight density
of water is 62
...

/2
/2

-series circuit in Problem 15

surface

0

A series circuit contains a resistor and a capacitor as
shown in Figure 1
...
15
...


0

()

Bobbing motion of floating barrel i
Problem 18

After a mass
is attached to a spring, it stretches it
units and then hangs at rest in the equilibrium position
as shown in Figure 1
...
18(b)
...
3
...
Determine a differential equation for the velocity ( ) of a falling body of
mass if air resistance is proportional to the square of
the instantaneous velocity
...
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...
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...
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...
As indicated in Figure 1
...
17(c), assume that the
downward direction is positive, that the motion takes
place in a vertical straight line through the center of
gravity of the mass, and that the only forces acting on
the system are the weight of the mass and the restoring
force of the stretched spring
...
Determine a differential equation for the
displacement ( ) at time Ͼ 0
In Problem 19, what is a differential equation for the
displacement ( ) if the motion takes place in a
medium that imparts a damping force on the spring/
mass system that is proportional to the instantaneous
velocity of the mass and acts in a direction opposite to
that of motion?

When the mass of a body is changing with time, Newton’s
second law of motion becomes
ϭ

(

),

(17)

where is the net force acting on the body and
momentum
...


is its

A small single-stage rocket is launched vertically as
shown in Figure 1
...
19
...
If it is assumed that the positive direction is upward, air resistance is proportional to the instantaneous
velocity of the rocket, and is the upward thrust or
force generated by the propulsion system, then construct a mathematical model for the velocity ( ) of the
rocket
...
3
...

Show that the rate at which the total mass ( ) of the
rocket changes is the same as the rate at which the
mass ( ) of the fuel changes
...
Then rewrite the differential equation in
Problem 21 in terms of l and the initial total mass
(0) ϭ 0
...


By
the free-fall
acceleration of a body, such as the satellite shown in
Figure 1
...
20, falling a great distance to the surface is
the constant Rather, the acceleration is inversely proportional to the square of the distance from the center of
the Earth, ϭ ͞ 2, where is the constant of proportionality
...
Davidson/Shutterstock
...
3
...
Construct a mathematical model
that describes the motion of the ball
...

surface

Hole through
Earth in Problem 24

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...
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...
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...


●

In the theory of learning, the rate at
which a subject is memorized is assumed to be proportional to the amount that is left to be memorized
...

In Problem 25 assume that the rate at
which material is
is proportional to the amount
memorized in time Ͼ 0 Determine a differential equation for the amount ( ) when forgetfulness is taken into
account
...

Simultaneously, the drug is removed at a rate proportional
to the amount ( ) of the drug present at time Determine
a differential equation for the amount ( )
...
3
...

The weight, initially located on the -axis at (0, ), is
pulled by a rope of constant length , which is kept taut
throughout the motion
...
Assume that the rope is always
tangent to

Reread Problem 41 in Exercises 1
...
Find a oneparameter family of solutions of (1)
...
Discuss why we would expect Ͻ 0 in (3) in both cases of cooling and warming
...

Reread the discussion leading up to equation (8)
...
Discuss how you might determine from
the DE, without actually solving it, the number of
pounds of salt in the tank after a long period of time
...
Discuss an interpretation for the solution of this
equation
...
3
...
Use the fact that the angle of incidence is equal
to the angle of reflection to determine a differential
tangent

(

equation that describes the shape of the curve Such a
curve is important in applications ranging from construction of telescopes to satellite antennas, automobile
headlights, and solar collectors
...
Why? Now use
an appropriate trigonometric identity
...
3
...
The rotating fluid forms a surface of
revolution To identify , we first establish a coordinate
system consisting of a vertical plane determined by the
-axis and an -axis drawn perpendicular to the -axis
such that the point of intersection of the axes (the origin)
is located at the lowest point on the surface We then
seek a function ϭ ( ) that represents the curve of intersection of the surface and the vertical coordinate
plane
...
See Figure 1
...
23(b)
...
3
...


Use part (a) to find a first-order differential equation
that defines the function ϭ ( )
...
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...
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...
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...
Assume that a
typical raindrop is spherical
...

If it is assumed that a raindrop evaporates in such a
manner that its shape remains spherical, then it also
makes sense to assume that the rate at which the raindrop evaporates—that is, the rate at which it loses
mass—is proportional to its surface area
...

Find ( )
...
1
...
[
:
Use the form of Newton’s second law given in
(17)
...
What does the differential equation
obtained in Problem 23 become when is very small in
comparison to ? [
: Think binomial series for
( ϩ )Ϫ2 ϭ

Ϫ2

(1 ϩ ͞ )Ϫ2
...
The symbol 1
represents a constant
...


In Problems 5 and 6 compute Ј and Љ and then combine
these derivatives with as a linear second-order differential
equation that is free of the symbols 1 and 2 and has the form
( , Ј Љ) ϭ 0
...


cosh

ϩ

1

ϩ

ϭ

2

1

cos ϩ

sin

2

In Problems 7 –12 match each of the given differential equations with one or more of these solutions:

)ϭ

In Problems 3 and 4 fill in the blank and then write this result
as a linear second-order differential equation that is free of
the symbols 1 and 2 and has the form ( , Љ) ϭ 0
...

(
2

Find the textbook
Ralph Palmer
Agnew, McGraw-Hill Book Co
...


ϭ

ϭ
1

?

2

sinh

)ϭ

ϭ 0,
Јϭ2

ϭ 2,

ϭ2 ,

ϭ 2 2
...

Љϭ Ј

Ј ϭ ( Ϫ 3)

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 15 and 16 interpret each statement as a differential equation
...

On the graph of ϭ ␾( ) the rate at which the slope
changes with respect to at a point ( , ) is the negative of the slope of the tangent line at ( , )
...

Give the largest interval of definition over which
ϭ 2/3 is solution of the differential equation
3 Ј Ϫ 2 ϭ 0
...

Find a member of the one-parameter family in
part (a) that satisfies the initial condition (0) ϭ 1
...
Give the
domain of the function ␾
...

ϩ

ϭ

1
2

3 3

;

΂ ΃ ϩ1ϭ 1;
2

(1 Ϫ

) Јϭ

3

ϩ1

( Ϫ 5)2 ϩ

2

Љ ϭ 2 ( Ј)3;

ϭ

2

ϭ1

3

2

ϩ3 ϭ1Ϫ3
;
ϭ

In Problems 31– 34, ϭ 1 3 ϩ 2 Ϫ Ϫ 2 is a twoparameter family of the second-order DE Љ Ϫ 2 Ј Ϫ 3 ϭ
6 ϩ 4
...

(0) ϭ 0, Ј(0) ϭ 0

(0) ϭ 1, Ј(0) ϭ Ϫ3

(1) ϭ 4, Ј(1) ϭ Ϫ2

(Ϫ1) ϭ 0, Ј(Ϫ1) ϭ 1

The graph of a solution of a second-order initial-value
problem 2 ͞ 2 ϭ ( , , Ј), (2) ϭ 0, Ј(2) ϭ 1, is
given in Figure 1
...
1
...

5

Given that ϭ Ϫ 2͞ is a solution of the DE Ј ϩ
ϭ 2 Find 0 and the largest interval for which ( ) is
a solution of the first-order IVP Ј ϩ ϭ 2 , ( 0 ) ϭ 1
...
In some neighborhood of ϭ 1 use the DE to determine whether
( ) is increasing or decreasing and whether the graph
( ) is concave up or concave down
...

Plot different members of the families
ϭ ␾1( ) ϭ 2 ϩ 1 and ϭ ␾ 2( ) ϭ Ϫ 2 ϩ 2
...

Construct a piecewise-defined function that is a
solution of the nonlinear DE in part (b) but is not a
member of either family of solutions in part (a)
...
Give an
interval of definition for each solution
...
If the tank
is initially full of water and water leaks from a circular
hole of radius 1 inch at its bottom, determine a differen2
tial equation for the height of the water at time Ͼ 0
Ignore friction and contraction of water at the hole
...
Determine a differential equation governing a
population of owls that feed on the mice if the rate at
which the owl population grows is proportional to the
difference between the number of owls at time and
number of field mice at time Ͼ 0
Suppose that
͞ ϭ Ϫ0
...
How
much of the radium sample remains at the time when
the sample is decaying at a rate of 0
...
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...
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...
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...
1 Solution Curves Without a Solution
2
...
1 Direction Fields
2
...
2 Autonomous First-Order DEs
2
...
3 Linear Equations
2
...
5 Solutions by Substitutions
2
...
In Sections 2
...
5 we will study some of the more
important analytical methods for solving first-order DEs
...
In Sections 2
...
6
we do not solve any DEs but show how to glean information about solutions
directly from the equation itself
...
1 we see how the DE yields
qualitative information about graphs that enables us to sketch renditions of solution
curves
...
6 we use the differential equation to construct a procedure,
called a numerical method, for approximating solutions
...
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...
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...
This is not as bad a predicament as one might think, since the differential equation itself can sometimes “tell” us specifics about how its solutions “behave
...
Both these ways enable us to determine, in an approximate sense, what a solution curve
must look like without actually solving the equation
...
2 that whenever ( , )
and Ѩ ͞Ѩ satisfy certain continuity conditions, qualitative questions about existence
and uniqueness of solutions can be answered
...
2

Because a solution ϭ ( ) of a first-order di ferential equation
(2, 3)

ϭ ( , )

solution
curve

(2, 3)
tangent

A solution curve is
tangent to lineal element at (2, 3)

(1)

is necessarily a differentiable function on its interval of definition, it must also be continuous on Thus the corresponding solution curve on must have no breaks and must
possess a tangent line at each point ( , ( ))
...
Now suppose that ( , )
represents any point in a region of the -plane over which the function is defined
...
2 , where ( , ) ϭ 0
...
2(2)(3) ϭ 1
...
Figure 2
...
1(a) shows a line segment with
slope 1
...
As shown in Figure 2
...
1(b), a solution curve also
passes through the point (2, 3), it does so tangent to this line segment; in other words,
the lineal element is a miniature tangent line at that point
...
Visually, the direction
field suggests the appearance or shape of a family of solution curves of the
differential equation, and consequently, it may be possible to see at a glance certain
qualitative aspects of the solutions — regions in the plane, for example, in which a

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
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...


●

solution exhibits an unusual behavior
...
Figure 2
...
2 shows a computer-generated direction field of the differential equation ͞ ϭ sin( ϩ ) over a region of the -plane
...
2 shown in Figure 2
...
3(a)
was obtained by using computer software in which a 5 ϫ 5 grid of points ( , ),
and integers, was defined by letting Ϫ5 Յ Յ 5, Ϫ5 Յ Յ 5, and ϭ 1
...
1
...
Moreover, observe in the first quadrant that for a fixed value
of the values of ( , ) ϭ 0
...
2 increase as increases
...
2 Ͼ 0 for Ͼ 0, Ͼ 0)
...
2
0 for
0, Ͼ 0)
...
From this it could be surmised that : ϱ
as : Ϯϱ
...
2 Ͼ 0 and
( , ) ϭ 0
...
We saw in (1) of Section 1
...
1 is an explicit solution of the differential equation ͞ ϭ 0
...
1
...
1
...
1
...


2
c=0 x
c<0
_2

Use a direction field to sketch an approximate solution curve for the initial-value
problem ͞ ϭ sin , (0) ϭ Ϫ3
...
2
...
Now we set our computer software again for a 5 ϫ 5 rectangular region and specify (because of the initial condition)
points in that region with vertical and horizontal separation of 1 unit — that is, at
2
points ( , ), ϭ 1 , and integers such that Ϫ10 Յ Յ 10, Ϫ10 Յ Յ 10
...
1
...
Because the right-hand side of ͞ ϭ sin is 0
at ϭ 0, and at ϭ Ϫ␲, the lineal elements are horizontal at all points whose second
coordinates are ϭ 0 or ϭ Ϫ␲
...
Another
telling property of the first derivative will be used next, namely, if ͞ Ͼ 0 (or
͞
0) for all in an interval , then a differentiable function ϭ ( ) is
increasing (or decreasing) on

Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

y
4
2

x
_2
_4
_4

_2

2

4

Sketching a direction fiel by hand is straightforward but time consuming; it is
probably one of those tasks about which an argument can be made for doing it
once or twice in a lifetime, but it is overall most efficientl carried out by means
of computer software
...
For the DE
͞ ϭ ( , ), any member of the family of curves ( , ) ϭ , a constant,
is called an
Lineal elements drawn through points on a specifi isocline, say, ( , ) ϭ 1 all have the same slope 1
...
1
you have your two opportunities to sketch a direction fiel by hand
...
1 we divided the class of ordinary differential equations into two types: linear and nonlinear
...
An ordinary differential equation in which the independent variable does
not appear explicitly is said to be
If the symbol denotes the independent variable, then an autonomous first-order differential equation can be written as
( , Ј) ϭ 0 or in normal form as
ϭ ( )
...
2

are autonomous and nonautonomous, respectively
...
As we have
already seen in Section 1
...
For example, if
represents time then inspection of
ϭ

,

ϭ

( ϩ 1 Ϫ ),

ϭ ( Ϫ

),

ϭ6Ϫ

1
,
100

where , , and
are constants, shows that each equation is time independent
...
3 are time
independent and so are autonomous
...
We
say that a real number is a
of the autonomous differential equation (2)
if it is a zero of — that is, ( ) ϭ 0
...
This means:
(2),

( )ϭ

A constant solution ( ) ϭ of (2) is called an
the
constant solutions of (2)
...
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...
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...
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...


The differential equation
ϭ ( Ϫ

Phase portrait of
DE in Example 3

),

where and are positive constants, has the normal form ͞ ϭ ( ), which is (2)
with and playing the parts of and , respectively, and hence is autonomous
...
The arrows on the line shown in Figure 2
...
5 indicate
the algebraic sign of ( ) ϭ ( Ϫ ) on these intervals and whether a nonconstant
solution ( ) is increasing or decreasing on an interval
...
1
...
The vertical line is called a

Without solving an autonomous differential equation, we
can usually say a great deal about its solution curves
...
Also, since and its derivative Ј are continuous functions of on some
interval of the -axis, the fundamental results of Theorem 1
...
1 hold in some horizontal strip or region in the -plane corresponding to , and so through any point
( 0, 0) in there passes only one solution curve of (2)
...
1
...
For the
sake of discussion, let us suppose that (2) possesses exactly two critical points 1 and
2 and that 1
2
...
1
...
Without proof here are some conclusions
that we can draw about a nonconstant solution ( ) of (2):

( ) ϭ 2 partition
subregions

Lines ( ) ϭ 1 and
into three horizontal

• If ( 0, 0) is in a subregion , ϭ 1, 2, 3, and ( ) is a solution whose
graph passes through this point, then ( ) remains in the subregion for all
As illustrated in Figure 2
...
6(b), the solution ( ) in 2 is bounded below
by 1 and above by 2, that is, 1
( )
The solution curve
2 for all
stays within 2 for all because the graph of a nonconstant solution of (2)
cannot cross the graph of either equilibrium solution ( ) ϭ 1 or ( ) ϭ 2
...
1
...
In other words, ( ) cannot change signs in a
subregion
...
1
...
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...
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...
Therefore ( ) cannot be oscillatory, nor
can it have a relative extremum (maximum or minimum)
...
1
...
If ( ) is
— that is, bounded above and below by two consecutive critical
points (as in subregion 2 where 1
( )
2 for all ) — then the graph
of ( ) must approach the graphs of the equilibrium solutions ( ) ϭ 1 and
( ) ϭ 2, one as : ϱ and the other as : Ϫϱ
...
See Problem 34 in Exercises 2
...

With the foregoing facts in mind, let us reexamine the differential equation in
Example 3
...
The phase portrait in Figure 2
...
7 tells us that ( ) is decreasing
in 1, increasing in 2, and decreasing in 3
...
Since ( ) is decreasing, ( )
decreases without bound for increasing , and so ( ) : 0 as : Ϫϱ
...

( ) For 0
͞ , ( ) is bounded
...
The graphs of the two equilibrium
solutions, ( ) ϭ 0 and ( ) ϭ ͞ , are horizontal lines that are horizontal
asymptotes for any solution curve starting in this subregion
...
Since ( ) is decreasing, ( ) : ͞
as : ϱ
...


In Figure 2
...
7 the phase line is the -axis in the -plane
...
1
...
The graphs of the equilibrium solutions
( ) ϭ ͞ and ( ) ϭ 0 (the -axis) are shown in the figure as blue dashed lines;
the solid graphs represent typical graphs of ( ) illustrating the three cases just
discussed
...
Do
interpret this last statement to
mean ( ) : Ϫϱ as : ϱ; we could have ( ) : Ϫϱ as : , where Ͼ 0 is a
finite number that depends on the initial condition ( 0) ϭ 0
...
A similar remark holds for the subregion 3
...
Moreover,
we now know that because the solution ( ) that passes through (0, Ϫ3) is bounded
2

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...
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...
Editorial review has
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...


●

above and below by two consecutive critical points (Ϫ␲
( ) 0) and is
decreasing (sin
0 for Ϫ␲
0), the graph of ( ) must approach the graphs
of the equilibrium solutions as horizontal asymptotes: ( ) : Ϫ␲ as : ϱ and
( ) : 0 as : Ϫϱ
...

From the phase portrait in Figure 2
...
8(a) we conclude that a solution ( ) is an
increasing function in the subregions defined by Ϫϱ
1 and 1
Ͻ ϱ, where
Ϫϱ
Ͻ ϱ
...

Now ( ) ϭ 1 Ϫ 1͞( ϩ ) is a one-parameter family of solutions of the differential equation
...
2
...
As shown in
2
Figures 2
...
8(b) and 2
...
8(c), the graph of each of these rational functions possesses

1

increasing

(0, 2)
1

1

1

(0, 1)

increasing
1
2

Ͻ

Ͼ

Behavior of solutions near ϭ 1 in Example 5

a vertical asymptote
...
They are, respectively,
( )ϭ1Ϫ

0

0

1
,
ϩ1
2

Ϫ1
2

Ͻϱ

and

( )ϭ1Ϫ

1
,
Ϫ1

Ϫϱ

1
...
1
...
1
...
As predicted by the phase portrait, for the solution curve
in Figure 2
...
8(b), ( ) : 1 as : ϱ; for the solution curve in Figure 2
...
8(c),
( ) : ϱ as : 1 from the left
...


Suppose that ( ) is a nonconstant solution of the
autonomous differential equation given in (1) and that is a critical point of the DE
...
1
...
When both arrowheads on
either side of the dot labeled point
, as in Figure 2
...
9(a), all solutions ( )
of (1) that start from an initial point ( 0, 0) sufficiently near exhibit the asymptotic behavior lim :ϱ ( ) ϭ
...
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...
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...
1
...
In this case the critical
point is said to be
An unstable critical point is also called a
for
obvious reasons
...
1
...
1
...
But since exhibits characteristics of both an
attractor and a repeller—that is, a solution starting from an initial point ( 0, 0) suffi
ciently near is attracted to from one side and repelled from the other side—we say
that the critical point is
In Example 3 the critical point ͞ is
asymptotically stable (an attractor) and the critical point 0 is unstable (a repeller)
...


slopes of lineal
elements on a
vertical line vary
y

y
x

Direction field for a
autonomous DE

If a first-order differential equation is autonomous, then we see from the right-hand side of its normal form ͞ ϭ
( ) that slopes of lineal elements through points in the rectangular grid used to construct a direction field for the DE depend solely on the -coordinate of the points
...
These facts are apparent from inspection of the horizontal yellow strip and vertical blue strip in Figure 2
...
10
...
The red lineal elements
in Figure 2
...
10 have zero slope because they lie along the graph of the equilibrium
solution ϭ 1
...
It turns out that under the conditions stipulated for (2), solution curves of an autonomous first-order DE are related
by the concept of translation
...
Because ϭ 0 and ϭ 3 are equilibrium solutions of the DE,
their graphs divide the -plane into three subregions 1, 2, and 3:
1:

y

2:

0

3,

and

3:

3

ϱ
...
1
...
The figure illustrates that all solution curves of the same color, that is, solution curves lying within a particular subregion , all look alike
...
That said, the following
of an automonous DE should make sense:

3

0

0,

Ϫϱ

x
1(

Translated solution
curves of an autonomous DE

͞

( )
)ϭ ( Ϫ )

ϭ ( )

,

Thus, if ( ) is a solution of the initial-value problem ͞ ϭ ( ), (0) ϭ 0, then
ϭ ( ), ( 0) ϭ 0
...


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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

y

4

In Problems 1 – 4 reproduce the given computer-generated
direction field
...

Use different colored pencils for each solution curve
...
01

(Ϫ6) ϭ 0
(0) ϭ Ϫ4

_4

(0) ϭ 1
(8) ϭ Ϫ4

_4

_2

2

4

Direction field for Problem
y

In Problems 5 –12 use computer software to obtain a direction field for the given differential equation
...


8
4
x

Јϭ
(0) ϭ 0
(0) ϭ Ϫ3

_4

Јϭ ϩ
(Ϫ2) ϭ 2
(1) ϭ Ϫ3

_8
_8

_4

4

8

Direction field for Problem

(Ϫ1) ϭ 0
(0) ϭ Ϫ4

ϭ

(1) ϭ 1
(0) ϭ 4
ϭ 0
...
5

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...
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...
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...


●

Јϭ

Ϫ cos

ϭ1Ϫ

2

(Ϫ1) ϭ 2
2
3
(2) ϭ 0

(2) ϭ 2
(Ϫ1) ϭ 0

In Problems 13 and 14 the given figure represents the graph
of ( ) and ( ), respectively
...


1
1

For a first-orde DE ͞ ϭ ( , ) a curve in the plane
define by ( , ) ϭ 0 is called a
of the equation, since a lineal element at a point on the curve has zero
slope
...
Discuss the behavior of solution
1 2
curves in regions of the plane defined by
and by
2
1 2
Ͼ 2
...
Try
to generalize your observations
...
With a colored pencil, circle
any lineal elements in Figures 2
...
12, 2
...
14, and
2
...
15 that you think may be a lineal element at a
point on a nullcline
...

By hand, sketch the graph of a typical solution ( )
when 0 has the given values
...
In
each case, use this rough direction field to sketch an approximate solution curve for the IVP consisting of the DE
and the initial condition (0) ϭ 1
...
Describe the slopes of the lineal
elements on the lines ϭ 0, ϭ 3, ϭ 4, and ϭ 5
...
Can a solution ( ) : ϱ as : ϱ?
Based on the information in part (a), discuss
...
By
hand, sketch the graph of a typical solution ( ) when 0
has the given values
...

Classify each critical point as asymptotically stable, unstable,
or semi-stable
...

ϭ

2

Ϫ3

ϭ

ϭ ( Ϫ 2)4
ϭ

2

(4 Ϫ

2

Ϫ

3

ϭ 10 ϩ 3 Ϫ
2

)

ϭ ln( ϩ 2)

2

ϭ (2 Ϫ )(4 Ϫ )
ϭ

Ϫ9

In Problems 29 and 30 consider the autonomous differential
equation ͞ ϭ ( ), where the graph of is given
...
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...
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...
Sketch a phase portrait of each differential equation
...


Graph for Problem 29

Using the autonomous equation (2), discuss how it is
possible to obtain information about the location of
points of inflection of a solution curve
Consider the autonomous DE ͞ ϭ 2 Ϫ Ϫ 6
...

Discuss why
solution curve of an initial-value
problem of the form ͞ ϭ 2 Ϫ Ϫ 6, (0) ϭ 0 ,
where Ϫ2
3, has a point of inflection with the
0
same -coordinate
...
Repeat
for (2) ϭ 2
...
Discuss the behavior of the solutions
...
Suppose the DE
is changed to
ϭ (

Graph for Problem 30

Ϫ ),

where and are positive constants
...

Another population model is given

Consider the autonomous DE ͞ ϭ (2͞␲) Ϫ sin
Determine the critical points of the equation
...

Classify the critical points as asymptotically stable,
unstable, or semi-stable
...
Can there exist an
autonomous DE of the form given in (2) for which
critical point is nonisolated? Discuss; do not think
profound thoughts
...
Discuss: Why can’t the graph
of ( ) cross the graph of the equilibrium solution
ϭ ? Why can’t ( ) change signs in one of the
subregions discussed on page 39? Why can’t ( ) be
oscillatory or have a relative extremum (maximum or
minimum)?
Suppose that ( ) is a solution of the autonomous equation ͞ ϭ ( ) and is bounded above and below by
two consecutive critical points 1
2, as in subregion
of Figure 2
...
6(b)
...
Discuss why there cannot exist a number
2 such that lim :ϱ ( ) ϭ
...


by
ϭ

Ϫ ,

where and are positive constants
...
3 we saw that the autonomous differential equation
ϭ

Ϫ

,

where is a positive constant and is the acceleration
due to gravity, is a model for the velocity of a body of
mass
that is falling under the influenc of gravity
...
Use a phase
portrait of the differential equation to fin the limiting, or
terminal, velocity of the body
...

Suppose the model in Problem 40 is modified so
that air resistance is proportional to 2, that is,
ϭ

Ϫ

2


...
3
...
Explain your
reasoning
...
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...
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...
Use a phase portrait
of the differential equation to predict the behavior
of ( ) as : ϱ when (0) ␣
...

Verify that an explicit solution of the DE in the case
when ϭ 1 and ␣ ϭ ␤ is ( ) ϭ ␣ Ϫ 1͞( ϩ )
...
Then fin
a solution that satisfies (0) ϭ 2␣
...
Does the behavior of the solutions as
: ϱ agree with your answers to part (b)?

ϭ (␣ Ϫ )( ␤ Ϫ ),
where Ͼ 0 is a constant of proportionality and
␤ Ͼ ␣ Ͼ 0
...


●
●
●

Basic integration formulas (See inside front cover)
Techniques of integration: integration by parts and partial fraction decomposition
See also the

We begin our study of how to solve differential equations with the simplest
of all differential equations: first-order equations with separable variables
...


Consider the first-order differential equation ͞ ϭ
( , )
...
If ( ) is a continuous function, then integrating both
sides of (1) gives ϭ ͐ ( )
ϭ ( ) ϩ , where ( ) is an antiderivative (indefi
nite integral) of ( )
...

ϭ ͐(1 ϩ )
Equation (1), as well as its method of solution, is just a special
case when the function in the normal form ͞ ϭ ( , ) can be factored into a
function of times a function of

A first-order di ferential equation of the form
ϭ ( ) ( )
is said to be

or to have

For example, the equations
ϭ

2

3 ϩ4

and

ϭ

ϩ sin

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

are separable and nonseparable, respectively
...
From this last form we can
see immediately that (2) reduces to (1) when ( ) ϭ 1
...


(3)

, and so (3) is the same as

͵

( )

ϭ

͵

( )

or

( )ϭ

( )ϩ ,

(4)

where ( ) and ( ) are antiderivatives of ( ) ϭ 1͞ ( ) and ( ), respectively
...
A one-parameter family of solutions, usually given implicitly, is obtained
by integrating both sides of ( ) ϭ ( )
There is no need to use two constants in the integration of a separable
equation, because if we write ( ) ϩ 1 ϭ ( ) ϩ 2, then the difference 2 Ϫ 1 can
be replaced by a single constant , as in (4)
...

For example, multiples of constants or combinations of constants can sometimes be
replaced by a single constant
...

͞ ϭ

Dividing by (1 ϩ ) , we can write
follows that

͵ ͵
ϭ

1ϩ

ln͉ ͉ ϭ ln͉ 1 ϩ ͉ ϩ
ϭ

ln͉1ϩ ͉ϩ

ϭ ͉1 ϩ ͉

1

ϭ

1
ln͉1ϩ ͉

1

ؒ

1

; laws of exponents

Ά͉͉ 1 ϩ
1ϩ

1

ϭ Ϯ 1(1 ϩ )
...


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...
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...
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...


ՆϪ1
<Ϫ1

●

Because each integral results in a logarithm, a judicious
choice for the constant of integration is ln͉ ͉ rather than Rewriting the second
line of the solution as ln͉ ͉ ϭ ln͉1 ϩ ͉ ϩ ln͉ ͉ enables us to combine the terms on
the right-hand side by the properties of logarithms
...
Even if the indefinite integrals are not
logarithms,
it may still be advantageous to use ln͉ ͉
...

In Section 1
...


Solve the initial-value problem

Rewriting the equation as

͵

(4, 3)

Solution curve for the
IVP in Example 2

(4) ϭ Ϫ3
...


We can write the result of the integration as 2 ϩ 2 ϭ 2 by replacing the constant
2 1 by 2
...

Now when ϭ 4, ϭ Ϫ3, so 16 ϩ 9 ϭ 25 ϭ 2
...
Because of its simplicity we can
solve this implicit solution for an explicit solution that satisfies the initial condition
...
1
...
In this case the
solution curve is the lower semicircle, shown in dark blue in Figure 2
...
1 containing
the point (4, Ϫ3)
...
Specificall , if
is a zero
of the function ( ), then substituting ϭ into ͞ ϭ ( ) ( ) makes both sides
zero; in other words, ϭ is a constant solution of the differential equation
...
Recall that such a solution is called a singular
solution
...


We put the equation in the form
2

Ϫ4

or

ϭ

΄

1
4

Ϫ2

Ϫ

1
4

ϩ2

΅


...
Integrating and using the laws of logarithms give

ln

͉

1
1
ln͉ Ϫ 2 ͉ Ϫ ln͉ ϩ 2 ͉ ϭ ϩ 1
4
4
Ϫ2
Ϫ2
ϭ4 ϩ 2
or
ϭϮ
ϩ2
ϩ2

͉

4 ϩ

2


...
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...
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...
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...
Finally, after replacing Ϯ 2 by
last equation for , we get the one-parameter family of solutions
1ϩ
1Ϫ

ϭ2

and solving the

4
4


...
1
that ϭ 2 and ϭ Ϫ2 are two constant (equilibrium) solutions
...

However, ϭ Ϫ2 is a singular solution; it cannot be obtained from (6) for any choice of
the parameter This latter solution was lost early on in the solution process
...


Solve (

2

Ϫ ) cos

(0) ϭ 0
...


)

ϩ

ϭ2
Ϫ

͵

sin

ϭ Ϫ2 cos ϩ

(7)

The initial condition ϭ 0 when ϭ 0 implies ϭ 4
...
Then

y

2

cos gives

x

( /2,0)

c=2

_1

1

Level curves
ϭ 2 and ϭ 4

2

Ϫ

ϩ

Ϫ

ϭ 4 Ϫ 2 cos
...
1 mentioned
that it may be difficult to use an implicit solution ( , ) ϭ 0 to find an explicit
solution ϭ ␾ ( )
...
The problem of “seeing” what an implicit solution looks like
can be overcome in some cases by means of technology
...
Recall from
multivariate calculus that for a function of two variables ϭ ( , ) the
curves defined by ( , ) ϭ where is constant, are called the
of the function
...
2
...
The
family of solutions defined by (7) is the level curves ( , ) ϭ Figure 2
...
3 illustrates the level curve ( , ) ϭ 4, which is the particular solution (8), in blue color
...
2
...

If an initial condition leads to a particular solution by yielding a specific value of
the parameter in a family of solutions for a first-order differential equation, there is
*

In Section 2
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

a natural inclination for most students (and instructors) to relax and be content
...
We saw in
Example 4 of Section 1
...
We are now in a position to solve the
equation
...


1
When ϭ 0, then ϭ 0, so necessarily, ϭ 0
...
The trivial solution
1/2
ϭ 0 was lost by dividing by

...
See Figure 2
...
4
...


You might recall that the foregoing result is one of the two forms of the fundamental
theorem of calculus
...

There are times when this form is convenient in solving DEs
...
Since
an antiderivative of a continuous function cannot always be expressed in terms of
elementary functions, this might be the best we can do in obtaining an explicit
solution of an IVP
...


Solve

ϭ

Ϫ

2

,

(3) ϭ 5
...
Using as dummy variable of integration, we can
write

͵

ϭ

3

͵
͵
͵

Ϫ2

3

]

()

3

ϭ

( ) Ϫ (3) ϭ

Ϫ2

3
Ϫ2

3

( ) ϭ (3) ϩ

͵

Ϫ2


...
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...
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...
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...


3

The procedure demonstrated in Example 5 works equally well on separable
equations > ϭ ( ) ( ) where, say, ( ) possesses an elementary antiderivative
but ( ) does not possess an elementary antiderivative
...
2
...
Integrals of these kinds of
2
functions are called
For example, ͵3 Ϫ
and ͵sin 2 are
nonelementary integrals
...
3
...
Also, it can easily happen that two individuals solving the
same equation correctly arrive at dissimilar expressions for their answers
...
See Problems 27 and 28 in Exercises 2
...


In Problems 1–22 solve the given differential equation by
separation of variables
...

ϭ 4(
2

ϭ0
3 Ϫ

ϩ

ϩ3 Ϫ Ϫ3
Ϫ2 ϩ4 Ϫ8

ϭ

ϭ0
Ϫ

2

Ϫ

ϭ 11 Ϫ

ϭ0

ϭ0

ϩ(

)

2

2 ϩ3
ϭ
4 ϩ5

2 Ϫ

ϩ 1)
(1 ϩ

΂

2

ϩ 2 cos 33

sin 3
(

ϩ2

3 ϩ2

ln

ϭ

ϭ ( ϩ 1)2

ϭ sin 5

ϭ (

ϭ

ϭ

2

ϭ0
2

ϭ

2

ϩ 1),

Ϫ1
,
Ϫ1
Ϫ

( >4) ϭ 1

(2) ϭ 2
,

(Ϫ1) ϭ Ϫ1

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...
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...
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...


●

11 Ϫ

Ϫ 11 Ϫ

(1 ϩ

2

4

2

ϭ 0,

ϩ (1 ϩ 4 2 )

)

13
2
(1) ϭ 0

(0) ϭ 5
2

ϩ 2 ϭ 1,

(0) ϭ

ϭ 0,

In Problems 29 and 30 proceed as in Example 5 and find an
explicit solution of the given initial-value problem
...
Determine the exact interval of defi
nition by analytical methods
...

ϭ

2 ϩ1
,
2

(2 Ϫ 2)
Ϫ
sin

(Ϫ2) ϭ Ϫ1

ϭ3
Ϫ

2

ϩ 4 ϩ 2,

ϭ 0,

ϩ

ϭ 0,

(1) ϭ Ϫ2

(0) ϭ 0
(0) ϭ 1

Find a solution of the initial-value problem consisting of the differential equation in Example 3 and
each of the initial-conditions: (0) ϭ 2, (0) ϭ Ϫ2,
and 1 ϭ 1
...


()

Find a solution of
ϭ
the indicated points
...
01,

(0) ϭ 1

ϭ ( Ϫ 1)2 Ϫ 0
...
Find explicit solutions 1( ), 2( ), 3( ),
and 4( ) of the differential equation ͞ ϭ Ϫ 3
that satisfy, in turn, the initial conditions 1(0) ϭ 2,
1
1
2(0) ϭ 2 , 3(0) ϭ Ϫ2 , and 4(0) ϭ Ϫ2
...
Compare these
graphs with those predicted in Problem 19 of Exercises 2
...

Give the exact interval of definitio for each solution
...

Nevertheless, place 3 on the phase line and obtain
a phase portrait of the equation
...
1)
...

Find explicit solutions 1( ), 2( ), 3( ), and 4( )
of the differential equation in part (a) that satisfy,
in turn, the initial conditions 1(0) ϭ 4, 2(0) ϭ 2,
3(1) ϭ 2, and 4(Ϫ1) ϭ 4
...
Give
the exact interval of definition for each solution
...

ϭ

ϭ

ϭ1 ϩ

1
1 ϩ sin
1

(1 ϩ )

Find a singular solution of Problem 21
...

Show that an implicit solution of

(0) ϭ 1
...

Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the
initial condition or the equation itself
...
Use a
graphing utility to plot the graph of each solution
...

ϭ ( Ϫ 1)2,

(0) ϭ 1

Explain why the interval of definition of the explicit
solution ϭ ␾ 2 ( ) of the initial-value problem in
Example 2 is the
interval (Ϫ5, 5)
...
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...
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...
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...
Does this fact help in the solution of the
initial-value problem

2

sin2 ,

(0) ϭ 1?
2

Discuss
...

Solve the two initial-value problems:
ϭ ,

(0) ϭ 1

and
ϭ

ϩ

ln

,

DE (10) is separable under the following conditions that
describe a suspension bridge
...
2
...
In the
case of a suspension bridge, the usual assumption is that
the vertical load in (10) is only a uniform roadbed distributed along the horizontal axis
...
Use this information to
set up and solve an appropriate initial-value problem
from which the shape (a curve with equation ϭ ␾( ))
of each of the two cables in a suspension bridge is
determined
...
2
...


( ) ϭ 1
...
65 million digits in
the -coordinate of the point of intersection of the
two solution curves in part (a)
...

1Ϫ
B1 Ϫ

/2

The differential equation in Problem 27 is equivalent to the normal form
ϭ

(span)

/2
roadbed (load)

Shape of a cable in Problem 57

2
2

in the square region in the -plane defined by
͉ ͉ Ͻ 1, ͉ ͉ Ͻ 1
...
Sketch all regions in the
-plane for
which this differential equation possesses real
solutions
...
Then find an implicit and an
explicit solution of the differential equation subject
to (2) ϭ 2
...
Experiment with different numbers
3 ϩ1
of level curves as well as various rectangular
regions defined by Յ Յ , Յ Յ
On separate coordinate axes plot the graphs of the
particular solutions corresponding to the initial
conditions: (0) ϭ Ϫ1; (0) ϭ 2; (Ϫ1) ϭ 4;
(Ϫ1) ϭ Ϫ3
...
3 we saw that
a mathematical model for the shape of a flexible cable
strung between two vertical supports is
,

ϭ

(10)

1

where
denotes the portion of the total vertical load
between the points 1 and 2 shown in Figure 1
...
7
...


Use part (a) to find an explicit solution ϭ ␾( ) of
the IVP
...

Use a graphing utility or a CAS to graph this function, and then use the graph to estimate its domain
...
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...
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...
With the aid of a root-finding application
of a CAS, determine the approximate largest interval of definition of the solution ␾
...
]
Repeat part (b) for the initial condition (0) ϭ Ϫ2
...
Use a
graphing utility or a CAS to graph the solution
curve for the IVP on this interval
...
Experiment with different
(Ϫ2 ϩ )
numbers of level curves as well as various rectangular regions in the
-plane until your result
resembles Figure 2
...
6
...
Use a colored pencil to mark off that
2
segment of the graph that corresponds to the solution curve of a solution ␾ that satisfies the initial

●

y

x

Level curves in Problem 60

Review the definitions of linear DEs in (6) and (7) of Section 1
...
Linear differential equations are an especially “friendly” family
of differential equations, in that, given a linear equation, whether first order or a higher-order kin,
there is always a good possibility that we can find some sort of solution of the equation that we can
examine
...
1
...


A first-order di ferential equation of the form
1(

)

ϩ

0(

) ϭ ( ),

(1)

is said to be a linear equation in the variable

By dividing both sides of (1) by the lead coefficient 1( ), we
obtain a more useful form, the
, of a linear equation:
ϩ ( ) ϭ ( )
...


(2)
and

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Before we examine a general procedure for solving equations of form (2) we
note that in some instances (2) can be solved by separation of variables
...

The method for solving (2) hinges on a remarkable fact
that the
of the equation can be recast into the form of the exact derivative of a product by multiplying the both sides of (2) by a special function ␮( )
...


ϩ␮

c
c
these must be equal

The equality is true provided that

␮

ϭ␮
...
Integrating

␮
ϭ
␮

᭤

and solving

ln͉ ␮( )͉ ϭ

͵

( )

ϩ

1

gives ␮( ) ϭ 2 ͐ ( )
...
Hence we can simplify life
and choose 2 ϭ 1
...

Here is what we have so far: We multiplied both sides of (2) by (3) and, by
construction, the left-hand side is the derivative of a product of the integrating factor
and :
͐ ( )

ϭ

͐ ( )

( )

[ ͐ () ]ϭ

͐ ( )

( )
...
We can integrate both

͐ ( )

( )ϩ

and solve for
...


( )

ϩ

Ϫ͐ ( )


...
The following proce-

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...
May not be copied, scanned, or duplicated, in whole or in part
...
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...


●

( ) Remember to put a linear equation into the standard form (2)
...
No constant need be used in evaluating the
indefinite integral ͐ ( )
...
The left-hand side of the resulting equation is automatically the
derivative of the product of the integrating factor ͐ ( ) and :

[

͐ ( )

]ϭ

͐ ( )

( )
...


Solve

Ϫ 3 ϭ 0
...

Alternatively, since the differential equation is already in standard form (2), we identify ( ) ϭ Ϫ3, and so the integrating factor is ͐(Ϫ3) ϭ Ϫ3
...


͵

0

Ͻ ϱ
...


This linear equation, like the one in Example 1, is already in standard
form with ( ) ϭ Ϫ3
...
This time multiplying the given equation by this factor gives
Ϫ3

Ϫ3

ϭ6

Ϫ3

Ϫ3

and so

[

Ϫ3

]ϭ6

Ϫ3


...


When 1, 0, and in (1) are constants, the differential equation is autonomous
...
Thus a solution curve with an

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...
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...
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...


●

y
1

initial point either above or below the graph of the equilibrium solution
ϭ Ϫ2 pushes away from this horizontal line as increases
...
3
...


x

_1

y =_2

_2
_3
_1

1

2

3

4

Solution curves of DE
in Example 2

Suppose again that the functions and in (2) are continuous on a common interval In the steps leading to (4) we showed that (2) has
a solution on , then it must be of the form given in (4)
...
1
...
From the continuity of and on the
interval we see that and Ѩ ͞Ѩ are also continuous on With Theorem 1
...
1 as
our justification, we conclude that there exists one and only one solution of the
initial-value problem
ϩ ( ) ϭ ( ),

( 0) ϭ

(5)

0

defined on
interval 0 containing 0
...
2
...


Dividing by , the standard form of the given DE is
Ϫ

4

ϭ

5


...
Hence the integrating factor is
we can use ln instead of ln Ϳ Ϳ since Ͼ 0
Ϫ4͐ /

ϭ

Here we have used the basic identity
and rewrite
Ϫ4

Ϫ4

Ϫ5

ϭ

Ϫ4ln
log

ϭ
ϭ ,

as

ln

Ϫ4

ϭ

Ϫ4
...
Now we multiply (6) by

[

Ϫ4

]ϭ

Ϫ4


...

ؕ

᭤

Except in the case in which the lead coefficient is 1, the recasting of equation
(1) into the standard form (2) requires division by 1( )
...
Singular points are poten1( ) ϭ 0 are called
tially troublesome
...


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...
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...
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...


●

Find the general solution of (

2

Ϫ 9)

ϭ 0
...
Although is continuous on (Ϫϱ, Ϫ3), (Ϫ3, 3), and
(3, ϱ), we shall solve the equation on the first and third intervals
...


After multiplying the standard form (7) by this factor, we get
Integrating both sides of the last equation gives 1
2

Ϫ9

΅ ϭ 0
...
Thus for either
2


...
On the other hand, ϭ 0 is a singular point of the differential equation
in Example 3, but the general solution ϭ 5 Ϫ 4 ϩ 4 is noteworthy in that
every function in this one-parameter family is continuous at ϭ 0 and is define
on the interval (Ϫϱ, ϱ) and not just on (0, ϱ), as stated in the solution
...
See
Problems 45 and 46 in Exercises 2
...


Solve

ϩ

ϭ ,

(0) ϭ 4
...
The integrating factor is ͐ ϭ , so integrating
[

y
4
c>0

2

c=5

x
_2
_4 c=0
_4

_2

2

4

Solution curves of DE
in Example 5

gives
ϭ
Ϫ ϩ Solving this last equation for yields the general solution
ϭ Ϫ 1 ϩ Ϫ
...

Substituting these values into the general solution implies that ϭ 5
...


(8)

Figure 2
...
2, obtained with the aid of a graphing utility, shows the graph of
the solution (8) in dark blue along with the graphs of other members of the oneparameter family of solutions ϭ Ϫ 1 ϩ Ϫ
...
The last solution corresponds to ϭ 0 in the family and is shown in

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...
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...
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...


●

dark green in Figure 2
...
2
...

that the contribution of Ϫ ,
We say that Ϫ is a
since Ϫ : 0 as : ϱ
...

In applications, the coefficients ( ) and ( ) in
(2) may be piecewise continuous
...

We solve the problem in two parts corresponding to the two intervals over which is
defined
...


Solve

ϭ ( ),

ϩ

(0) ϭ 0 where

( )ϭ

Ά1,
0,

0Յ

Յ 1,
Ͼ 1
...
3
...
We
solve the DE for ( ) first on the interval [0, 1] and then on the interval (1, ϱ)
...


Integrating this last equation and solving for gives ϭ 1 ϩ 1 Ϫ
...
Then for Ͼ 1 the
equation
ϩ
leads to ϭ

2

Ϫ

ϭ0

,

0Յ


...


By appealing to the definition of continuity at a point, it is possible to determine
ϭ 1
...
As seen in
Figure 2
...
4, the function
1

ϭ
Graph of (9) in

Example 6

Ά

1Ϫ Ϫ,
( Ϫ 1) Ϫ ,

0Յ

Յ 1,
Ͼ1

(9)

is continuous on (0, ϱ)
...
3
...
3
...
2 we discussed
the fact that some simple continuous functions do not possess antiderivatives that
are elementary functions and that integrals of these kinds of functions are called
2
For example, you may have seen in calculus that ͐ Ϫ
and
͐sin 2 are nonelementary integrals
...
Two such
are
the
and
erf( ) ϭ

1

2

͵

0

Ϫ

2

and

erfc( ) ϭ

2
1

͵

ϱ
Ϫ

2


...
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...
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...
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...

ϱ
ϱ
Then from ͐0 ϭ ͐ 0 ϩ ͐ it is seen from (10) that the complementary error function erfc( ) is related to erf( ) by erf( ) ϩ erfc( ) ϭ 1
...
Note that erf(0) ϭ 0 is one obvious function
value
...


●

2

2

Solve the initial-value problem

ing factor is
y

Ϫ2

ϭ 2,

(0) ϭ 1
...


(11)

[1 ϩ 1 erf( )]
...
Hence the solution of the
problem is
ϭ2

2

͵

Ϫ

2

2

ϩ

or

ϭ

0

Solution curves of DE
in Example 7

2

The graph of this solution on the interval (Ϫϱ, ϱ), shown in dark blue in Figure 2
...
5
among other members of the family defined in (11), was obtained with the aid of a
computer algebra system
...
†

( ) A linear first-order di ferential equation
1(

is said to be

)

ϩ

0(

) ϭ0

whereas the equation
1(

)

ϩ

0(

) ϭ ( )

with ( ) not identically zero is said to be
For example, the
linear equations Ј ϩ ϭ 0 and Ј ϩ ϭ are, in turn, homogeneous and
nonhomogeneous
...
Store this terminology in the
back of your mind because it becomes important when we study linear higherorder ordinary differential equations in Chapter 4
...

Certain commands have the same spelling, but in
commands begin with a capital letter
whereas in
the same command begins with a lower case letter
When
discussing such common syntax, we compromise and write, for example,
See the
for the complete input commands used to solve a linear first-order DE
†

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...
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...


●

( ) Occasionally, a first-order differential equation is not linear in one variable
but in linear in the other variable
...
But its reciprocal
ϭ

ϩ

2

or

Ϫ

ϭ

2

is recognized as linear in the variable
...
This expression is then an
implicit solution of the first equation
( ) Mathematicians have adopted as their own certain words from engineering, which they found appropriately descriptive
...
In future discussions the words
and
will
occasionally pop up
...
3
...
More special functions are
studied in Section 6
...


In Problems 1–24 find the general solution of the given differential equation
...
Determine whether there are any
transient terms in the general solution
...

Give the largest interval over which the solution is defined
ϭ

ϩ5 ,

ϭ2 Ϫ3 ,
Јϩ

ϭ

,

(0) ϭ 3
(0) ϭ 1
3
(1) ϭ 2

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...
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...
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...


●

ϭ 2 2,

Ϫ

(1) ϭ 5

ϭ ,

ϩ

ϭ ( Ϫ

(0) ϭ 0,
),

(0) ϭ

ϭ 4 ϩ 1,

ϩ
Јϩ4

3

ϭ

( ϩ 1)

ϩ

( ϩ 1)

2

,

0,

,

0 constants

,

0 constants

(1) ϭ 8
(0) ϭ Ϫ1

ϭ ln ,

ϩ

, , ,

(1) ϭ 10

ϭ 1,

( )ϭ1

( >2) ϭ 1

Ј Ϫ (sin ) ϭ 2 sin ,
Ј ϩ (tan ) ϭ cos2 ,

(0) ϭ Ϫ1

In Problems 37– 40 proceed as in Example 6 to solve the
given initial-value problem
...

ϩ 2 ϭ ( ), (0) ϭ 0, where
( )ϭ

0Յ

Յ3
Ͼ3

ϭ ( ), (0) ϭ 1, where

ϩ

( )ϭ
ϩ2

Ά1,
Ϫ1,

0Յ

Յ1
Ͼ1

ϭ ( ), (0) ϭ 2, where
( )ϭ

(1 ϩ

Ά1,
0,

2

)

ϩ2
( )ϭ

Ά0,,

0Յ

1
Ն1

ϭ ( ), (0) ϭ 0, where

ΆϪ, ,

0Յ

1
Ն1

Proceed in a manner analogous to Example 6 to solve the
initial-value problem Ј ϩ ( ) ϭ 4 , (0) ϭ 3, where
( )ϭ

2,
ΆϪ2> ,

0 Յ Յ 1,
Ͼ 1
...

Consider the initial-value problem Ј ϩ
ϭ ( ),
(0) ϭ 1
...
What is the solution when ( ) ϭ 0? When ( ) ϭ ?
Express the solution of the initial-value problem
Ј Ϫ 2 ϭ 1, (1) ϭ 1, in terms of erf( )
...
Construct a
linear first-order differential equation for which all
nonconstant solutions approach the horizontal asymptote ϭ 4 as : ϱ
...
2
...

(0) ϭ 0
(0) ϭ 0 , 0 Ͼ 0
( 0) ϭ 0 , 0 Ͼ 0, 0 Ͼ 0
Reread Example 4 and then find the general solution of
the differential equation on the interval (Ϫ3, 3)
...
Construct a
linear first-order differential equation for which all solutions are asymptotic to the line ϭ 3 Ϫ 5 as : ϱ
...

Construct a linear first-order differential equation of
the form Ј ϩ 0 ( ) ϭ ( ) for which ϭ ͞ 3
and
ϭ 3
...

Give an initial condition ( 0) ϭ 0 for the DE
found in part (a) so that the solution of the IVP
is
ϭ 3 Ϫ 1͞ 3
...
Give an interval of definition of
each of these solutions
...
Is
there an initial-value problem whose solution is
defined on Ϫϱ, ϱ)?
Is each IVP found in part (b) unique? That is, can
there be more than one IVP for which, say,
ϭ 3 Ϫ 1͞ 3, in some interval , is the solution?
In determining the integrating factor (3), we did not use
a constant of integration in the evaluation of ͐ ( )
...

Suppose ( ) is continuous on some interval and is a
number in What can be said about the solution of the
initial-value problem Ј ϩ ( ) ϭ 0, ( ) ϭ 0?

The following system
of differential equations is encountered in the study of the
decay of a special type of radioactive series of elements:
ϭ Ϫ␭1
ϭ ␭1 Ϫ ␭2 ,
where ␭1 and ␭2 are constants
...
Carry out your
ideas
...
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...
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...
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...

During the time the heart is being stimulated, the voltage
across the heart satisfies the linear differential equation
ϭϪ

1

Solve the DE, subject to (4) ϭ

defined to be 1 at ϭ 0
...

Use a CAS to graph the solution curve for the IVP
for Ͼ 0
...

The
is defined by
( ) ϭ ͐ 0 sin(p 2>2)
...

Use a CAS to graph the solution curve for the IVP
on (Ϫϱ, ϱ)
...
What does the solution ( ) approach
as : ϱ? As : Ϫϱ?
Use a CAS to find the values of the absolute
maximum and the absolute minimum of the
solution ( )
...

0
...

Use tables or a CAS to find the value of (2)
...

The
Si( ) ϭ ͐ 0 (sin > )

●
●
●

is defined by
, where the integrand is

Multivariate calculus
Partial differentiation and partial integration
Differential of a function of two variables
Although the simple first-order equation
ϩ

ϭ0

is separable, we can solve the equation in an alternative manner by recognizing that the expression
on the left-hand side of the equality is the differential of the function ( , ) ϭ ; that is,
( )ϭ

ϩ

In this section we examine first-order equations in differential form ( , ) ϩ ( , ) ϭ 0
...
If the answer is yes, we can construct by partial integration
...


(1)

In the special case when ( , ) ϭ , where is a constant, then (1) implies
Ѩ
Ѩ

ϩ

Ѩ
Ѩ

ϭ 0
...
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...
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...
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...
For example, if 2 Ϫ 5 ϩ 3 ϭ , then (2) gives the first-order D
ϩ (Ϫ5 ϩ 3 2 )

(2 Ϫ 5 )

ϭ 0
...
So for our
purposes it is more important to turn the foregoing example around; namely, if
we are given a first-order DE such as (3), is there some way we can recognize
is the differential
that the differential expression (2 Ϫ 5 )
ϩ (Ϫ5 ϩ 3 2)
2
3
( Ϫ 5 ϩ )? If there is, then an implicit solution of (3) is 2 Ϫ 5 ϩ 3 ϭ
We answer this question after the next definition

A differential expression ( , )
ϩ ( , )
is an
in a
region of the -plane if it corresponds to the differential of some function
( , ) defined in A first-order di ferential equation of the form
( , )
is said to be an
exact differential
...


Notice that if we make the identifications ( , ) ϭ 2 3 and ( , ) ϭ 3 2, then
Ѩ ͞Ѩ ϭ 3 2 2 ϭ Ѩ ͞Ѩ Theorem 2
...
1, given next, shows that the equality of the
partial derivatives Ѩ ͞Ѩ and Ѩ ͞Ѩ is no coincidence
...

Ѩ
Ѩ

(4)

For simplicity let us assume that ( , ) and
( , ) have continuous first partial derivatives for all ( , )
...


Ѩ
,
Ѩ

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...
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...
Editorial review has
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...


●

΂ѨѨ

Ѩ
Ѩ
ϭ
Ѩ
Ѩ

and

Ѩ2
Ѩ
ϭ
Ѩ Ѩ
Ѩ

ϭ

΂ѨѨ

ϭ

Ѩ

...

The sufficiency part of Theorem 2
...
1 consists of showing that there exists a
function for which Ѩ ͞Ѩ ϭ ( , ) and Ѩ ͞Ѩ ϭ ( , ) whenever (4) holds
...

Given an equation in the differential form
( , )
ϩ ( , )
ϭ 0, determine whether the equality in (4) holds
...


( , ) with respect to while holding constant:

( , )ϭ

͵

( , )

ϩ ( ),

(5)

where the arbitrary function ( ) is the “constant” of integration
...


Ј( ) ϭ ( , ) Ϫ

This gives

Ѩ
Ѩ

͵

( , )


...
The implicit
solution of the equation is ( , ) ϭ
Some observations are in order
...

Ѩ
Ѩ

Second, we could just as well start the foregoing procedure with the assumption that
Ѩ ͞Ѩ ϭ ( , )
...


( , )ϭ2

and ( , ) ϭ

2

Ϫ 1 we have

Ѩ
Ѩ

...
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...
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...
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...


●

Thus the equation is exact, and so by Theorem 2
...
1 there exists a function ( , )
such that
Ѩ
ϭ2
Ѩ

Ѩ
ϭ
Ѩ

and

2

Ϫ 1
...


Taking the partial derivative of the last expression with respect to
result equal to ( , ) gives
Ѩ
ϭ
Ѩ

2

ϩ Ј( ) ϭ

2

Ϫ 1
...

The solution of the DE in Example 1 is
( , ) ϭ 2 Ϫ Rather, it
is ( , ) ϭ ; if a constant is used in the integration of Ј( ), we can then write the
solution as ( , ) ϭ 0
...


Solve (

2

Ϫ cos

)

2

ϩ (2

Ϫ cos

ϩ2 )

ϭ 0
...

Ѩ

Hence a function ( , ) exists for which
( , )ϭ

Ѩ
Ѩ

and

( , )ϭ

Ѩ

...


Remember, the reason can come out in front of the symbol ͐ is that in the integration with respect to , is treated as an ordinary constant
...


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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

2

Solve

ϭ

Ϫ cos sin
,
(1 Ϫ 2)

(0) ϭ 2
...

Ѩ
)
2

)ϩ ( )

ϩ Ј( ) ϭ cos sin Ϫ

The last equation implies that Ј( ) ϭ cos sin

͵

( ) ϭ Ϫ (cos )(Ϫsin
2

Thus
y

x

Solution curves of DE
in Example 3

2

(1 Ϫ

2

)Ϫ

1
cos2 ϭ
2

1

2


...

2
2

or

(1 Ϫ

2

) Ϫ cos2 ϭ ,

(7)

where 2 1 has been replaced by The initial condition ϭ 2 when ϭ 0 demands
that 4(1) Ϫ cos 2 (0) ϭ , and so ϭ 3
...

The solution curve of the IVP is the curve drawn in blue in Figure 2
...
1;
it is part of an interesting family of curves
...
2) and
using a graphing utility to carefully graph the explicit functions obtained for various values of by solving 2 ϭ ( ϩ cos 2 )͞(1 Ϫ 2) for
Recall from Section 2
...
The same basic idea sometimes works for a
nonexact differential equation ( , )
ϩ ( , )
ϭ 0
...
In an attempt to find ␮, we turn to the criterion (4) for exactness
...
By the Product Rule of differentiation the last equation is
the same as ␮ ϩ ␮
ϭ ␮ ϩ ␮ or
␮

Ϫ␮

ϭ(

Ϫ

)␮

(9)

Although , ,
, and
are known functions of and , the difficulty here in
determining the unknown ␮( , ) from (9) is that we must solve a partial differential

Copyright 2012 Cengage Learning
...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

equation
...

Suppose ␮ is a function of one variable; for example, say that ␮ depends only on In
this case, ␮ ϭ ␮͞ and ␮ ϭ 0, so (9) can be written as

␮

Ϫ

ϭ

␮
...
We can finally determine ␮ because (10) is
as well as
It follows from either Section 2
...
3 that
␮( ) ϭ ͐(( Ϫ )/ )
...

(11)
In this case, if ( Ϫ )͞ is a function of only, then we can solve (11) for ␮
...


)͞

͵

Ϫ


...


(14)

The nonlinear first-order di ferential equation
ϩ (2

2

ϩ3

2

Ϫ 20)

ϭ0

is not exact
...


3

3

ϭ

Ϫ 20 3)


...


You should verify that the last equation is now exact as well as show, using the
method of this section, that a family of solutions is 1 2 4 ϩ 1 6 Ϫ 5 4 ϭ
...
Sometimes a differential equation
is written
( , )
ϭ ( , )
In this case, first rewrite it as
( , )
Ϫ ( , )
ϭ 0 and then identify
( , ) ϭ ( , ) and
( , ) ϭ Ϫ ( , ) before using (4)
...
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...
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...
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...
Then the method for finding integrating factors
just discussed can be used to derive an integrating factor for
Ј ϩ ( ) ϭ ( )
...

͐ ( )

From (13) we arrive at the already familiar integrating factor
Section 2
...


In Problems 1–20 determine whether the given differential
equation is exact
...

(2 Ϫ 1)

ϩ (3 ϩ 7)

ϭ0

(2 ϩ )

Ϫ( ϩ6 )

ϭ0

ϩ (4 Ϫ 8 3)

(5 ϩ 4 )

(sin Ϫ sin )
(2

2

΂2

Ϫ

(

2

Ϫ 3)
1
)

΂1 ϩ ln
3

(

3

ϩ

2

3

)

sin )

)

)

ϩ

Ϫ

2

3

ϩ 3 sin 3 ϭ 0

ϩ6

1
1ϩ9

ϭ

ϩ 2 cos )

ϭ0

ln
Ϫ2 )

3

(2 sin cos Ϫ

2

ϩ

Ϫ )
2

ϭ0

2

ϭ0

2

ϩ

ϩ

2

Ϫ

Ϫ 1)

ϭ 0,

)

(1) ϭ 1

ϭ 0,

ϩ (6 ϩ 4 Ϫ 1)

(0) ϭ 1
ϭ 0,

3

(Ϫ1) ϭ 2

2

ϩ

5

2

2

2
3

2

4

ϭ 0,

(1) ϭ 1

Ϫ2 )
ϩ ln )

ϭ 0,

(0) ϭ

ϭ ( ϩ sin ), (0) ϭ 1

ϩ cos Ϫ 2

ϩ

4

3

ϩ cos )

Ϫ2 )

2

ϩ (3
ϩ (2

2 2

ϩ 20

so that the given
2 3

Ϫ sin )

)

ϭ0
ϭ0

In Problems 29 and 30 verify that the given differential
equation is not exact
...
Solve
...


΂1 ϩ
ϩ

ϩ (2

( 2 cos Ϫ 3
ϩ (2 sin Ϫ

ϭ0

3

ϩ(
Ϫ

2

΂

ϩ

2

ϩ

(4 ϩ 2 Ϫ 5)

ϭ0

ϭ (3

ϩ

΂1 Ϫ 3 ϩ
2 3

Ϫ4

Ϫ2 )

ϩ3
Ϫ

ϭ0

ϭ0

2

2

( ϩ )2

ϭ (1 Ϫ ln )

2

ϩ

ϭ2

΂

2

ϩ

( ln Ϫ
(3

ϩ(

ϩ

΂1 ϩ 1 Ϫ

ϩ( 4ϩ3

In Problems 21–26 solve the given initial-value problem
...


ϭ0

2

ϩ 2 cos

2

)

2

ϩ3 )

( ϩ ϩ 1)

ϩ2

ϭ0

ϩ( ϩ2 )

ϭ0

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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
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...


●

ϩ (4 ϩ 9 2)

6

΂

cos

Ϫ3

(10 Ϫ 6 ϩ
(

2

2

ϩ 1ϩ

3

)

ϩ

ϭ0
sin

ϭ0

Ϫ2

ϭ0

)

ϩ (5

2

Ϫ

3

ϩ

sin )

ϭ0

In Problems 37 and 38 solve the given initial-value problem
by finding as in Example 4, an appropriate integrating factor
...
See
Figure 2
...
2
...
Starting at ϭ 0
seconds, the weight of the overhanging portion causes
the chain on the table to uncoil smoothly and to fall to
the floo
...
When all resistive forces are ignored, it can
be shown that a mathematical model relating to is
given by

ϭ0

ϩ

2

ϭ 32
...

Find explicit solutions 1( ) and 2( ) of the differential equation in part (a) such that 1(0) ϭ Ϫ2
and 2(1) ϭ 1
...


Rewrite this model in differential form
...

Find an explicit solution ( )
...

peg

Consider the concept of an integrating factor used in
Problems 29–38
...


platform edge
()

Uncoiling chain in Problem 45

Reread Example 3 and then discuss why we can conclude that the interval of definition of the explicit
solution of the IVP (the blue curve in Figure 2
...
1) is
(Ϫ1, 1)
...
Carry
out your ideas
...
Here is a little exercise in
cleverness: Although the differential equation
( Ϫ 1 2 ϩ 2)
ϩ
ϭ 0 is not exact, show how
the rearrangement (
ϩ
) ͞1 2 ϩ 2 ϭ
and
1
2
2
the observation 2 ( ϩ ) ϭ
ϩ
can lead to
a solution
...


The solution of the differential equation
2
( 2ϩ

2 2

)

΄

ϩ 1ϩ

2

(

2

Ϫ 2
ϩ 2) 2

΅

ϭ0

is a family of curves that can be interpreted as
streamlines of a fluid flow around a circular object
whose boundary is described by the equation
2
ϩ 2 ϭ 1
...

Use a CAS to plot the streamlines for
ϭ 0, Ϯ0
...
4, Ϯ0
...
8 in three
different ways
...

Second, solve for in terms of the variable Plot
the resulting two functions of for the given values
of , and then combine the graphs
...
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...
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...
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...
But it is not uncommon to be
stumped by a differential equation because it does not fall into one of the classes of equations that
we know how to solve
...


Often the first step in solving a differential equation consists
of transforming it into another differential equation by means of a
For example, suppose we wish to transform the first-order differential equation
͞ ϭ ( , ) by the substitution ϭ ( , ), where is regarded as a function of
the variable If possesses first-partial derivatives, then the Chain Rul
ϭ

for

͞

ϩ

͞

If we replace
the DE

Ѩ
Ѩ

Ѩ
Ѩ

gives

( , )ϩ

ϭ

( , )


...
If we can determine a solution ϭ ␾( ) of this

last equation, then a solution of the original differential equation is ϭ ( , ␾( ))
...

If a function possesses the property ( , ) ϭ
( , ) for some real number ␣, then is said to be a
of
degree ␣
...
A first-order DE in differential
( , )

ϩ ( , )

ϭ0

(1)

*

is said to be
functions of the

if both coefficient functions and are homogeneous
degree
...
3
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

( , 1), where ϭ >
...
5
...
Specificall ,
of the substitutions ϭ
or ϭ
where and are new dependent variables, will reduce a
homogeneous equation to a
first-orde differential equation
...

(1, ) ϩ (1, )

ϩ

At this point we offer the same advice as in the preceding sections: Do not memorize
anything here (especially the last formula); rather,
The proof that the substitutions ϭ
and
ϭ
ϩ
also lead to a
separable equation follows in an analogous manner from (3)
...


Inspection of ( , ) ϭ 2 ϩ 2 and ( , ) ϭ 2 Ϫ
these coefficients are homogeneous functions of degree 2
...


ϩ

2
΄Ϫ1 ϩ 1 ϩ ΅

; long division

After integration the last line gives
Ϫ ϩ 2 ln͉ 1 ϩ ͉ ϩ ln͉ ͉ ϭ ln͉ ͉

͉

Ϫ ϩ 2 ln 1 ϩ

͉ ϩ ln͉

͉ ϭ ln͉ ͉
...


Although either of the indicated substitutions can be used for every homogeneous differential equation, in practice we try ϭ whenever the function ( , )
is simpler than ( , )
...


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...
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...
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...


●

The differential equation
ϩ ( ) ϭ ( ) ,
where is any real number, is called
ϭ 1, equation (4) is linear
...


Solve

ϩ

ϭ

(4)

Note that for ϭ 0 and
1 the substitution ϭ 1Ϫ reduces

2 2


...
We then substitute

Ϫ2

; Chain Rule

into the given equation and simplify
...


The integrating factor for this linear equation on, say, (0, ϱ) is
Ϫ͐

/

ϭ

Integrating

Ϫln

[

ln

ϭ

Ϫ1


...

A differential equation of the form
ϭ (

ϩ

ϩ

(5)

)

can always be reduced to an equation with separable variables by means of the substitution ϭ
ϩ
ϩ ,
0
...


Solve

ϭ (Ϫ2 ϩ ) 2 Ϫ 7,

(0) ϭ 0
...


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...
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...
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...


●

The last equation is separable
...

ϭ0

ϩ

ϩ( Ϫ2 )
(

2

ϩ

)

(

2

ϩ

)

ϭ

Ϫ
ϩ

ϭ

ϩ
ϭ

ϭ0

2

ϩ

ϩ 1
2

(

2

3

ϭ
ϩ 2 2)

( ϩ

/

)

Ϫ

3

ϭ
Ϫ

or

6

ϭ


...


(6)

In Problems 15 – 20 solve the given differential equation by
using an appropriate substitution
...

2

)

ϭ0
2

Ϫ

,

,

3/2

ϭ 1,

(1) ϭ 1
2
(0) ϭ 4
( )

ϭ 0,

ϩ (ln Ϫ ln Ϫ 1)

In Problems 23 – 28 solve the given differential equation by
using an appropriate substitution
...

2

)

ϭ0

ϭ0

ϩ1

(

6
6

ϭ0

ϩ

1

and then resubstituting gives the solution

ϭ 2( ϩ )

Ϫ

ϩ3
3 ϩ

Ϫ

( ϩ )

2

6 ϩ6

Finally, applying the initial condition (0) ϭ 0 to the last equation in (6) gives
ϭ Ϫ1
...
5
...


Example 3

( Ϫ )

3(1 ϩ
1Ϫ

Ϫ3
ϭ
ϩ3

or

1

ϩ 1) 2

ϭ2ϩ 1 Ϫ2 ϩ3
ϭ tan2 ( ϩ )

(1) ϭ 0

ϭ 0,

(1) ϭ

ϭ

1Ϫ

Ϫ
ϩ

ϭ sin( ϩ )
ϭ1ϩ

Ϫ ϩ5

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...
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...
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...


●

In Problems 29 and 30 solve the given initial-value problem
...


a

and

( , )ϭ

Put the homogeneous differential equation
(5

2

Ϫ 2 2)

Ϫ

a

(1, > )
...

Determine two singular solutions of the DE in
Problem 10
...

In Example 3 the solution ( ) becomes unbounded as
: Ϯϱ
...
What are the
equations of these curves?
The differential equation
is known as

͞

ϭϪ
where

1

4
2

Ϫ

1

ϩ

2

ϭ 2͞ is a known solution of the equation
...


You might start by proving that
( , )ϭ

particular solution 1 of the equation
...
The
Bernoulli equation can then be reduced to a linear
equation by the substitution ϭ Ϫ1
...
4 we saw
that a mathematical model for the velocity of a chain
slipping off the edge of a high horizontal platform is
ϩ

2

ϭ 32
...
This time solve the DE using the fact that it is a
Bernoulli equation
...
Although we
will come back to this equation and solve it by an
alternative method in Section 3
...


A first-order differential equation ͞ ϭ ( , ) is a source of information
...
Then in Sections 2
...
5 we
examined first-order DEs
—that is, we developed some procedures for obtaining explicit
and implicit solutions
...
So to round out the picture of the different types of analyses of differential
equations, we conclude this chapter with a method by which we can “solve” the differential equation
— this means that the DE is used as the cornerstone of an algorithm for approximating the unknown solution
...
A more extensive treatment of numerical methods for ordinary differential equations is given in Chapter 9
...
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...
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...
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...
One way of approximating this solution is to use tangent lines
...
11 ϩ 0
...
The nonlinear differential equation in
this IVP cannot be solved directly by any of the methods considered in Sections 2
...
4, and 2
...
Specificall , suppose we wish to know the value of (2
...
The IVP
has a solution, and as the flow of the direction field of the DE in Figure 2
...
1(a) suggests, a solution curve must have a shape similar to the curve shown in blue
...
6
...
As the solution curve passes
through the initial point (2, 4), the lineal element at this point is a tangent line with
slope given by (2, 4) ϭ 0
...
4(2) 2 ϭ 1
...
As is apparent in Figure 2
...
1(a)
and the “zoom in” in Figure 2
...
1(b), when is close to 2, the points on the solution
curve are close to the points on the tangent line (the lineal element)
...
8, and the point-slope form of a line, we find that an equation of the tangent line is ϭ ( ), where ( ) ϭ 1
...
4
...
If 1 ϭ ( 1) denotes the -coordinate on the
tangent line and ( 1) is the -coordinate on the solution curve corresponding to an
-coordinate 1 that is close to ϭ 2, then ( 1) Ϸ 1
...
1,
then 1 ϭ (2
...
8(2
...
4 ϭ 4
...
1) Ϸ 4
...

y
solution
curve

4

(2, 4)

slope
1
...


(2)

The graph of this linearization is a straight line tangent to the graph of ϭ ( ) at
the point ( 0, 0)
...
6
...
Then by replacing by 1 ϭ 0 ϩ in (2), we get
( 1) ϭ

( )

0

0

ϩ ( 0,

0)( 0

ϩ

Ϫ

0)

or

1

ϭ

0

ϩ

( 1,

1),

where 1 ϭ ( 1)
...
Of course, the accuracy of the approximation ( 1) Ϸ ( 1) or 1 Ϸ ( 1) depends heavily on the size of the increment
Usually, we must choose this
to be “reasonably small
...
* By identifying the new starting
*

This is not an actual tangent line, since ( 1,

1)

lies on the first tangent and not on the solution curve

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

point as ( 1, 1) with ( 0, 0) in the above discussion, we obtain an approximation
from 0, that is, 2 ϭ 1 ϩ ϭ
2 Ϸ ( 2) corresponding to two steps of length
ϩ 2 , and
0
( 2) ϭ (

0

ϩ2 )ϭ (

Continuing in this manner, we see that
the general formula
ϩ1

where ϭ 0 ϩ
lines” is called

ϭ 0
...
00
2
...
20
2
...
40
2
...
0000
4
...
3768
4
...
8244
5
...
00
2
...
10
2
...
20
2
...
30
2
...
40
2
...
50

4
...
0900
4
...
2826
4
...
4927
4
...
7210
4
...
9686
5
...



...
This procedure of using successive “tangent

Consider the initial-value problem Ј ϭ 0
...
4 2, (2) ϭ 4
...
5) using first ϭ 0
...
05
...
11 ϩ 0
...
1,
1

ϭ 0
...
11

ϩ 0
...


ϭ 0 we fin

(0
...
4 2) ϭ 4 ϩ 0
...
114 ϩ 0
...
18,
0

which, as we have already seen, is an estimate to the value of (2
...
However, if we
use the smaller step size ϭ 0
...
1
...
05 0
...
4(2)2 ϭ 4
...
09 ϩ 0
...
114
...
4(2
...
18416187

(

)

we have 1 Ϸ (2
...
1)
...
The results are summarized in Tables 2
...
1 and 2
...
2,
where each entry has been rounded to four decimal places
...
6
...
6
...
1 and 10 steps with ϭ 0
...
5
...
0997 corresponding to
ϭ 0
...
5) than the value 5 ϭ 5
...
1
...
We do this to compare the values of the approximations at each step with the true or actual values of the solution ( ) of the initialvalue problem
...
2 , (1) ϭ 1
...
5) using first ϭ 0
...
05
...
2 , (3) becomes
ϩ1

ϭ

ϩ (0
...
Again with the aid of computer software we obtain the
values in Tables 2
...
3 and 2
...
4 on page 78
...
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...
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...
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...
05
Actual value
1
...
05
1
...
15
1
...
25
1
...
35
1
...
45
1
...
1
Actual value
1
...
10
1
...
30
1
...
50

1
...
0200
1
...
0675
1
...
1259

Abs
...
error

1
...
0212
1
...
0714
1
...
1331

0
...
0012
0
...
0040
0
...
0073

0
...
12
0
...
37
0
...
64

ϭ

% Rel
...
0000
1
...
0212
1
...
0450
1
...
0714
1
...
1008
1
...
1331

0
...
0003
0
...
0009
0
...
0016
0
...
0024
0
...
0032
0
...
00
0
...
06
0
...
12
0
...
19
0
...
25
0
...
32

1
...
0100
1
...
0318
1
...
0562
1
...
0833
1
...
1133
1
...
1( 2Ϫ1)

...
) The
is defined to b
͉

The

Abs
...


Ϫ

and

͉

are, in turn,

͉

and

͉

͉

ϫ 100
...
6
...
6
...
Also, we see that even though the percentage
relative error is growing with each step, it does not appear to be that bad
...
If we simply change the coefficient of the
right side of the DE in Example 2 from 0
...
5 the percentage
relative errors increase dramatically
...
6
...
Although attractive for its simplicity,
It was introduced
here simply to give you a first taste of numerical methods
...

Regardless of whether we can actually fin an explicit
or implicit solution, if a solution of a differential equation exists, it represents a
smooth curve in the Cartesian plane
...
, ( , ) whose -coordinates approximate the -coordinates ( ) of points ( 1, ( 1 )), ( 2, ( 2 )),
...
By taking the -coordinates
close together (that is, for small values of ) and by joining the points ( 1, 1),
( 2, 2 ),
...

Drawing curves is something that is well suited to a computer
...
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...
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...
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...
Some software packages
simply plot the generated numerical approximations, whereas others generate hard
numerical data as well as the corresponding approximate or
By way of illustration of the connect-the-dots nature of the graphs produced by a numerical solver, the two colored polygonal graphs in Figure 2
...
3 are
the numerical solution curves for the initial-value problem Ј ϭ 0
...
The blue smooth curve is the graph of the exact solution ϭ 0
...
Notice in Figure 2
...
3 that, even with the ridiculously large step size of
ϭ 1, the RK4 method produces the more believable “solution curve
...
1
is used
...
A solver usually requires that the differential equation be expressed in normal form ͞ ϭ ( , )
...
If the idea is to approximate the numerical value of ( ), then a solver may additionally require that you state a value for
or, equivalently, give the number of steps that you want to take to get from ϭ 0
to ϭ For example, if we wanted to approximate (4) for the IVP illustrated in
Figure 2
...
3, then, starting at ϭ 0 it would take four steps to reach ϭ 4 with a step
size of ϭ 1; 40 steps is equivalent to a step size of ϭ 0
...
Although we will not
delve here into the many problems that one can encounter when attempting to approximate mathematical quantities, you should at least be aware of the fact that a numerical solver may break down near certain points or give an incomplete or misleading picture when applied to some first-order differential equations in the normal
form
...
6
...
Equivalent results
were obtained using three different commercial numerical solvers, yet the graph is
hardly a plausible solution curve
...


y

6
5
4
3
2
1

x
_1
_2 _1

1

2

3

4

5

A not-very helpful
numerical solution curve

In Problems 1 and 2 use Euler’s method to obtain a fourdecimal approximation of the indicated value
...
1 and then using
ϭ 0
...


In Problems 5 –10 use a numerical solver and Euler’s
method to obtain a four-decimal approximation of the indicated value
...
1 and then use ϭ 0
...

Јϭ

Ј ϭ 2 Ϫ 3 ϩ 1, (1) ϭ 5;
Јϭ ϩ

2

, (0) ϭ 0;

(1
...
2)

In Problems 3 and 4 use Euler’s method to obtain a fourdecimal approximation of the indicated value
...
1 and then use ϭ 0
...
Find an explicit solution for
each initial-value problem and then construct tables similar to
Tables 2
...
3 and 2
...
4
...
0)
(1
...
5)

ϩ 1 , (0) ϭ 1;
2

, (0) ϭ 1;

ϩ

2

(0
...
5;

(0
...
5)

Јϭ

2

Јϭ Ϫ

Ϫ , (1) ϭ 1;
2

, (0) ϭ 0
...
5)
(0
...

First use Euler’s method and then the RK4 method
...
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...
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...
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...
25 in each case
...
If possible, use a different color
for each curve
...
1 and ϭ 0
...

Ј ϭ 2(cos ) ,

Use a numerical solver and the RK4 method to
graph the solution of the initial-value problem
Ј ϭ Ϫ2 ϩ 1, (0) ϭ 0
...

Use the analytic solution ( ) found in part (b)
and a CAS to find the coordinates of all relative
extrema
...
0), where ( ) is the solution to
Ј ϭ 2 2, (0) ϭ 1
...
1 and then use
ϭ 0
...
Repeat, using the RK4 method
...
0) to
differ so greatly
...

Fill in the blanks or answer true or false
...
The critical point
of the equation is a(n)
(attractor or repeller) for Ͼ 0
and a(n)
(attractor or repeller) for
0
...


1(

) Јϩ

0(

) ϭ 0 is also separable
...

The first-order DE
ϭ u ϩ ϩ u ϩ 1 is not separau
ble
...


An example of an autonomous linear first-order DE with
a single critical point Ϫ3 is
, whereas an
autonomous nonlinear first-order DE with a single critical point Ϫ3 is

...

The linear DE,

ϭ

In Problems 13 and 14 construct an autonomous first-orde
differential equation ͞ ϭ ( ) whose phase portrait is
consistent with the given figure

3
1

Graph for Problem 13

ϭ ( ) is separable
...

Ј ϩ ͉ ͉ ϭ 2 are

2

If Ј ϭ

0

, then ϭ

If a differentiable function
(Ϫ1) ϭ 2, then ( ) ϭ


...


Ј ϭ ͉ ͉,
Graph for Problem 14

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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

The number 0 is a critical point of the autonomous differential equation ͞ ϭ , where is a positive integer
...

Consider the differential equation
( ) ϭ Ϫ0
...
7 ϩ 3
...


2

ϩ

2

2

ϭ (3 Ϫ ln

Ϫ

Јϩ

2

ϭ2

2

)

ϩ1

ϩ

2

ϩ

2

3

ϩ

2

ϭ0

In Problems 19– 26 solve the given differential equation
...
R
...
Without attempting to solve the differential equation, estimate the value of lim :ϱ ( )
...
R
...
By hand, sketch two
different solution curves —one that is tangent to the lineal element shown in black and one that is tangent to the
lineal element shown in red
...
Some equations may
be more than one kind
...

Ϫ

ϭ
( ϩ 1)

2

ϭ

2

ϭ
ϭ Ϫ ϩ 10

ϩ
ϩ

1
Ϫ

0

has no solution for 0 0
...


Find an implicit solution of the initial-value problem
2

1
ϭ
( Ϫ )
ϭ5 ϩ

( 0) ϭ

2

ϭ

Ϫ

2

,

Find an explicit solution of the problem in part (a) and
give the largest interval over which the solution is
defined
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Graphs of some members of a family of solutions for a
first-orde differential equation
͞ ϭ ( , ) are
shown in Figure 2
...
5
...

Reproduce the figur on a piece of paper
...

Estimate the intervals on which the solutions ϭ 1( )
and ϭ 2( ) are defined

integers, Ϫ7 Յ Յ 7, Ϫ7 Յ Յ 7
...
Discuss: Does
it appear that the DE possesses critical points in the interval
Ϫ3
...
5? If so, classify the critical points as asymptotically stable, unstable, or semi-stable
...
1 to approximate (1
...

In Problems 33 and 34 each figure represents a portion of a
direction field of an autonomous first-order differential equation ͞ ϭ ( )
...

The points of the grid are ( , ), where ϭ 1, and
2

3
2
1
x
_1
_2
_3
_3 _2 _1

1

2

3

Portion of a direction field for Problem 3

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


Modeling with First-Order
Differential Equations

3
...
2 Nonlinear Models
3
...
3 we saw how a first-order di ferential equation could be used as a
mathematical model in the study of population growth, radioactive decay,
continuous compound interest, cooling of bodies, mixtures, chemical reactions,
fluid draining from a tank, velocity of a falling bod , and current in a series circuit
...
1 and nonlinear DEs in Section 3
...

The chapter concludes with the natural next step
...
3 we examine how
systems of first-order di ferential equations can arise as mathematical models in
coupled physical systems (for example, electrical networks, and a population of
predators such as foxes interacting with a population of prey such as rabbits)
...
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...
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...
3
Reread “Solving a Linear First-Order Equation” on page 56 in Section 2
...
3
...
We saw in Section 1
...
The constant of proportionality in (1) can be determined from the solution of the initial-value problem, using a
subsequent measurement of at a time 1 Ͼ 0
...
At ϭ 1 h the number of bacteria is measured to be 3 0
...

We first solve the differential equation in (1), with the symbol replaced
by With 0 ϭ 0 the initial condition is (0) ϭ 0
...

When it is put in the standard form of a linear first-order DE
Ϫ
()
3

0

0
...


Therefore ( ) ϭ
At ϭ 0 it follows that 0 ϭ 0 ϭ , so ( ) ϭ 0
At
3
ϭ 1 we have 2 0 ϭ 0
or
ϭ 3
...
4055, so ( ) ϭ 0 0
...
To find the time at which the number of bac2
teria has tripled, we solve 3 0 ϭ 0 0
...
4055 ϭ ln 3, or

0

2
...
71 h
...
4055

See Figure 3
...
1
...
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...

The time necessary for an initial population of, say, 100 or 1,000,000 bacteria to
triple is still approximately 2
...

As shown in Figure 3
...
2, the exponential function
increases as increases
for Ͼ 0 and decreases as increases for Ͻ 0
...
Accordingly, we say that is either a
( Ͼ 0) or
a
( Ͻ 0)
...
The half-life is simply the time it takes for one-half of the atoms in an
initial amount 0 to disintegrate, or transmute, into the atoms of another element
...
For example, the halflife of highly radioactive radium, Ra-226, is about 1700 years
...
The most
commonly occurring uranium isotope, U-238, has a half-life of approximately
4,500,000,000 years
...
5 billion years, one-half of a quantity of U-238 is
transmuted into lead, Pb-206
...
After 15 years it is determined that 0
...
Find the half-life of this isotope if the rate of disintegration is proportional to the amount remaining
...
If 0
...
957% of the
substance remains
...
99957 0 ϭ (15) —that is,
1
0
...
Solving for then gives ϭ 15 ln 0
...
00002867
...
00002867
Hence ( ) ϭ 0

...
Solving for gives 2 0 ϭ 0 Ϫ0
...
00002867
...

0
...
The theory of
is based on the fact that the radioisotope carbon-14
is produced in the atmosphere by the action of cosmic radiation on nitrogen-14
...
When a living organism dies, the
absorption of C-14, by breathing, eating, or photosynthesis, ceases
...

The method is based on the knowledge of the half-life of C-14
...
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...
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...
For his work, Libby was
awarded the Nobel Prize for chemistry in 1960
...
See Figure 3
...
3 and
Problem 12 in Exercises 3
...


A fossilized bone is found to contain 0
...
Determine
the age of the fossil
...
To determine the value of
the decay constant we use the fact that 1 0 ϭ (5730) or 1 0 ϭ 0 5730
...
00012097
...
00012097
...
001 0 we have
0
...
00012097 and Ϫ0
...
001) ϭ Ϫln 1000
...

0
...

The usual carbon-14 technique is limited to about 10 half-lives of the isotope, or
roughly 60,000 years
...
001 0
...
If this measurement is accomplished indirectly, based on the actual radioactivity of the specimen, then it is very difficult to
distinguish between the radiation from the specimen and the normal background
radiation
...
When the precise value of the ratio
of C-14 to C-12 is computed, the accuracy can be extended to 70,000 to 100,000
years
...
Nonisotopic methods based on the use of amino acids
are also sometimes possible
...
3 we saw
that the mathematical formulation of Newton’s empirical law of cooling/warming of
an object is given by the linear first-order di ferential equation
ϭ ( Ϫ

),

(2)

where is a constant of proportionality, ( ) is the temperature of the object for Ͼ 0,
and is the ambient temperature — that is, the temperature of the medium around the
object
...


When a cake is removed from an oven, its temperature is measured at 300° F
...
How long will it take for the cake to cool off
to a room temperature of 70° F?
*

The number of disintegrations per minute per gram of carbon is recorded by using a Geiger counter
...
1 disintegrations per minute per gram
...
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...
We must then solve the

(0) ϭ 300

(3)

and determine the value of so that (3) ϭ 200
...
If we separate variables,
Ϫ 70

ϭ

,

yields ln ͉ Ϫ 70͉ ϭ ϩ 1, and so ϭ 70 ϩ 2
...
Finally, the measurement
(3) ϭ 200 leads to 3 ϭ 13, or ϭ 1 ln 13 ϭ Ϫ0
...
Thus
23
3
23

Њ
Њ
Њ
Њ
Њ

( ) ϭ 70 ϩ 230
Њ

Temperature of cooling
cake in Example 4

(4)

Ϫ0
...

Yet we intuitively expect the cake to reach room temperature after a reasonably long
period of time
...
Parts (a) and (b) of
Figure 3
...
4 clearly show that the cake will be approximately at room temperature in
about one-half hour
...
1
The mixing of two fluids sometimes gives rise to a linear first-orde
differential equation
...
3, we assumed that the rate Ј( ) at which the amount of salt in the mixing
tank changes was a net rate:
ϭ (input rate of salt) Ϫ (output rate of salt) ϭ

Ϫ


...
3
...
3 held 300 gallons of a brine
solution
...
The concentration of the salt
in the inflo , or solution entering, was 2 lb/gal, so salt was entering the tank at the
rate
ϭ (2 lb/gal) ؒ (3 gal/min) ϭ 6 lb/min and leaving the tank at the rate
ϭ
( ͞300 lb/gal) ؒ (3 gal/min) ϭ ͞100 lb/min
...
3
...


Note here that the side condition is the initial amount of salt (0) ϭ 50 in the tank
and
the initial amount of liquid in the tank
...


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...
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...
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...


●

600

Integrating the last equation and solving for
gives the general solution
( ) ϭ 600 ϩ Ϫ /100
...
Thus the
amount of salt in the tank at time is given by
( ) ϭ 600 Ϫ 550

Ϫ /100


...
1
...
Also, it can be
seen from (6) and Figure 3
...
5(a) that ( ) : 600 as : ϱ
...


500

In Example 5 we assumed that the rate at which the solution was pumped in was
the same as the rate at which the solution was pumped out
...
The next
example illustrates the case when the mixture is pumped out at rate that is
than the rate at which the brine solution is being pumped into the tank
...
After minutes,
(1 gal/min) ( min) ϭ gal
will accumulate, so the tank will contain 300 ϩ gallons of brine
...

300 ϩ

Hence equation (5) becomes
ϭ6Ϫ

2
300 ϩ

or

ϩ

2
300 ϩ

ϭ 6
...


Integrating the last equation gives (300 ϩ )2 ϭ 2(300 ϩ )3 ϩ
...
95 ϫ 10 7)(300 ϩ ) Ϫ2
...
1
...

For a series circuit containing only a resistor and an inductor,
Kirchhoff’s second law states that the sum of the voltage drop across the inductor
( ( ͞ )) and the voltage drop across the resistor ( ) is the same as the impressed
voltage ( ( )) on the circuit
...
1
...

Thus we obtain the linear differential equation for the current ( ),
ϩ

-series circuit

ϭ ( ),

(7)

where and are constants known as the inductance and the resistance, respectively
...


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...
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...
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...


●

The voltage drop across a capacitor with capacitance is given by ( )͞ ,
where is the charge on the capacitor
...
1
...

(8)
-series circuit

But current and charge are related by ϭ ͞ , so (8) becomes the linear differential equation
1
ϩ
ϭ ( )
...
Determine the current if the initial current is zero
...
First, we multiply the differential equation by 2 and read off the
integrating factor 20
...


Integrating each side of the last equation and solving for gives ( ) ϭ 6 ϩ Ϫ20
...
Therefore the response is
5
5
( ) ϭ 6 Ϫ 6 Ϫ20
...
3 we can write a general solution of (7):
1

2

1

Ϫ( / )

()ϭ
In particular, when ( ) ϭ

0

͵

( / )

()

ϩ

Ϫ( / )


...


(11)

Note that as : ϱ, the second term in equation (11) approaches zero
...
In this case 0 ͞ is also called the
for large values of time
it appears that the current in the circuit is simply governed by Ohm’s law ( ϭ )
...
4055 of the initial-value problem in Example 1
described the population of a colony of bacteria at any time Ͼ 0
...
But since we are talking about a population, common sense
0
dictates that can take on only positive integer values
...
Perhaps, then, the graph
shown in Figure 3
...
9(a) is a more realistic description of than is the graph
of an exponential function
...
However,
for some purposes we may be satisfie if our model describes the system
fairly closely when viewed macroscopically in time, as in Figures 3
...
9(b)
and 3
...
9(c), rather than microscopically, as in Figure 3
...
9(a)
...
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...
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...

Find the amount of money accrued at the end of
5 years when $5000 is deposited in a savings
3
account drawing 5 4% annual interest compounded
continuously
...
0575)) 5(4)
4
that is accrued when interest is compounded
quarterly
...
What was the
initial population 0? What will be the population in
10 years? How fast is the population growing at ϭ 10?
The population of a town grows at a rate proportional to
the population present at time The initial population of
500 increases by 15% in 10 years
...
See Figure 3
...
10
...
5% of the C-14 found in
living trees of the same type had decayed
...

After 10 hours 2000 bacteria are present
...
3 hours
...
After 6 hours the mass had decreased by 3%
...


Cave wall painting in Problem 11

The Shroud of Turin, which shows the negative image of
the body of a man who appears to have been crucified, is
believed by many to be the burial shroud of Jesus of
Nazareth
...
1
...
In 1988 the Vatican granted
permission to have the shroud carbon-dated
...


Determine the half-life of the radioactive substance
described in Problem 6
...
Show that, in general, the half-life
of the substance is ϭ Ϫ(ln 2)͞
Show that the solution of the initial-value problem
in part (a) can be written ( ) ϭ 0 2 Ϫ /
...
In clear seawater,
the intensity 3 feet below the surface is 25% of the initial
intensity 0 of the incident beam
...
For more information o
this fascinating mystery see the Shroud of Turin home page at
http://www
...
com/
...
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...
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...
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...


A thermometer is removed from a room where the
temperature is 70° F and is taken outside, where the air
temperature is 10° F
...
What is the reading of the thermometer at ϭ 1 min? How long will it take for the
thermometer to reach 15° F?
A thermometer is taken from an inside room to the outside, where the air temperature is 5° F
...
What is the initial temperature of the inside
room?
A small metal bar, whose initial temperature was 20° C,
is dropped into a large container of boiling water
...
The fluids in containers and
are maintained at 0° C and 100° C, respectively
...
After 2 minutes the bar is removed
and instantly transferred to the other container
...
How long, measured from the start of the entire
process, will it take the bar to reach 99
...
Through a glass
window in the oven door, an observer records that the
thermometer reads 110° F after 1 minute and 145° F
2
after 1 minute
...
The initial temperature of
the chemical in the test tube is 80° F
...
1 , Ն 0,
where is measured in minutes
...
1 in (2)
...
In the long term
...
Use a graphing utility to plot the graph of ( ) on time intervals of various lengths
...
At the time
of discovery the core temperature of the body was
determined to be 85° F
...
Assume that the time of death corresponds to
ϭ 0 and that the core temperature at that time was
98
...
Determine how many hours elapsed before the
body was found
...
]
The rate at which a body cools also depends on its
exposed surface area
...
Suppose that two cups
and are filled with coffee at the same time
...
The exposed
surface area of the coffee in cup is twice the surface
area of the coffee in cup
...
If ϭ 70° F, then what
is the temperature of the coffee in cup after 30 min?

A tank contains 200 liters of fluid in which 30 grams of
salt is dissolved
...
Find
the number ( ) of grams of salt in the tank at time
Solve Problem 21 assuming that pure water is pumped
into the tank
...
Brine containing 2 pounds of salt per gallon is
pumped into the tank at a rate of 5 gal/min
...
Find the
number ( ) of pounds of salt in the tank at time
In Problem 23, what is the concentration ( ) of the salt
in the tank at time ? At ϭ 5 min? What is the concentration of the salt in the tank after a long time, that is, as
: ϱ? At what time is the concentration of the salt in
the tank equal to one-half this limiting value?
Solve Problem 23 under the assumption that the solution is pumped out at a faster rate of 10 gal/min
...

Without actually graphing, conjecture what the solution
curve of the IVP should look like
...
Repeat for the interval [0, 600] and compare
your graph with that in Figure 3
...
5(a)
...
Brine containing

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...
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...


●

1
2

pound of salt per gallon is pumped into the tank at a
rate of 6 gal/min
...
Find the number of pounds of salt in the tank after 30 minutes
...
Suppose, as in the discussion
following Example 5, that the rate at which brine is
pumped into the tank is 3 gal/min but that the wellstirred solution is pumped out at a rate of 2 gal/min
...
Now suppose that the tank has an open
top and has a total capacity of 400 gallons
...
Devise a
method for determining the number of pounds of
salt in the tank at ϭ 150 minutes
...
Does your answer agree with your intuition?
Use a graphing utility to plot the graph of ( ) on
the interval [0, 500)
...
1 henry and the
resistance is 50 ohms
...

Determine the current as : ϱ
...

A 100-volt electromotive force is applied to an
series circuit in which the resistance is 200 ohms and
the capacitance is 10 Ϫ4 farad
...
Find the current ( )
...
Find the charge ( ) on the
capacitor if (0) ϭ 0
...
Determine the charge and current
at ϭ 0
...
Determine the charge as : ϱ
...
If the
resistance at time is given by ϭ 1 ϩ 2 , where 1 and
2 are known positive constants, then (9) becomes
(

1

ϩ

2

)

If ( ) ϭ 0 and (0) ϭ
constants, show that
()ϭ

0

ϩ(

1

ϩ

0

Ϫ

0,

ϭ ( )
...


In (14) of Section 1
...
The positive
direction is downward
...

Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass
...
1
...

Suppose a small
cannonball weighing 16 pounds is shot vertically
upward, as shown in Figure 3
...
12, with an initial velocity 0 ϭ 300 ft/s
...

Suppose air resistance is ignored
...
3)
...
Find the
current ( ) if (0) ϭ 0
...
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...
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...
Find the velocity ( )
of the cannonball at time
Use the result obtained in part (a) to determine the
height ( ) of the cannonball measured from ground
level
...

Repeat Problem 36, but this time assume that air resistance is
proportional to instantaneous velocity
...

Show this by supposing that the constant of proportionality is ϭ 0
...
[
: Slightly modify the DE in
Problem 35
...
After exiting from a plane at an altitude of
15,000 feet, she waits 15 seconds and opens her parachute
...
5 during
free fall and ϭ 10 after the parachute is opened
...
What is her velocity and how far has she traveled
20 seconds after leaving the plane? See Figure 3
...
13
...
]
free fall
air resistance is 0
...
If we make the
further assumptions that the rate at which the raindrop
evaporates is proportional to its surface area and that air
resistance is negligible, then a model for the velocity
( ) of the raindrop is
ϩ

3( /␳)
( /␳) ϩ

ϭ
...

Solve for ( ) if the raindrop falls from rest
...
3 and then
show that the radius of the raindrop at time is
( ) ϭ ( ͞r) ϩ 0
...
01 ft and ϭ 0
...

The differential equation
͞ ϭ ( cos ) , where is a positive constant, is a
mathematical model for a population ( ) that undergoes yearly seasonal fluctuations
...
Use a graphing utility to graph the
solution for different choices of 0
...

Solve for ( ) if ͞ ϭ 1 and
͞ ϭ 2
Analyze the cases 1 Ͼ 2, 1 ϭ 2, and 1 Ͻ 2
...

Solve the DE subject to (0) ϭ 0
...

Use the results from part (b) to determine whether
the fish population will ever go extinct in finit
time, that is, whether there exists a time Ͼ 0
such that ( ) ϭ 0
...

A mathematical model for the
rate at which a drug disseminates into the bloodstream
is given by
ϭ Ϫ

,

where and are positive constants
...
1 to find the limiting value of
( ) as : ϱ
...
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...
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...
Sketch the graph
of ( ) and verify your prediction in part (a)
...
Suppose 1 ϭ 4 s, 2 ϭ 2 s,
(0) ϭ 0, (4) ϭ 12, (6) ϭ 0,
0 ϭ 12 V, and
(10) ϭ 12, (12) ϭ 0, and so on
...

Suppose for the sake of illustration that ϭ ϭ 1
...


where 1 Ͼ 0, 2 Ͼ 0, ( ) is the amount memorized
in time ,
is the total amount to be memorized, and
Ϫ is the amount remaining to be memorized
...
1 to find the limiting value of
( ) as : ϱ
...

Solve the DE subject to (0) ϭ 0
...


A box of mass
slides down an
inclined plane that makes an angle u with the horizontal as shown in Figure 3
...
15
...
1
...
When the switch is at , the
capacitor charges; when is at , the capacitor discharges, sending an electrical stimulus to the heart
...
3 we saw that during this
time the electrical stimulus is being applied to the heart,
the voltage across the heart satisfies the linear D

In cases ( ) and ( ), use the fact that the force of
friction opposing the motion of the box is m ,
where m is the coefficient of sliding friction and
is the normal component of the weight of the box
...

In part (a), suppose that the box weighs 96 pounds,
that the angle of inclination of the plane is u ϭ 30°,
that the coefficient of sliding friction is ϭ 13͞4,
and that the additional retarding force due to air
resistance is numerically equal to 1
...


ϭϪ

1


...
1
...
When the switch is moved to position
at time 1 the capacitor discharges, sending an
impulse to the heart over the time interval of
length 2: 1
Ͻ 1 ϩ 2
...


Box sliding down inclined plane in
Problem 46

heart

switch

0

Model of a pacemaker in
Problem 45

In Problem 46 let ( ) be
the distance measured down the inclined plane
from the highest point
...
A rootfinding application of a CAS may be useful here
...
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...
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...
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...
Moreover,
the magnitude of the impact velocity will be the
same as the initial velocity 0 of the cannonball
...

Then, using the model in Problem 37 that takes air
resistance into account, compare the value of
with and the value of the magnitude of with 0
...


point above ground when the inclination angle u
satisfies tan u m
...
Suppose that
ϭ 13͞4 and
u ϭ 23°
...
How far will the
box slide down the plane if 0 ϭ 1 ft /s?

Using the values ϭ 13͞4 and u ϭ 23°, approximate the smallest initial velocity 0 that can be given
to the box so that, starting at the highest point 50 ft
above ground, it will slide completely down the inclined plane
...


●
●

Equations (5), (6), and (10) of Section 1
...
3
Separation of variables in Section 2
...


If ( ) denotes the size of a population at time , the
model for exponential growth begins with the assumption that ͞ ϭ
for some
Ͼ 0
...
Thus for other models, (1) can be expected
to decrease as the population increases in size
...
3) can be stated as
>

( )

ϭ ( )

or

ϭ

( )
...
The quantity is called the
of the environment
...
2
...
The simplest assumption that we can make is that ( )
is linear — that is, ( ) ϭ 1 ϩ 2
...
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...
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...
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...


Ϫ

(3)

With constants relabeled, the nonlinear equation (3) is the same as
ϭ ( Ϫ

)
...
One of the equations he studied was (4), where Ͼ 0 and Ͼ 0
...
Overcrowded conditions, with the resulting detrimental effects on the environment such as pollution and
excessive and competitive demands for food and fuel, can have an inhibiting effect
on population growth
...

If we rewrite (4) as ͞ ϭ
Ϫ 2, the nonlinear term Ϫ 2, Ͼ 0, can be interpreted as an “inhibition” or “competition” term
...

One method of solving (4) is separation
͞ ( Ϫ ) ϭ into partial fractions

of variables
...


It follows from the last equation that
()ϭ

1

1ϩ

If (0) ϭ 0, 0
͞ , we find
simplifying, the solution becomes

1

ϭ

0 ͞(

0

1

1

ϭ
1

ϩ( Ϫ

Ϫ

ϩ
0),

0

()ϭ

0)

Ϫ

Ϫ


...


(5)

The basic shape of the graph of the logistic function ( ) can be
obtained without too much effort
...


0

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

/

The dashed line ϭ ͞2 shown in Figure 3
...
2 corresponds to the ordinate of a
point of inflection of the logistic curve
...


From calculus recall that the points where 2 ͞ 2 ϭ 0 are possible points of inflec
tion, but ϭ 0 and ϭ ͞ can obviously be ruled out
...
For
0 Ͻ Ͻ ͞2 it follows that Љ Ͼ 0, and ͞2 Ͻ Ͻ ͞ implies that Љ Ͻ 0
...
2
...
For ͞2 Ͻ 0 Ͻ ͞ the graph is still S-shaped, but the point of
inflection occurs at a negative value of , as shown in Figure 3
...
2(b)
...
3 in the form
͞ ϭ ( ϩ 1 Ϫ ), Ͼ 0
...
The solution ( ) represents the
number of individuals infected with the disease at time

Suppose a student carrying a flu virus returns to an isolated college campus of 1000
students
...

Assuming that no one leaves the campus throughout the duration of the
disease, we must solve the initial-value problem

500

ϭ

5

10

(1000 Ϫ ),

By making the identification
(5) that

ϭ 1000 and

()ϭ

(0) ϭ 1
...

ϭ
ϩ 999 Ϫ1000
1 ϩ 999 Ϫ1000

Now, using the information (4) ϭ 50, we determine from
50 ϭ

1000

...
9906
...

1 ϩ 999 Ϫ0
...

1 ϩ 999 Ϫ5
...
2
...
Note that
the number of infected students ( ) approaches 1000 as increases
...
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...
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...
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...
For example, the differential equations
ϭ ( Ϫ

and

)Ϫ

ϭ ( Ϫ

)ϩ

(6)

could serve, in turn, as models for the population in a fisher where fis are
or are
at rate When Ͼ 0 is a constant, the DEs in (6) can be readily analyzed qualitatively or solved analytically by separation of variables
...
The rate in (6) could be a function of
time or could be population dependent; for example, harvesting might be done periodically over time or might be done at a rate proportional to the population at time
In the latter instance, the model would look like Ј ϭ ( Ϫ ) Ϫ , Ͼ 0
...

See Problem 24 in Exercises 3
...
Another equation of the form given in (2),
ϭ ( Ϫ

ln ),

(7)

is a modification of the logistic equation known as the
named after the English mathematician
(1779–1865)
...
See
Problem 8 in Exercises 3
...

Suppose that grams of chemical are combined with
grams of chemical If there are parts of and parts of formed in the compound and ( ) is the number of grams of chemical formed, then the number of
grams of chemical and the number of grams of chemical remaining at time are,
respectively,
Ϫ

and

ϩ

Ϫ


...


(8)

If we factor out ͞( ϩ ) from the first factor and ͞( ϩ ) from the second
and introduce a constant of proportionality Ͼ 0, (8) has the form
ϭ (␣ Ϫ )(␤ Ϫ ),

(9)

where a ϭ ( ϩ )͞ and b ϭ ( ϩ )͞ Recall from (6) of Section 1
...
The resulting
reaction between the two chemicals is such that for each gram of , 4 grams of is
used
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Determine the amount of at time if the rate of the reaction is proportional to the
amounts of and remaining and if initially there are 50 grams of and 32 grams
of How much of the compound is present at 15 minutes? Interpret the solution
as : ϱ
...

If, for example, 2 grams of compound
is present, we must have used,
say, grams of and grams of , so ϩ ϭ 2 and ϭ 4 Thus we must use
ϭ 2 ϭ 2 1 g of chemical and ϭ 8 ϭ 2 4 g of In general, for grams of
5
5
5
5
we must use
time

()

()

1
5
The amounts of

and

grams of

4
5

and

grams of
...

Now we know that the rate at which compound

΂

ϰ 50 Ϫ

is formed satisfie

4
΃΂32 Ϫ 5 ΃
...


By separation of variables and partial fractions we can write
40

Ϫ

1
210

ϩ

250 Ϫ

1
210

40 Ϫ

ϭ


...


(10)

When ϭ 0, ϭ 0, so it follows at this point that 2 ϭ 25
...
1258
...
1258
...
1258

(11)

From (11) we find (15) ϭ 34
...
The behavior of as a function of time
is displayed in Figure 3
...
4
...
This means that 40 grams of compound is formed, leaving

compound

Number of grams of
in Example 2

1
50 Ϫ (40) ϭ 42 g of
5

and

4
32 Ϫ (40) ϭ 0 g of
...
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...
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...
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...
For example,
of the two results

͵
͵

2

2

Ϫ

2

1
ϭ tanhϪ1

Ϫ

2

ϭ

1
ln
2

͉

ϩ , ͉ ͉Ͻ
ϩ
Ϫ

͉ϩ

, ͉ ͉

(12)
,

(13)

(12) may be convenient in Problems 15 and 26 in Exercises 3
...


The number ( ) of supermarkets throughout the country
that are using a computerized checkout system is
described by the initial-value problem
ϭ

(1 Ϫ 0
...


Use the phase portrait concept of Section 2
...
By
hand, sketch a solution curve of the given initialvalue problem
...

How many companies are expected to adopt the new
technology when ϭ 10?
The number ( ) of people in a community who are
exposed to a particular advertisement is governed by
the logistic equation
...
Solve for ( ) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000
...
What is the limiting
value of the population? At what time will the population be equal to one-half of this limiting value?
Census data for the United States between 1790 and
1950 are given in Table 3
...
1
...


Construct a table comparing actual census population with the population predicted by the model in
part (a)
...


Year

Population (in millions)

1790
1800
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950

3
...
308
7
...
638
12
...
069
23
...
433
38
...
156
62
...
996
91
...
711
122
...
669
150
...

Suppose ϭ 5, ϭ 1, and ϭ 4
...
1 to sketch representative solution curves

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...
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...
Editorial review has
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...


●

corresponding to the cases 0 Ͼ 4, 1 Ͻ 0 Ͻ 4, and
0 Ͻ 0 Ͻ 1
...

Solve the IVP in part (a)
...

Use the information in parts (a) and (b) to determine
whether the fishery population becomes extinct in
finite time
...
Determine whether the population becomes
4
extinct in finite time
...


Suppose ϭ ϭ 1 in the Gompertz differential
equation (7)
...
1 to sketch representative solution curves corresponding to the
cases 0 Ͼ and 0 Ͻ 0 Ͻ
Suppose ϭ 1, ϭ Ϫ1 in (7)
...

Find an explicit solution of (7) subject to (0) ϭ 0
...
It is observed that 10 grams of is formed in
5 minutes
...
At what time is chemical half-formed?

A tank in the form of a
right-circular cylinder standing on end is leaking water
through a circular hole in its bottom
...
3, when friction and contraction of water at
the hole are ignored, the height of water in the tank is
described by
ϭϪ

of definition in terms of the symbols , , and
Use ϭ 32 ft/s 2
...
If the
2
tank is initially full, how long will it take to empty?

ϭϪ

where 0 Ͻ Ͻ 1
...
6? See Problem 13 in
Exercises 1
...

A tank in the form of a rightcircular cone standing on end, vertex down, is leaking
water through a circular hole in its bottom
...
In
Problem 14 in Exercises 1
...


In this model, friction and contraction of the water
at the hole were taken into account with ϭ 0
...
See Figure 1
...
12
...
Determine the differential equation governing the height of water
...
6 and ϭ 32 ft/s 2
...
2
...
Is the time it takes to
empty a full tank the same as for the tank with vertex
down in Problem 13? Take the friction/contraction coefficient to be ϭ 0
...


12 ,

where
and
are the cross-sectional areas of the
water and the hole, respectively
...
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...
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...
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...
The positive direction is downward
...

Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass
...
1
...

Consider the
16-pound cannonball shot vertically upward in Problems 36 and 37 in Exercises 3
...
Determine the maximum height attained by
the cannonball if air resistance is assumed to be proportional to the square of the instantaneous velocity
...
0003
...
]
Determine a differential
equation for the velocity ( ) of a mass sinking
in water that imparts a resistance proportional to
the square of the instantaneous velocity and also
exerts an upward buoyant force whose magnitude is
given by Archimedes’ principle
...
3
...

Solve the differential equation in part (a)
...

Ϫ ϩ 1

The differential equation
ϭ

2

ϩ

2

describes the shape of a plane curve that will reflect all
incoming light beams to the same point and could be a
model for the mirror of a reflecting telescope, a satellite
antenna, or a solar collector
...
3
...

Verify that the differential equation is homogeneous
(see Section 2
...
Show that the substitution ϭ
yields
11 ϩ

2

(1 Ϫ 11 ϩ 2)

ϭ

Use a CAS (or another judicious substitution) to
integrate the left-hand side of the equation
...

Show that the first differential equation can also be
solved by means of the substitution ϭ 2 ϩ 2
...
By inspection, find all constant solutions of
the DE
...
A CAS
may be useful for integration
...

An outdoor decorative pond in the shape
of a hemispherical tank is to be filled with water pumped
into the tank through an inlet in its bottom
...
See Figure 3
...
6
...
Assume that the rate of evaporation
is proportional to the area of the surface of the water
and that the constant of proportionality is ϭ 0
...

The rate of change ͞ of the volume of the water
at time is a net rate
...
Express the area
3
of the surface of the water ϭ p 2 in terms of
Solve the differential equation in part (a)
...

If there were no evaporation, how long would it take
the tank to fill
With evaporation, what is the depth of the water at
the time found in part (c)? Will the tank ever be
filled? Prove your assertion
Output: water evaporates
at rate proportional
to area of surface

Input: water pumped in
at rate ft3/min


...
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...
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...
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...
In Section 3
...
See
Example 1 on page 84
...
01 that the nonlinear differential
equation
ϭ

1
...
Solve
the differential equation subject to the initial condition (0) ϭ 10 and the fact that the animal population has doubled in 5 months
...
Find
...


( ) If Ͼ 0, Ͼ 0 show by means of a phase portrait
(see page 39) that, depending on the initial condition (0) ϭ 0, the mathematical model could include a doomsday scenario ( ( ) : ϱ) or an extinction scenario ( ( ) : 0)
...
0005 Ϫ 0
...

Show that this model predicts a doomsday for the
population in a finite time
...
Show that this
model predicts extinction for the population as
: ϱ
...
One way of
1
doing this is to approximate the left-hand side
of
the first equation in (2), using the forward difference
quotient in place of ͞ :
()ϭ

1 ( ϩ )Ϫ ()

...
, 160 and ϭ 10
...
With (0) ϭ 3
...
308,
(0) ϭ

1 (10) Ϫ (0)
ϭ 0
...

(0)
10

Note that (160) depends on the 1960 census population (170)
...

Use a CAS to obtain a scatter plot of the data
( ( ), ( )) computed in part (a)
...

Construct a logistic model ͞ ϭ ( ), where ( )
is the equation of the regression line found in part (b)
...
929
...
Use your CAS to superimpose the graph of
the solution in part (d) on the scatter plot
...
S
...
What population does the logistic model in part
(c) predict for these years? What does the model predict for the U
...
population ( ) as : ϱ?
In Examples 3 and 4 of
Section 2
...
Use a root-finding application of
a CAS (or a graphic calculator) to approximate the
equilibrium solution of the immigration model
ϭ (1 Ϫ ) ϩ 0
...


Use a graphing utility to graph the function
( ) ϭ (1 Ϫ ) ϩ 0
...
Explain how this graph
can be used to determine whether the number found
in part (a) is an attractor
...
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...
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...
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...
2 and
for the IVPs

0

(0) ϭ

0

ϭ 1
...
3

Ϫ

,

(0) ϭ

0

for 0 ϭ 0
...
2
...
Over a long period of time, what percentage
increase does the immigration model predict in the
population compared to the logistic model?
In Problem 16 let be the time it
takes the cannonball to attain its maximum height and
let be the time it takes the cannonball to fall from the
maximum height to the ground
...
See
Problem 48 in Exercises 3
...
A root-finding application
of a CAS might be useful here
...
]
A skydiver is equipped with a stopwatch
and an altimeter
...
2
...
Assume that air resistance is proportional to
the square of the instantaneous velocity, his initial velocity on leaving the plane is zero, and ϭ 32 ft/s 2
...
Use the
expression for terminal velocity obtained in part
(b) of Problem 15 to eliminate from the IVP
...
]
How far does the skydiver fall and what is his
velocity at ϭ 15 s?

14,800 ft

A helicopter hovers 500 feet above a
large open tank full of liquid (not water)
...
Assume
that air resistance is proportional to instantaneous velocity while the object is in the air and that viscous
damping is proportional to 2 after the object has entered the liquid
...
1
...
If the tank is 75 feet high, determine the
time and the impact velocity when the object hits the
bottom of the tank
...
If you use (13), be careful in removing the absolute value sign
...
]
In Figure 3
...
8(a) suppose that the
-axis and the dashed vertical line ϭ 1 represent, respectively, the straight west and east beaches of a river
that is 1 mile wide
...
A man
enters the current at the point (1, 0) on the east shore and
swims in a direction and rate relative to the river given by
the vector , where the speed | | ϭ mi/h is a constant
...
Use
Figure 3
...
8(b) as an aid in showing that a mathematical
model for the path of the swimmer in the river is
Ϫ

ϭ

1

2

ϩ

2


...
2
...

Resolve
and
into components in the - and
swimmer

west
beach

east
beach
current

(0, 0)

(1, 0)

()

( ( ), ( ))
()

25 s

(0, 0)

Skydiver in Problem 26

()

(1, 0)

Path of swimmer in Problem 28

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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

-directions
...
]
Solve the DE in Problem 28 subject to (1) ϭ 0
...

Determine the values of for which the swimmer
will reach the point (0, 0) by examining limϩ ( ) in
:0
the cases ϭ 1, Ͼ 1, and 0 Ͻ Ͻ 1
...
Assume that the
speed | | ϭ mi/h is a constant
...
2
...
, 12 hr
...


1
2

1 hour
2 hours

ϭϪ
...
Rather, an approximation to the current speed (measured in miles per hour)
could be a function such as ( ) ϭ 30 (1 Ϫ ),
0
1, whose values are small at the shores (in this
case, (0) ϭ 0 and (1) ϭ 0) and largest in the middle of
the river
...
When the swimmer makes it across the river, how far will he have to
walk along the beach to reach the point (0, 0)?
When a bottle of liquid
refreshment was opened recently, the following factoid
was found inside the bottle cap:
A quick search of the Internet found that meteorologist
Jeff Haby offers the additional information that an
“average” spherical raindrop has a radius of 0
...
and
an approximate volume of 0
...
Use this data
and, if need be, dig up other data and make other reasonable assumptions to determine whether
is consistent with the models in Problems
35 and 36 in Exercises 3
...
Also see Problem 36 in Exercises 1
...

The
or water clock, was a
device that the ancient Egyptians, Greeks, Romans, and
Chinese used to measure the passage of time by observing the change in the height of water that was permitted
to flow out of a small hole in the bottom of a container
or tank
...
Assume that
(0) ϭ 2 ft corresponds to water filled to the top of
the tank, a hole in the bottom is circular with radius
1
ϭ 32 ft/s2, and ϭ 0
...
Use the differential
32 in
...


Clepsydra in Problem 33

Suppose that a glass tank has the shape of a cone with
circular cross section as shown in Figure 3
...
10
...
,
2
ϭ 32 ft/s , and ϭ 0
...
Use the differential equation in Problem 12 to find the height ( ) of the
water
...


1
2

Clepsydra in Problem 34

Suppose that ϭ ( ) defines the shape of a water clock
for which the time marks are equally spaced
...
Assume
that the cross-sectional area
of the hole is constant
...
]

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

●

Section 1
...
3 in that we are just going to discuss certain mathematical models, but instead of a single differential equation the models will be systems of
first-order differential equations
...
There are reasons for this: First, we do not possess the necessary mathematical
tools for solving systems at this point
...
We shall examine
solution methods for systems of
DEs in Chapters 4, 7, and 8
...
But if
there are, say, two interacting and perhaps competing species living in the same
environment (for example, rabbits and foxes), then a model for their populations ( )
and ( ) might be a system of two first-order di ferential equations such as
ϭ

1(

, , )
(1)

ϭ
When
1(

1

and

2

, , )ϭ

2(

, , )
...
3
and 3
...
This process, called a
continues until a stable element is reached
...

The half-lives of the various elements in a radioactive series can range from
billions of years (4
...
Suppose a
Ϫ␭1
Ϫ␭2
radioactive series is described schematically by : : , where 1 ϭ Ϫl1 Ͻ 0
and 2 ϭ Ϫl 2 Ͻ 0 are the decay constants for substances and , respectively,
and is a stable element
...
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...
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...
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...
Since is a stable element, it is simply gaining atoms from
the decay of element
ϭ ␭2
...


Consider the two tanks shown in Figure 3
...
1
...
Suppose tank contains 50 gallons of pure water
...

We wish to construct a mathematical model that describes the number of pounds 1( )
and 2( ) of salt in tanks and , respectively, at time

pure water
3 gal/min

mixture
1 gal/min

mixture
4 gal/min

mixture
3 gal/min

Connected mixing tanks

By an analysis similar to that on page 24 in Section 1
...
1 we see that the net rate of change of 1( ) for tank is
input rate
of salt

output rate
of salt

(

)

(

1
2
1
––– ϭ (3 gal/min) ؒ (0 lb/gal) ϩ (1 gal/min) ؒ ––– lb/gal Ϫ (4 gal/min) ؒ ––– lb/gal
50
50

2
ϭ Ϫ –––
25

1

1
ϩ ––– 2
...

25 2

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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


)

●

Thus we obtain the linear system
1

ϭϪ

2

ϭ

2
25
2
25

1

1

1
50
2
Ϫ
25
ϩ

2

(3)
2
...


1(0)

ϭ 25,

Suppose that two different species of animals interact
within the same environment or ecosystem, and suppose further that the firs species
eats only vegetation and the second eats only the first species
...
For example, wolves hunt grass-eating
caribou, sharks devour little fish, and the snowy owl pursues an arctic rodent called
the lemming
...

Let ( ) and ( ) denote the fox and rabbit populations, respectively, at time
If there were no rabbits, then one might expect that the foxes, lacking an adequate
food supply, would decline in number according to
ϭϪ ,

Ͼ 0
...
Adding this last rate to (4) gives a model for the fox population:
ϭϪ


...


(6)

But when foxes are present, a model for the rabbit population is (6) decreased by
, Ͼ 0 — that is, decreased by the rate at which the rabbits are eaten during their
encounters with the foxes:
ϭ


...
This famous system of equations is known
as the
Except for two constant solutions, ( ) ϭ 0, ( ) ϭ 0 and ( ) ϭ ͞ , ( ) ϭ ͞ ,
the nonlinear system (8) cannot be solved in terms of elementary functions
...
See Chapter 9,
“Numerical Solutions of Ordinary Differential Equations,” and Chapter 10, “Plane
Autonomous Systems
...
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...
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...
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...
16 ϩ 0
...
5 Ϫ 0
...
Because we are dealing with populations, we
have ( ) Ն 0, ( ) Ն 0
...
3
...
The initial conditions used were (0) ϭ 4, (0) ϭ 4
...
Observe that the model seems to
predict that both populations ( ) and ( ) are periodic in time
...

Now suppose two different species of animals occupy
the same ecosystem, not as predator and prey but rather as competitors for the same
resources (such as food and living space) in the system
...

Since the two species compete, another assumption might be that each of these
rates is diminished simply by the influence, or existence, of the other population
...

On the other hand, we might assume, as we did in (5), that each growth rate in
(9) should be reduced by a rate proportional to the number of interactions between
the two species:
ϭ

Ϫ

ϭ

Ϫ

(11)

...
Finally, it might be more realistic to replace the rates in (9), which
indicate that the population of each species in isolation grows exponentially, with
rates indicating that each population grows logistically (that is, over a long time the
population is bounded):
ϭ

1

Ϫ

1

2

and

ϭ

2

Ϫ

2

2


...
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...
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...
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...
The linear system (10) and the nonlinear systems
(11) and (13) are, of course, called
1

1

1

An electrical network having more than one loop also gives rise to
simultaneous differential equations
...
3
...
By
we can write

3

1
1

2

1

2

) ϭ 2( ) ϩ 3( )
...
For loop
summing the voltage drops across each part of the loop gives

2
2

1(

2

2

Network whose model
is given in (17)

()ϭ
Similarly, for loop

1 1 1

2

2

1 1

2

1

ϩ

2

1

1 1 2

2

1,

2 2
...


(16)

we fin

()ϭ

1 1

ϩ

2

3

Using (14) to eliminate 1 in (15) and (16) yields two linear first-order equations for
the currents 2( ) and 3( ):
1

2

ϩ(

1

ϩ

2) 2

ϩ

ϭ

1 3

()
(17)

2

1

3
2

3

ϩ

1 2

ϩ

1 3

ϭ ( )
...
3) to show that the system of differential equations describing the currents 1( ) and 2( ) in the network containing a resistor, an inductor, and a capacitor shown in Figure 3
...
4 is
1

ϩ

ϭ ()

2

(18)
Network whose
model is given in (18)

We have not discussed methods by which systems
of first-order differential equations can be solved
...
Find a solution of (2) subject to the
initial conditions (0) ϭ 0, (0) ϭ 0, (0) ϭ 0
...
138629 and
2 ϭ Ϫ0
...
Use a graphing utility
to obtain the graphs of the solutions ( ), ( ), and ( )

2

ϩ

2

Ϫ

1

ϭ 0
...
Use the graphs to
approximate the half-lives of substances and
Use the graphs in Problem 2 to approximate the times
when the amounts ( ) and ( ) are the same, the
times when the amounts ( ) and ( ) are the same, and
the times when the amounts ( ) and ( ) are the same
...


Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Consider two tanks and , with liquid being pumped in
and out at the same rates, as described by the system of
equations (3)
...
3
...

pure water
4 gal/min

mixture
2 gal/min

2( ), and 3( ) at time in tanks , , and , respectively
...


pure water
4 gal/min

200 gal

150 gal

mixture
4 gal/min

mixture
1 gal/min

100 gal

mixture
4 gal/min

mixture
4 gal/min

Mixing tanks in Problem 8

100 gal

100 gal

mixture
6 gal/min

Consider the Lotka-Volterra predator-prey model
defined b

100 gal

mixture
5 gal/min

mixture
4 gal/min

Mixing tanks in Problem 6

Two very large tanks and are each partially filled
with 100 gallons of brine
...
3
...


ϭ Ϫ0
...
02
ϭ 0
...
025

,

where the populations ( ) (predators) and ( ) (prey)
are measured in thousands
...
Use a numerical solver to graph ( ) and ( )
...
Use the graphs to
approximate the period of each population
...
4 Ϫ 0
...
1 Ϫ 0
...

Find a relationship between the variables 1( )
and 2( ) that holds at time Explain why this
relationship makes intuitive sense
...

Three large tanks contain brine, as shown in Figure 3
...
7
...
Use a numerical solver to
analyze the populations over a long period of time for
each of the following cases:
(0) ϭ 1
...
5
(0) ϭ 1, (0) ϭ 1
(0) ϭ 2, (0) ϭ 7
(0) ϭ 4
...
5
Consider the competition model defined by
ϭ (1 Ϫ 0
...
05 )
ϭ (1
...
1 Ϫ 0
...
Use a numerical solver to

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...
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...
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...


●

analyze the populations over a long period of time for
each of the following cases:
(0) ϭ 1, (0) ϭ 1
(0) ϭ 4, (0) ϭ 10
(0) ϭ 9, (0) ϭ 4
(0) ϭ 5
...
5

let ( ), ( ), and ( ) denote, in turn, the number of people in the community (measured in hundreds) who are
to the disease but not yet infected with it,
the number of people who are
with the disease, and the number of people who have
from the disease
...


2

where

2

ϭ

ϭ ()

1

ϭϪ

Show that a system of differential equations that
describes the currents 2( ) and 3( ) in the electrical
network shown in Figure 3
...
8 is

,

2

ϩ

1

(called the
) and 2 (called the
) are positive constants, is a reasonable
mathematical model, commonly called a
for the spread of the epidemic throughout the community
...

1

In Problem 15, explain why it is sufficient to
analyze only

1

ϭϪ

1

ϭϪ

2

Network in Problem 12

Determine a system of first-order differential equations
that describes the currents 2( ) and 3( ) in the electrical
network shown in Figure 3
...
9
...


Suppose 1 ϭ 0
...
7, and ϭ 10
...
Use a
numerical solver to determine what the model predicts about the epidemic in the two cases 0 Ͼ 2 ͞ 1
and 0
2 ͞ 1
...


2

1

2

2

3

Network in Problem 13

Suppose compartments
and shown in Figure 3
...
10 are filled with fluids and
are separated by a permeable membrane
...
Suppose, too, that a nutrient necessary
for cell growth passes through the membrane
...
3
...
[
: ͞ ϭ 3
...
Suppose that everyone
is initially susceptible to the disease and that no one leaves
the community while the epidemic is spreading
...
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...
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...
Let (0) ϭ 0 and
(0) ϭ 0 denote the initial concentrations of the nutrient
...
Explain your reasoning
...


solver to obtain numerical solution curves of (3) subject
to the initial conditions 1(0) ϭ 25, 2(0) ϭ 0
...
3
...
As
container cools, the temperature of the medium inside container does not change significantly and can
be considered to be a constant
...
Find a solution of
the system subject to the initial conditions (0) ϭ 0,
(0) ϭ 1
...
Solve for ( ) and
( ) and compare their graphs with your sketches in
Problem 17
...
Explain why the answer to the last question makes intuitive sense
...
3
...
What is the behavior of each function over a
long period of time? Sketch possible graphs of 1( ) and
2( )
...

Fill in the blank or answer true or false
...
15 gives the population in an environment
at time , then a differential equation satisfied by ( )
is

...

In March 1976 the world population reached 4 billion
...
8%, the world
population would be 8 billion in 45 years
...
06% carbon dioxide is pumped into a
room whose volume is 8000 ft 3
...
If there is an initial concentration of 0
...
See Problem 28 in Exercises 1
...
Assume
that the initial point on the -axis in (0, 10) and that the
length of the rope is ϭ 10 ft
...
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...
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...
Suppose further
that the cell has constant volume and that the area of
its permeable membrane is the constant By
the
rate of change of its mass is directly proportional to the
area and the difference Ϫ ( ), where ( ) is the concentration of the solute inside the cell at time Find ( ) if
ϭ ؒ ( ) and (0) ϭ 0
...
R
...


concentration
()

concentration

Find the current ( ) if the resistance is 0
...
Graph ( )
...
See Figure 3
...
2
...

First solve for
in terms of and , and then use the
substitution ϭ sin 2u to obtain a parametric form of
the solution
...


molecules of solute
diffusing through
cell membrane

(0, 0)
bead

Cell in Problem 6

Suppose that as a body cools, the temperature of the
surrounding medium increases because it completely
absorbs the heat being lost by the body
...
If the initial temperature of the
body is 1 and the initial temperature of the medium
is 2, then it can be shown in this case that Newton’s
law of cooling is
͞ ϭ ( Ϫ ), Ͻ 0, where
ϭ 2 ϩ ( 1 Ϫ ), Ͼ 0 is a constant
...
Use the phase
portrait concept of Section 2
...

What is the limiting value of ( ) as : ϱ?
Verify your answers in part (a) by actually solving
the differential equation
...

According to
the absolute
temperature of a body cooling in a medium at constant
absolute temperature
is given by
ϭ (

4

Ϫ

4

),

An
-series circuit has a variable inductor with the
inductance defined by

()ϭ

Ά

0,

1
,
10

0

Ͻ 10
Ն 10
...


Solve for and in terms of
Initially, two large tanks and each hold 100 gallons
of brine
...
R
...
Use the information
given in the figure to construct a mathematical model
for the number of pounds of salt 1( ) and 2( ) at time
in tanks and , respectively
...
Stefan’s law can be used over a
greater temperature range than Newton’s law of cooling
...

is small in comparison to
Show that when Ϫ
then Newton’s law of cooling approximates
Stefan’s law
...
]

1Ϫ

( 1,

mixture
5 gal/min

100 gal

mixture
3 gal/min

100 gal

mixture
1 gal/min

mixture
4 gal/min

Mixing tanks in Problem 12

When all the curves in a family ( , , 1) ϭ 0 intersect
orthogonally all the curves in another family ( , , 2) ϭ 0,
the families are said to be
of each
other
...
R
...
If ͞ ϭ ( , ) is the differential
equation of one family, then the differential equation for the

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...
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...
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...


●

orthogonal trajectories of this family is ͞ ϭ Ϫ1͞ ( , )
...
Find the orthogonal trajectories of this family
...

( , ,

1)

argon-40 (Ar-40)
...

From the foregoing system of differential equations
find ( ) if (0) ϭ 0
...

It is known that l1 ϭ 4
...
5874 ϫ 10Ϫ10
...

Use your solutions for ( ) and ( ) to determine the
percentage of an initial amount 0 of K-40 that decays into Ca-40 and the percentage that decays into
Ar-40 over a very long period of time
...
Although potassium occurs naturally in the form
of three isotopes, only the isotope potassium-40 (K-40)
is radioactive
...
Over time,
by emitting a beta particle, a great percentage of an initial amount of K-40 decays into the stable isotope calcium-40 (Ca-40), whereas by electron capture a smaller
percentage of K-40 decays into the stable isotope

*

The knowledge of how K-40 decays is the basis for the

...
Fossils can sometimes be dated indirectly by dating the
igneous rocks in the substrata in which the fossils are found
...
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...
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...
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...
1

Preliminary Theory—Linear Equations
Initial-Value and Boundary-Value Problems
4
...
2 Homogeneous Equations
4
...
3 Nonhomogeneous Equations
4
...
3 Homogeneous Linear Equations with Constant Coefficients
4
...
5 Undetermined Coefficients—Annihilator Approach
4
...
7 Cauchy-Euler Equation
4
...
8
...
8
...
9 Solving Systems of Linear DEs by Elimination
4
...
1
...

In the first seven sections of this chapter we examine the underlying theory an
solution methods for certain kinds of linear equations
...
8 we build on the material of Section 4
...
The elimination
method of solving systems of linear equations is introduced in Section 4
...
The chapter concludes with a brief examination of nonlinear
higher-order equations in Section 4
...


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...
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...
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...


●

●
●

Reread the
at the end of Section 1
...
3 (especially page 57)

In Chapter 2 we saw that we could solve a few first-order differential equations by recognizing them as separable, linear, exact, homogeneous, or perhaps Bernoulli equations
...
Only
in the case of
first-order differential equations were we able to obtain general solutions, by
paying attention to certain continuity conditions imposed on the coefficients
...
2 we defined an initial-value problem for
th-order differential equation
...
,

1(

)

( Ϫ1)

( 0) ϭ

ϩ

0(

) ϭ ( )

(1)

Ϫ1
...
, ( Ϫ1)( 0) ϭ Ϫ1
...

In Section 1
...
The theorem that follows gives sufficien
conditions for the existence of a unique solution of the problem in (1)
...
, 1( ), 0( ) and ( ) be continuous on an interval and
let ( ) 0 for every in this interval
...


The initial-value problem
3 ٞ ϩ 5 Љ Ϫ Ј ϩ 7 ϭ 0,

(1) ϭ 0,

Ј(1) ϭ 0,

Љ(1) ϭ 0

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

possesses the trivial solution ϭ 0
...
1
...


2

You should verify that the function ϭ 3
value problem
Љ Ϫ 4 ϭ 12 ,

Ϫ2

ϩ

Ϫ 3 is a solution of the initial-

(0) ϭ 4,

Ј(0) ϭ 1
...
We conclude from
Theorem 4
...
1 that the given function is the unique solution on
The requirements in Theorem 4
...
1 that ( ), ϭ 0, 1, 2,
...
Specificall , if ( ) ϭ 0 for some
in the interval, then the solution of a linear initial-value problem may not be unique
or even exist
...
Although most of the conditions of Theorem 4
...
1
are satisfied, the obvious difficulties are that 2( ) ϭ 2 is zero at ϭ 0 and that the
initial conditions are also imposed at ϭ 0
...
See Figure 4
...
1
...
These three pairs of conditions are just
special cases of the general boundary conditions

␣1 ( ) ϩ ␤1 Ј( ) ϭ

1

␣2 ( ) ϩ ␤ 2 Ј( ) ϭ

2
...
1
...
1
...


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...
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...
Editorial review has
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...


●

In Example 7 of Section 1
...


Suppose we now wish to determine the solution of the equation that further
satisfies the boundary conditions (0) ϭ 0, (p͞2) ϭ 0
...
Hence the
boundary-value problem

1
4

Љ ϩ 16 ϭ 0,

( 2, 0)

Solution curves for
BVP in part (a) of Example 3

΂␲΃ ϭ 0
2

(0) ϭ 0,

(3)

has infinitely many solutions
...
1
...

If the boundary-value problem in (3) is changed to
Љ ϩ 16 ϭ 0,

΂␲΃ ϭ 0,
8

(0) ϭ 0,

(4)

then (0) ϭ 0 still requires 1 ϭ 0 in the solution (2)
...
Hence ϭ 0 is a solution of this
new boundary-value problem
...

Finally, if we change the problem to
Љ ϩ 16 ϭ 0,

΂␲΃ ϭ 1,
2

(0) ϭ 0,

(5)

we find again from (0) ϭ 0 that 1 ϭ 0, but applying (p͞2) ϭ 1 to ϭ 2 sin 4 leads
to the contradiction 1 ϭ 2 sin 2p ϭ 2 ؒ 0 ϭ 0
...


A linear th-order differential equation of the form
Ϫ1

( )

ϩ

Ϫ1(

is said to be

)

Ϫ1

1(

ϩиииϩ

)

) ϭ0

(6)

) ϭ ( ),

(7)

0(

ϩ

whereas an equation
Ϫ1

( )

ϩ

Ϫ1(

)

Ϫ1

ϩиииϩ

1(

)

ϩ

0(

with ( ) not identically zero, is said to be
For example,
2 Љ ϩ 3 Ј Ϫ 5 ϭ 0 is a homogeneous linear second-order differential equation,
whereas 3 ٞ ϩ 6 Ј ϩ 10 ϭ is a nonhomogeneous linear third-order differential equation
...
5
...

To avoid needless repetition throughout the remainder of this text, we
shall, as a matter of course, make the following important assumptions when

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...
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...
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...


●

stating definitions and theorems about linear equations (1)
...
,
•
( ) 0 for every in the interval
...
For example,
(cos 4 ) ϭ Ϫ4 sin 4 and (5 3 Ϫ 6 2 ) ϭ 15 2 Ϫ 12 Higher-order derivatives
can be expressed in terms of in a natural manner:

΂ ΃

2

ϭ

2

(

ϭ

2

)ϭ

and, in general,

ϭ

,

where represents a sufficiently differentiable function
...
In general, we define an
or
to be
ϭ

( )

ϩ

Ϫ1(

)

Ϫ1

ϩиииϩ

1(

)

ϩ

0(

)
...
In symbols this means that
{a ( ) ϩ b ( )} ϭ a ( ( )) ϩ b ( ( )),

(9)

where a and b are constants
...
For example, the differential equation Љ ϩ 5 Ј ϩ 6 ϭ 5 Ϫ 3
can be written as 2 ϩ 5 ϩ 6 ϭ 5 Ϫ 3 or ( 2 ϩ 5 ϩ 6) ϭ 5 Ϫ 3
...

In the next theorem we see that the sum, or
of two or more solutions of a homogeneous linear differential equation is
also a solution
...
, be solutions of the homogeneous th-order differential equation
(6) on an interval Then the linear combination
ϭ
where the
interval
...
,

)ϩ

2 2(

)ϩиииϩ

( ),

are arbitrary constants, is also a solution on the

We prove the case ϭ 2
...
If
we define ϭ 1 1( ) ϩ 2 2( ), then by linearity of we have
( )ϭ {

1 1(

)ϩ

2 2(

)} ϭ

1

( 1) ϩ

2

( 2) ϭ

1

ؒ0ϩ

2

ؒ 0 ϭ 0
...
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...
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...
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...

A homogeneous linear differential equation always possesses the trivial
solution ϭ 0
...
By the superposition principle
the linear combination
ϭ

1

2

ϩ

2

2

ln

The function ϭ 7 is a solution of Љ Ϫ 9 Ј ϩ 14 ϭ 0
...
For various values of we see that ϭ 9 7 , ϭ 0, ϭ Ϫ 15 7 ,
...

is also a solution of the equation on the interval
...


A set of functions 1( ), 2( ),
...
, , not all zero, such that
1 1(

)ϩ

2 2(

)ϩиииϩ

on an

( )ϭ0

for every in the interval
...

It is easy to understand these definitions for a set consisting of two functions
( ) and 2( )
...
Therefore if we assume that 1 0, it follows that
1 ( ) ϭ (Ϫ 2 ͞ 1) 2 ( ); that is,
Conversely, if 1( ) ϭ 2 2( )
for some constant 2, then (Ϫ1) ؒ 1( ) ϩ 2 2( ) ϭ 0 for every in the interval
...
We conclude that
1( ) and 2( )
on
the interval
...
Recall
from the double-angle formula for the sine that sin 2 ϭ 2 sin cos On the other
hand, the set of functions 1( ) ϭ , 2( ) ϭ ͉ ͉ is linearly independent on (Ϫϱ, )
...
1
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

It follows from the preceding discussion that the quotient 2( )͞ 1( ) is not a constant on an interval on which the set 1( ), 2( ) is linearly independent
...


The set of functions 1( ) ϭ cos 2 , 2( ) ϭ sin 2 , 3( ) ϭ sec 2 ,
linearly dependent on the interval (Ϫp͞2, p͞2) because
1

when 1 ϭ 2 ϭ 1,
1 ϩ tan2 ϭ sec2

cos2 ϩ
3

2

ϭ Ϫ1,

sin2 ϩ
4

3

sec2 ϩ

4

4(

) ϭ tan 2 is

tan2 ϭ 0

ϭ 1
...
, ( ) is linearly dependent on an interval if
at least one function can be expressed as a linear combination of the remaining
functions
...
Observe that
2(

) ϭ 1 ؒ 1( ) ϩ 5 ؒ 3( ) ϩ 0 ؒ 4( )

for every in the interval (0, )
...
Although we could always appeal directly to Definition 4
...
1,
it turns out that the question of whether the set of solutions 1, 2 ,
...


Suppose each of the functions 1( ), 2( ),
...
The determinant

( 1, 2,
...


Ј
и
и
и

( Ϫ1)

Ϫ1

͉

,

of the

Let 1, 2 ,
...
, ) 0 for every in the
interval
...
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...
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...
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...
1
...
, are solutions of (6) on
an interval , the Wronskian ( 1, 2 ,
...

A set of linearly independent solutions of a homogeneous linear th-order
differential equation is given a special name
...
, of linearly independent solutions of the homogeneous
linear th-order differential equation (6) on an interval is said to be a
on the interval
...


There exists a fundamental set of solutions for the homogeneous linear th-order
differential equation (6) on an interval
Analogous to the fact that any vector in three dimensions can be expressed as a
linear combination of the
vectors
any solution of an thorder homogeneous linear differential equation on an interval can be expressed as a
linear combination of linearly independent solutions on In other words, linearly
are the basic building blocks for the general
independent solutions 1, 2 ,
...


Let 1, 2,
...
, are arbitrary constants
...
1
...
,
can always be found so that
( )ϭ
We will prove the case when

1 1(

)ϩ

2 2(

)ϩиииϩ

( )
...


Let be a solution and let 1 and 2 be linearly independent solutions of
Љ ϩ 1 Ј ϩ 0 ϭ 0 on an interval Suppose that ϭ is a point in for which
( 1( ), 2( )) 0
...
If we now examine
the equations
2

1 1(

2 2(

)ϭ

1

1 Ј(
1

it follows that we can determine
the coefficients satisfi

)ϩ
)ϩ

2 Ј(
2

)ϭ

2,

1

and

͉ Ј(( ))
1

1

2

uniquely, provided that the determinant of
2(

)
Ј( )
2

͉

0
...
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...
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...
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...
If we define ( ) ϭ 1 1( ) ϩ 2 2( ), we observe that ( ) satisfies the
differential equation since it is a superposition of two known solutions; ( ) satisfie
the initial conditions
()ϭ

1 1(

)ϩ

2 2(

)ϭ

and

1

Ј( ) ϭ

1 Ј(
1

)ϩ

2 Ј(
2

)ϭ

2;

and ( ) satisfies the
linear equation and the
initial conditions
...
1
...


The functions 1 ϭ 3 and 2 ϭ Ϫ3 are both solutions of the homogeneous linear
equation Љ Ϫ 9 ϭ 0 on the interval (Ϫϱ, )
...
This fact can be corroborated by observing that the
Wronskian
(

3

,

Ϫ3

)ϭ

͉3

3

͉ ϭ Ϫ6

Ϫ3

3

Ϫ3

Ϫ3

0

for every We conclude that 1 and 2 form a fundamental set of solutions, and
consequently, ϭ 1 3 ϩ 2 Ϫ3 is the general solution of the equation on the
interval
...
(Verify this
...
1
...
Observe that if we choose
3
Ϫ 7 Ϫ3 can be rewritten as
1 ϭ 2 and 2 ϭ Ϫ7, then ϭ 2
ϭ2

3

Ϫ2

Ϫ3

Ϫ5

Ϫ3

ϭ4

΂

3

Ϫ3

Ϫ
2

The last expression is recognized as ϭ 4 sinh 3 Ϫ 5

The functions 1 ϭ , 2 ϭ 2 , and
ٞ Ϫ 6 Љ ϩ 11 Ј Ϫ 6 ϭ 0
...



...
We conclude that ϭ 1 ϩ 2 2 ϩ 3 3 is the general solution
of the differential equation on the interval
...
For example, it is a straightforward
task to show that the constant function
ϭ 3 is a particular solution of the
nonhomogeneous equation Љ ϩ 9 ϭ 27
...
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...
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...
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...
, are solutions of (6) on an interval and
solution of (7) on , then the linear combination
ϭ

1 1(

)ϩ

2 2(

)ϩиииϩ

is any particular
(10)

( )ϩ

is also a solution of the nonhomogeneous equation (7)
...
If we use ϭ linearly independent solutions of the th-order
equation (6), then the expression in (10) becomes the general solution of (7)
...
, be a fundamental set of solutions of the associated homogeneous differential equation (6) on Then the
of the equation on the interval is
ϭ

1 1(

)ϩ

2 2(

)ϩиииϩ

( )ϩ

,

where the , ϭ 1, 2,
...

Let be the differential operator defined in (8) and let ( ) and ( )
be particular solutions of the nonhomogeneous equation ( ) ϭ ( )
...


This shows that ( ) is a solution of the homogeneous equation ( ) ϭ 0
...
1
...


We see in Theorem 4
...
6 that the general solution of a nonhomogeneous linear equation consists of the sum of two functions:
ϭ

1 1(

)ϩ

2 2(

)ϩиииϩ

( )ϩ

( )ϭ

( )ϩ

( )
...
In
other words, to solve a nonhomogeneous linear differential equation, we first solve
the associated homogeneous equation and then find any particular solution of the
nonhomogeneous equation
...


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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


(11)

●

To write the general solution of (11), we must also be able to solve the associated
homogeneous equation
ٞ Ϫ 6 Љ ϩ 11 Ј Ϫ 6 ϭ 0
...
Hence the general solution of (11)
on the interval is
ϭ

ϩ

ϭ

ϩ

1

2

2

ϩ

3

3

Ϫ

11 1
Ϫ
...
4 when we consider a method for finding particular solutions
of nonhomogeneous equations
...
, be particular solutions of the nonhomogeneous linear thorder differential equation (7) on an interval corresponding, in turn, to distinct functions 1, 2,
...
,

)

( Ϫ1)

1(

ϩиииϩ

) Јϩ

0(

) ϭ

( ),

(12)

Then
ϭ

1

( )ϩ

2

( )ϩиииϩ

(13)

( )

is a particular solution of
( )

( )

ϩ

ϭ

1(

)ϩ

Ϫ1(
2(

)

( Ϫ1)

1(

ϩиииϩ

)ϩиииϩ

) Јϩ

0(

)
(14)

( )
...
Let be the differential operator defined in (8)
and let 1( ) and 2( ) be particular solutions of the nonhomogeneous equations
( ) ϭ 1( ) and ( ) ϭ 2( ), respectively
...
The result
follows again by the linearity of the operator :
( )ϭ {

( )ϩ

1

2

( )} ϭ (

( )) ϩ (

1

( )) ϭ

2

1(

)ϩ

2(

)
...
1
...


, and

2

,

3

,

is a solution of
Љ Ϫ 3 Ј ϩ 4 ϭ Ϫ16

2

ϩ 24 Ϫ 8 ϩ 2
1(

)

ϩ2

2

2(

)

Ϫ
...
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...
, , then the linear
ϭ

1

1

ϩ

2

2

,

ϩиииϩ

where the are constants, is also a particular solution of (14) when the right-hand
member of the equation is the linear combination
1 1(

)ϩ

2 2(

)ϩиииϩ

( )
...


This remark is a continuation of the brief discussion of dynamical systems
given at the end of Section 1
...

A dynamical system whose rule or mathematical model is a linear th-order
differential equation
()

( )

ϩ

Ϫ1(

)

( Ϫ1)

ϩиииϩ

1(

) Јϩ

0(

) ϭ ()

is said to be an th-order
The time-dependent functions ( ),
of the system
...
, ( Ϫ1)( ) are the
ues at some time give the
The function is variously
called the
or
A solution ( ) of the differential equation is said to be the
or
Under the conditions stated in Theorem 4
...
1, the output or response
( ) is uniquely determined by the input and the state of the system prescribed
at a time 0 —that is, by the initial conditions ( 0), Ј( 0),
...

For a dynamical system to be a linear system, it is necessary that the superposition principle (Theorem 4
...
7) holds in the system; that is, the response of
the system to a superposition of inputs is a superposition of outputs
...
1 (linear first-orde
equations); in Section 5
...


In Problems 1 – 4 the given family of functions is the general
solution of the differential equation on the indicated interval
...

ϭ 1 ϩ
Љ Ϫ ϭ 0,

2

Ϫ

, (Ϫϱ, );
(0) ϭ 0, Ј(0) ϭ 1

ϭ 1 4 ϩ 2 Ϫ , (Ϫϱ, );
Љ Ϫ 3 Ј Ϫ 4 ϭ 0, (0) ϭ 1,
ϭ 1 ϩ 2 ln , (0, );
Љ Ϫ Ј ϩ ϭ 0, (1) ϭ 3,

2

Ј(0) ϭ 2
Ј(1) ϭ Ϫ1

ϭ 1 ϩ 2 cos ϩ 3 sin , (Ϫϱ, );
ٞ ϩ Ј ϭ 0,
(p) ϭ 0, Ј(p) ϭ 2,

Љ(p) ϭ Ϫ1

Given that ϭ 1 ϩ 2 2 is a two-parameter family of
solutions of
Љ Ϫ Ј ϭ 0 on the interval (Ϫϱ, ),
show that constants 1 and 2 cannot be found so that a
member of the family satisfies the initial conditions
(0) ϭ 0, Ј(0) ϭ 1
...
1
...

Find two members of the family of solutions in
Problem 5 that satisfy the initial conditions (0) ϭ 0,
Ј(0) ϭ 0
...


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...
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...
Editorial review has
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...


●

Use the general solution of Љ ϩ v 2 ϭ 0 given in
Problem 7 to show that a solution satisfying the initial
conditions ( 0) ϭ 0 , Ј( 0) ϭ 1 is the solution given in
Problem 7 shifted by an amount 0:
()ϭ

sin ␻ ( Ϫ 0 )
...

0

cos ␻ ( Ϫ

( Ϫ 2) Љ ϩ 3 ϭ ,
Љ ϩ (tan ) ϭ

0)

1

ϩ

(0) ϭ 0,

,

(0) ϭ 1,

In Problems 13 and 14 the given two-parameter family is a
solution of the indicated differential equation on the interval
(Ϫϱ, )
...


ЉϪ2 Јϩ2 ϭ0

)ϭ

,

2(

2(

)ϭ

)ϭ ,
Ϫ

3(

,

3(

2

)ϭ

) ϭ sinh

In Problems 23 – 30 verify that the given functions form a
fundamental set of solutions of the differential equation on
the indicated interval
...

Љ Ϫ Ј Ϫ 12 ϭ 0;

Ϫ3

,

4

, (Ϫϱ, )

Љ Ϫ 4 ϭ 0; cosh 2 , sinh 2 , (Ϫϱ, )
Љ Ϫ 2 Ј ϩ 5 ϭ 0;
4 Љ Ϫ 4 Ј ϩ ϭ 0;

cos 2 ,
/2

/2

,
3

sin 2 , (Ϫϱ, )

, (Ϫϱ, )

2

Љ Ϫ 6 Ј ϩ 12 ϭ 0;

2

Љϩ

Ј ϩ ϭ 0; cos(ln ), sin(ln ), (0, )

3

ٞϩ6

2

(4)

ϩ Љ ϭ 0; 1, , cos , sin , (Ϫϱ, )

,

4

, (0, )

Љ ϩ 4 Ј Ϫ 4 ϭ 0;

,

Ϫ2

,

Ϫ2

ln , (0, )

In Problems 31 – 34 verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval
...


1

)ϭ1ϩ ,

Ј(0) ϭ 1

Use the family in Problem 1 to find a solution of
Љ Ϫ ϭ 0 that satisfies the boundary conditions
(0) ϭ 0, (1) ϭ 1
...
Use this
family to find a solution that satisfies the boundary
conditions in part (a)
...


Use part (a) to find particular solutions o
ЉϪ6 Јϩ5 ϭ5
In Problems 15 – 22 determine whether the given set of functions is linearly independent on the interval (Ϫϱ, )
...


By inspection find a particular solution of
Љ ϩ 2 ϭ 10
...

Find a particular solution of Љ ϩ 2 ϭ 8 ϩ 5
...
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...
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...
1
...
Discuss whether one, both, or neither of
the linear combinations is a general solution of the
differential equation on the interval (Ϫϱ, )
...
Discuss how the observations
ϭ 0 and
ϭ ! can be used to find the general solutions of the given differential equations
...


Suppose that 1 ϭ and 2 ϭ Ϫ are two solutions of
a homogeneous linear differential equation
...


Suppose 1, 2,
...
By
Theorem 4
...
2 it follows that ϩ1 ϭ 0 is also a solution
of the differential equation
...
,
pendent on (Ϫϱ, )? Discuss
...

Show that ( 1, 2) ϭ 0 for every real number
Does this result violate Theorem 4
...
3? Explain
...

Find a solution of the differential equation satisfying (0) ϭ 0, Ј(0) ϭ 0
...
, are nontrivial solutions of
a homogeneous linear th-order differential equation
with constant coefficients and that ϭ ϩ 1
...
, linearly dependent or linearly
independent on (Ϫϱ, )? Discuss
...
5 (using a substitution)
Section 4
...
This method,
which is a straightforward exercise in algebra, breaks down in a few cases and yields only a single
solution 1 of the DE
...
The basic idea described in this section is that
involving the known solution 1
...


Suppose that 1 denotes a nontrivial solution of (1) and
that 1 is defined on an interval We seek a second solution 2 so that the set consisting of 1 and 2 is linearly independent on Recall from Section 4
...
The function ( ) can be found by substituting
2 ( ) ϭ ( ) 1( ) into the given differential equation
...
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...
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...

If ϭ ( ) 1( ) ϭ ( ) , then the Product Rule gives
Јϭ

ϩ2

ϩ

Ј,

Љϭ

ЉϪ

ϭ

( Љ ϩ 2 Ј) ϭ 0
...
If we make the substitution
Since
ϭ Ј, this linear second-order equation in becomes Ј ϩ 2 ϭ 0, which is a
linear first-order equation in
Using the integrating factor 2 , we can write
[

2

] ϭ 0
...
Integrating

Thus

2
...


2

(2)

By picking 2 ϭ 0 and 1 ϭ Ϫ2, we obtain the desired second solution, 2 ϭ Ϫ
...

Since we have shown that 1 ϭ and 2 ϭ Ϫ are linearly independent solutions of a linear second-order equation, the expression in (2) is actually the general
solution of Љ Ϫ ϭ 0 on (Ϫϱ, )
...
If
1( ) is a known solution of (3) on and that 1( )
we define ϭ ( ) 1( ), it follows that
Јϭ
Љϩ

Јϩ
1

Ј,

1

ϭ [ 1ϩ
Љ

Јϩ

Ј
1

Љϭ
1]

ϩ

Љ
1
ϩ

ϩ2 Ј Јϩ
1

1

Љ

Љ ϩ (2 1 ϩ
Ј

1)

Ј ϭ 0
...
Observe that the last equation in (4) is both linear and
separable
...


ϭ Ј, and integrate again:
Ϫ͐
2
1

ϩ

2
...
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...
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...


)

(5)

It makes a good review of differentiation to verify that the function 2( ) defined in
(5) satisfies equation (3) and that 1 and 2 are linearly independent on any interval
on which 1( ) is not zero
...
Find the general solution of the differential equation on the interval (0, )
...


) is given by

ϭ

2 2;

that is,

( ) The derivation and use of formula (5) have been illustrated here because this
formula appears again in the next section and in Sections 4
...
3
...
Your instructor will tell you
whether you should memorize (5) or whether you should know the first principles of reduction of order
...
See Problems 17 – 20 in
Exercises 4
...


In Problems 1 –16 the indicated function 1( ) is a solution
of the given differential equation
...

Љ Ϫ 4 Ј ϩ 4 ϭ 0;
Љ ϩ 2 Ј ϩ ϭ 0;
Љ ϩ 16 ϭ 0;

1

Љ ϩ 9 ϭ 0;
Љ Ϫ ϭ 0;
Љ Ϫ 25 ϭ 0;

1
1

1
1

ϭ

ϭ

9 Љ Ϫ 12 Ј ϩ 4 ϭ 0;
6 Љ ϩ Ј Ϫ ϭ 0;

Љ Ϫ 7 Ј ϩ 16 ϭ 0;

2

Љ ϩ 2 Ј Ϫ 6 ϭ 0;
Љ ϩ Ј ϭ 0;

ϭ cos 4

ϭ sin 3

2

4

Љ ϩ ϭ 0;

1

1

ϭ

ϭ cosh

ЉϪ

5

2

Љ Ϫ 3 Ј ϩ 5 ϭ 0;

ϭ

1
1

4

ϭ

ϭ

2

ϭ ln

2

1

/3

ϭ

2

2
Ϫ

1

2 /3

ϭ

1

Ј ϩ 2 ϭ 0;

1/2
1

ln
ϭ sin(ln )

1

ϭ

2

cos(ln )

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

(1 Ϫ 2 Ϫ
(1 Ϫ

2

) Љ ϩ 2(1 ϩ ) Ј Ϫ 2 ϭ 0;

2

) Љ ϩ 2 Ј ϭ 0;

1

1

2
1
or of the form 2 ϭ
, 1 and 2
2 ϭ
constants
...
Can you explain why the
statements in parts (a) and (b) above are not
contradicted by the answers to Problems 3 – 5?

ϭ ϩ1

ϭ1

In Problems 17 –20 the indicated function 1( ) is a solution
of the associated homogeneous equation
...


Verify that 1( ) ϭ is a solution of Љ Ϫ Ј ϩ ϭ 0
...
Conjecture an interval of
definition for 2( )
...

Use (5) to find a second solution 2( )
...

Explain, using Corollary (A) of Theorem 4
...
2, why
the second solution can be written compactly as

Give a convincing demonstration that the secondorder equation Љ ϩ Ј ϩ ϭ 0,
and constants, always possesses at least one solution of the
form 1 ϭ 1 , 1 a constant
...
1 and Theorem 4
...
5
Review the algebra of solving polynomial equations (see the

1
!


...
This type of equation can be solved either by
separation of variables or with the aid of an integrating factor, but there is another solution method,
one that uses only algebra
...
This observation reveals the
nature of the unknown solution ; the only nontrivial elementary function whose derivative is a

...


Since
is never zero for real values of , the last equation is satisfied only when is a solution or
root of the first-degree polynomial equation
ϩ ϭ 0
...
To illustrate, consider the constant-coefficient equation 2 Ј ϩ 5 ϭ 0
...

In this section we will see that the foregoing procedure can produce exponential solutions for
homogeneous linear higher-order DEs,
( )

ϩ

Ϫ1

( Ϫ1)

where the coefficients , ϭ 0, 1,
...


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...
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...
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...


●

We begin by considering the special case of the secondorder equation
Љϩ

(2)

ϭ 0,

Јϩ

where , , and are constants
...


ϩ

As in the introduction we argue that because
0 for all , it is apparent that the
only way ϭ
can satisfy the differential equation (2) is when is chosen as a
root of the quadratic equation

This last equation is called the
of the differential equation (2)
...


ϩ

real and distinct ( 2 Ϫ 4 Ͼ 0),
2
Ϫ 4 ϭ 0), and
2 real and equal (
conjugate complex numbers ( 2 Ϫ 4
2
2

Ͻ 0)
...

Under the assumption that the auxiliary equation
(3) has two unequal real roots 1 and 2, we find two solutions, 1 ϭ 1 and 2 ϭ 2
...
It follows that the general solution of (2) on this interval is
ϭ

ϩ

1

1

2

(4)


...
From the quadratic formula we find that
2
Ϫ 4 ϭ 0
...
2 that a second solution of the equation is
2

ϭ

1

͵

2
2

1

͵

1

ϭ

In (5) we have used the fact that Ϫ ͞ ϭ 2
ϭ

1

1

ϩ

1

1
...


The general solution is then
1

2

(6)


...

Formally, there is no difference between this case and Case I, and hence
ϭ

1

(aϩ ␤)

ϩ

2

(aϪ ␤)


...
To this end we use
␪

ϭ cos ␪ ϩ sin ␪,

*

where u is any real number
...
, and then separating the series into real and imaginary
parts
...

2

ϭ

3

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...
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...
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...


●

where we have used cos(Ϫb ) ϭ cos b and sin(Ϫb ) ϭ Ϫsin b Note that by firs
adding and then subtracting the two equations in (7), we obtain, respectively,
␤

ϭ 2 cos ␤

Ϫ␤

ϩ

Since ϭ 1 (aϩ b) ϩ 2
and 2, the choices 1 ϭ
1

(aϩ ␤)

ϭ

and

␤

ϭ 2 sin ␤
...


cos ␤

a
a

sin ␤
...
1
...
Moreover, these solutions form a fundamental set on (Ϫϱ, )
...


(8)

Solve the following differential equations
...

2

2

Ϫ 5 Ϫ 3 ϭ (2 ϩ 1)( Ϫ 3) ϭ 0,

From (4), ϭ
2

Ϫ 10

ϩ

3

2

ϭ Ϫ2 ϩ 23 ,

ϩ 25 ϭ ( Ϫ 5) 2 ϭ 0,

2

y

ϩ4

1

5

ϩ

2

ϩ 7 ϭ 0,

5

1

ϭ Ϫ1,
2

2

ϭ3


...


Ϫ2

ϭ

ϭ Ϫ2 Ϫ 23

2

ϭ5

( 1 cos 23
2

ϩ

2

sin 23
...

By the quadratic formula we find that the roots of the auxiliary
equation 4 2 ϩ 4 ϩ 17 ϭ 0 are 1 ϭ Ϫ1 ϩ 2 and 2 ϭ Ϫ1 Ϫ 2
...
Applying the condition (0) ϭ Ϫ1,
0
we see from
( 1 cos 0 ϩ 2 sin 0) ϭ Ϫ1 that
Differentiating
1 ϭ Ϫ1
...

3
Ϫ /2
Hence the solution of the IVP is ϭ
(Ϫcos 2 ϩ 4 sin 2 )
...
3
...

The two differential equations
Љϩ

2

ϭ0

and

ЉϪ

2

ϭ 0,

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...
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...
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...


●

where is real, are important in applied mathematics
...


(9)

On the other hand, the auxiliary equation 2 Ϫ 2 ϭ 0 for Љ Ϫ 2 ϭ 0 has distinct
real roots 1 ϭ and 2 ϭ Ϫ , and so by (4) the general solution of the DE is
ϭ

ϩ

1

2


...
Since
2
2
cosh and sinh are linearly independent on any interval of the -axis, an alternative form for the general solution of Љ Ϫ 2 ϭ 0 is
1 cosh

ϭ

2 sinh

ϩ

(11)


...
3
...
, are real constants, we must solve an th-degree
polynomial equation
ϩ

Ϫ1

Ϫ1

ϩиииϩ

2

2

ϩ

1

ϩ

0

ϭ 0
...


ϩиииϩ

2

2

It is somewhat harder to summarize the analogues of Cases II and III because the
roots of an auxiliary equation of degree greater than two can occur in many combinations
...
When 1 is a
root of multiplicity of an th-degree auxiliary equation (that is, roots are equal
to 1), it can be shown that the linearly independent solutions are
1

,

1

2

,

,
...


Finally, it should be remembered that when the coefficients are real, complex
roots of an auxiliary equation always appear in conjugate pairs
...


Solve ٞ ϩ 3 Љ Ϫ 4 ϭ 0
...
By division we fin
3

ϩ3

2

Ϫ4ϭ(

so the other roots are
ϭ 1 ϩ 2 Ϫ2 ϩ 3

2ϭ
Ϫ2

Ϫ 1)(
3

2

ϩ4

ϩ 4) ϭ (

Ϫ 1)(

ϩ 2)2,

ϭ Ϫ2
...
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...
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...
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...


ϩ

The auxiliary equation 4 ϩ 2 2 ϩ 1 ϭ ( 2 ϩ 1) 2 ϭ 0 has roots
ϭ and 2 ϭ 4 ϭ Ϫ Thus from Case II the solution is
3
ϭ

1

ϩ

2

By Euler’s formula the grouping

Ϫ

ϩ

ϩ

1
1 cos

2

ϩ

ϩ

3

Ϫ
2

1 cos

2 sin

ϩ

ϩ


...
Similarly, ( 3 ϩ
( 3 cos ϩ 4 sin )
...


Example 4 illustrates a special case when the auxiliary equation has repeated
complex roots
...
,

Ϫ1 (aϩ ␤)

,

(aϪ ␤)

,

(aϪ ␤)

,

2 (aϪ ␤)

,


...
,

Ϫ1 a

cos ␤ ,

sin ␤ ,

2 a

sin ␤ ,


...


In Example 4 we identify ϭ 2, a ϭ 0, and b ϭ 1
...
For example,
to solve 3 ٞ ϩ 5 Љ ϩ 10 Ј Ϫ 4 ϭ 0, we must solve 3 3 ϩ 5 2 ϩ 10 Ϫ 4 ϭ 0
...
Recall that
if 1 ϭ ͞ is a rational root (expressed in lowest terms) of an auxiliary equation
ϩ и и и ϩ 1 ϩ 0 ϭ 0 with integer coefficients, then is a factor of 0 and is
a factor of
...
Each of these numbers can then be tested —say, by
3
3
3
synthetic division
...
Therefore the general solution of 3 ٞ ϩ 5 Љ ϩ 10 Ј Ϫ 4 ϭ 0 is
ϭ 1 /3 ϩ Ϫ ( 2 cos 23 ϩ 3 sin 23 )
...


Finding roots or approximation of roots of auxiliary equations is a routine problem with an appropriate calculator or computer software
...
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...
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...

In the classic text
by Ralph Palmer Agnew* (used by the
author as a student) the following statement is made:

4

4
...
179

ϩ 1
...
295

2

ϩ 3
...


(13)

Although it is debatable whether computing skills have improved in the intervening
years, it is a certainty that technology has
...
After simplification and some
relabeling of output,
yields the (approximate) general solution
ϭ

1

ϩ

Ϫ0
...
476478

cos(0
...
728852

2

cos(0
...
618605 )

0
...
759081 )
...
Using a CAS to solve the system can save lots of time
...
3 and Problem 41 in Chapter 4 in Review
...


In Problems 1 –14 find the general solution of the given
second-order differential equation
...

ٞϪ4 ЉϪ5 Јϭ0

2

ϩ 24

Ϫ2

4

5

2

3

5

4

ϩ 12

Ϫ 10

2

3

4

Ϫ7

2
3

2
3

ϩ8

ϩ

2

ϩ5 ϭ0

ϭ0

ٞϪ ϭ0
ٞϪ5 Љϩ3 Јϩ9 ϭ0

In Problems 29 – 36 solve the given initial-value problem
...
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...
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...

Љ Ϫ 10 Ј ϩ 25 ϭ 0,
Љ ϩ 4 ϭ 0,
Љϩ

ϭ 0,

(0) ϭ 1, (1) ϭ 0

(0) ϭ 0, (p) ϭ 0
Graph for Problem 46

Ј(0) ϭ 0, Ј(p>2) ϭ 0

Љ Ϫ 2 Ј ϩ 2 ϭ 0,

(0) ϭ 1, (p) ϭ 1

In Problems 41 and 42 solve the given problem first using
the form of the general solution given in (10)
...

Љ Ϫ 3 ϭ 0,
Љ Ϫ ϭ 0,

(0) ϭ 1, Ј(0) ϭ 5
(0) ϭ 1, Ј(1) ϭ 0

In Problems 43 –48 each figure represents the graph of a
particular solution of one of the following differential
equations:
ЉϪ3 ЈϪ4 ϭ0
Љϩ4 ϭ0
Љϩ2 Јϩ ϭ0
Љϩ ϭ0
Љϩ2 Јϩ2 ϭ0
ЉϪ3 Јϩ2 ϭ0
Match a solution curve with one of the differential equations
...


Graph for Problem 47

Graph for Problem 48

Graph for Problem 43

In Problems 49–58 find a homogeneous linear differential equation with constant coefficients whose general solution is given
...
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...
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...
What is the corre2
sponding homogeneous linear differential equation?
Discuss: Is your answer unique?

In Problems 65– 68 use a computer either as an aid in solving
the auxiliary equation or as a means of directly obtaining the
general solution of the given differential equation
...


Find the general solution of 2 ٞ ϩ 7 Љ ϩ 4 Ј Ϫ 4 ϭ 0
if 1 ϭ 1 is one root of its auxiliary equation
...


6
...
59 Љ ϩ 7
...
778 ϭ 0
3
...
This is a trivial problem using a CAS
but can also be done by hand working with complex
numbers
...

How does this help? Solve the differential equation
...
34 Љ ϩ 6
...
03 ϭ 0

ϩ2 ЉϪ Јϩ2 ϭ0

In Problems 69 and 70 use a CAS as an aid in solving the auxiliary equation
...
Then use a CAS as an aid in solving the system of
equations for the coefficients , ϭ 1, 2, 3, 4 that results when
the initial conditions are applied to the general solution
...
Reconcile this particular solution with the general solution of the DE
...
Discuss: Is it possible to
determine values of l so that the problem possesses
trivial solutions?
nontrivial solutions?

●

(4)

(4)

Ϫ 3 ٞ ϩ 3 Љ Ϫ Ј ϭ 0,
(0) ϭ Ј(0) ϭ 0, Љ(0) ϭ ٞ(0) ϭ 1

Review Theorems 4
...
6 and 4
...
7 (Section 4
...

Then, as was discussed in Section 4
...
The complementary function is the general solution of the associated homogeneous DE of (1), that is,
( )

ϩ

Ϫ1

( Ϫ1)

ϩиииϩ

1

Јϩ

0

ϭ 0
...
3 we saw how to solve these kinds of equations when the coefficients were constants
...


*

In this section the method of undetermined coefficients is developed from th
viewpoint of the superposition principle for nonhomogeneous equations (Theorem 4
...
1)
...
5
an entirely different approach will be presented, one utilizing the concept of differential annihilator
operators
...


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...
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...
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...


●

The first of two ways we shall
consider for obtaining a particular solution for a nonhomogeneous linear DE is
called the
The underlying idea behind
this method is a conjecture about the form of , an educated guess really, that is
motivated by the kinds of functions that make up the input function ( )
...
, are constants and
• ( ) is a constant , a polynomial function, an exponential function a ,
a sine or cosine function sin b or cos b , or finite sums and product
of these functions
...
Since a
constant function is probably not the first thing that comes to mind when you think
of polynomial functions, for emphasis we shall continue to use the redundancy
“constant functions, polynomials,
...


That is, ( ) is a linear combination of functions of the type
( )ϭ

ϩ

Ϫ1

Ϫ1

ϩиииϩ

1

ϩ

0,

( )

a

,

( )

a

sin ␤ ,

and

( )

a

cos ␤ ,

where is a nonnegative integer and a and b are real numbers
...
Differential equations in which the input ( ) is a function of this last kind
will be considered in Section 4
...

The set of functions that consists of constants, polynomials, exponentials
a
, sines, and cosines has the remarkable property that derivatives of their sums
and products are again sums and products of constants, polynomials, exponentials a , sines, and cosines
...

The next two examples illustrate the basic method
...
From the quadratic formula we find that the roots of the auxiliary equation 2 ϩ 4 Ϫ 2 ϭ 0 are 1 ϭ Ϫ2 Ϫ 16 and 2 ϭ Ϫ2 ϩ 16
...


ϭ

1

(

Ϫ 2ϩ16

) ϩ

2

(Ϫ2ϩ16 )
...


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...
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...


●

We seek to determine
coefficients , , and
of (2)
...


Because the last equation is supposed to be an identity, the coefficients of like powers
of must be equal:
equal

2

Ϫ2
That is,

ϩ 8 Ϫ2

Ϫ2 ϭ 2,

2 ϩ4 Ϫ2

ϩ

8 Ϫ 2 ϭ Ϫ3,

2

Ϫ

2

Ϫ3 ϩ6

2 ϩ 4 Ϫ 2 ϭ 6
...


5
Ϫ 9
...

2

Find a particular solution of Љ Ϫ Ј ϩ ϭ 2 sin 3
A natural first guess for a particular solution would be sin 3 But
because successive differentiations of sin 3 produce sin 3
cos 3 , we are
prompted instead to assume a particular solution that includes both of these terms:
ϭ

cos 3 ϩ

sin 3
...


From the resulting system of equations,
Ϫ8 Ϫ 3 ϭ 0,
we get

ϭ

6
73

and

ϭ

Ϫ16
...

73
73

As we mentioned, the form that we assume for the particular solution is an
educated guess; it is not a blind guess
...


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...
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●

2

Solve Љ Ϫ 2 Ј Ϫ 3 ϭ 4 Ϫ 5 ϩ 6

(3)


...

Next, the presence of 4 Ϫ 5 in ( ) suggests that the particular solution
includes a linear polynomial
...
In other words, is the sum of two basic kinds of functions:
( )ϭ

1(

)ϩ

2(

)ϭ


...
1
...
Substituting

ϩ

2

2

ϩ

into the given equation (3) and grouping like terms gives
ЉϪ 2 ЈϪ 3

Ϫ2 Ϫ3 Ϫ3

ϭ Ϫ3

2

ϩ (2 Ϫ 3 )

2

ϭ4 Ϫ5ϩ6

2


...


The last equation in this system results from the interpretation that the coefficient of
2
in the right member of (4) is zero
...
Consequently,
3
ϭϪ

4
23
ϩ
Ϫ2
3
9

2

Ϫ

4
3

2


...


In light of the superposition principle (Theorem 4
...
7) we can also approach
Example 3 from the viewpoint of solving two simpler problems
...
A particular solution of (3)
3
9
3
is then ϭ 1 ϩ 2
...


Find a particular solution of Љ Ϫ 5 Ј ϩ 4 ϭ 8
...
Therefore proceeding
as we did in the earlier examples, we can reasonably assume a particular solution of
the form ϭ

...
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...

The difficulty here is apparent on examining the complementary function
ϭ 1 ϩ 2 4
...
This
means that is a solution of the associated homogeneous differential equation, and
a constant multiple
when substituted into the differential equation necessarily
produces zero
...
3, let’s see
whether we can find a particular solution of the for

...


From the last equality we see that the value of is now determined as
Therefore a particular solution of the given equation is ϭ Ϫ8
...

3

The difference in the procedures used in Examples 1 – 3 and in Example 4
suggests that we consider two cases
...

No function in the assumed particular solution is a solution of the associated homogeneous differential equation
...
4
...
We are, of course, taking for granted
that no function in the assumed particular solution is duplicated by a function in
the complementary function
...
Using entry 9 in Table 4
...
1 as
a model, we assume a particular solution of the form
ϭ(

3

ϩ

2

ϩ

Note that there is no duplication between the terms in
mentary function ϭ 4 ( 1 cos 3 ϩ 2 sin 3 )
...


and the terms in the comple-

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...
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●

The function ( ) ϭ cos is similar to entry 11 in Table 4
...
1 except, of course,
that we use a linear rather than a quadratic polynomial and cos and sin instead of
cos 4 and sin 4 in the form of :
ϭ(

ϩ ) cos ϩ (

ϩ ) sin
...
,
corresponding to these terms:
ϭ

1

ϩ

2


...

( )

Determine the form of a particular solution of
Љ Ϫ 9 Ј ϩ 14 ϭ 3

Corresponding to 3

2

2

we assume

1

Corresponding to Ϫ 5 sin 2 we assume
Corresponding to 7

6

6

Ϫ 5 sin 2 ϩ 7

2

we assume

3

2

ϭ


...


ϩ

cos 2 ϩ

ϭ
ϭ(

6

)

ϩ

sin 2
...


The assumption for the particular solution is then
ϭ

1

ϩ

2

ϩ

3

ϭ

2

ϩ

ϩ

cos 2 ϩ

ϩ

No term in this assumption duplicates a term in

ϭ

1

sin 2 ϩ (
2

ϩ

2

7

ϩ

)

6


...


A function in the assumed particular solution is also a solution of the
associated homogeneous differential equation
...


Find a particular solution of Љ Ϫ 2 Ј ϩ ϭ


...
As in Example 4,
the assumption ϭ
will fail, since it is apparent from that is a solution of
the associated homogeneous equation Љ Ϫ 2 Ј ϩ ϭ 0
...
We next try
ϭ

2


...

2

ϭ

, so

ϭ 1
...
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...
4
...
, are the trial particular solution forms corresponding to
these terms
...

,

,

Solve Љ ϩ ϭ 4 ϩ 10 sin , (p) ϭ 0, Ј(p) ϭ 2
...
4
...


(5)

But there is an obvious duplication of the terms cos and sin in this assumed form
and two terms in the complementary function
...


(6)

Differentiating this expression and substituting the results into the differential
equation gives
Љϩ

ϭ

Ϫ 2 sin ϩ 2 cos ϭ 4 ϩ 10 sin ,

ϩ

and so ϭ 4, ϭ 0, Ϫ2 ϭ 10, and 2 ϭ 0
...
Therefore from (6) we obtain
ϭ 4 Ϫ 5 cos The general solution of the given equation is
ϭ

ϩ

ϭ

1 cos

ϩ

2 sin

ϩ 4 Ϫ 5 cos
...
First, (p) ϭ 1 cos p ϩ 2 sin p ϩ 4p Ϫ 5p cos p ϭ 0 yields 1 ϭ 9p,
since cos p ϭ Ϫ1 and sin p ϭ 0
...
The solution of the initial-value is then
ϭ 9␲ cos ϩ 7 sin ϩ 4 Ϫ 5 cos
...


The complementary function is ϭ 1 3 ϩ 2 3
...
4
...


2

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...
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...


●

Inspection of these functions shows that the one term in 2 is duplicated in
...
But multiplying
by 2 eliminates all duplications
...


Differentiating this last form, substituting into the differential equation, and collecting
like terms gives
ЉϪ 6 Ј ϩ 9

ϭ9

2

3

ϩ (Ϫ12 ϩ 9 ) ϩ 2 Ϫ 6 ϩ 9 ϩ 2

2
It follows from this identity that ϭ 3 ,
solution ϭ ϩ is ϭ 1 3 ϩ 2

Solve ٞ ϩ Љ ϭ

ϭ6

2

ϩ 2 Ϫ 12

3


...
Hence the general
3
8
2
ϩ9 ϩ2Ϫ6 2 3
...
Hence the complementary function of the equation is
ϭ 1 ϩ 2 ϩ 3 Ϫ
...
4
...


Because there are no functions in that duplicate functions in the complementary
solution, we proceed in the usual manner
...
This system gives ϭ Ϫ10 and ϭ 1,
5
1
1
so a particular solution is ϭ Ϫ10 cos ϩ 5 sin
...


2 Ϫ


...

is ϭ 3 ϩ 3 Ϫ ϩ 2 Ϫ ϩ

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...
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...


●

( ) In Problems 27 –36 in Exercises 4
...
As illustrated in Example 8, be sure to apply the initial conditions or
the boundary conditions to the general solution ϭ ϩ
...

1, 2,
...
For example, if ( ) is a polynomial, then continued differentiation of ( ) a sin b will generate an independent set containing only a
number of functions —all of the same type,
namely, a polynomial times a sin b or a polynomial times a cos b On
the other hand, repeated differentiation of input functions such as ( ) ϭ ln
or ( ) ϭ tanϪ1 generates an independent set containing an
number of
functions:
ln :
tanϪ1 :

In Problems 1 – 26 solve the given differential equation by
undetermined coefficients
Љϩ3 Јϩ2 ϭ6

1 , Ϫ1 , 2 ,
...

(1 ϩ 2 ) 2 (1 ϩ 2 ) 3

Ϫ ϩ6

cos 2
2

(cos Ϫ 3 sin )

Љ ϩ 2 Ј ϩ ϭ sin ϩ 3 cos 2

Љ Ϫ 10 Ј ϩ 25 ϭ 30 ϩ 3

Љ ϩ 2 Ј Ϫ 24 ϭ 16 Ϫ ( ϩ 2)

Љϩ ЈϪ6 ϭ2

ٞ Ϫ 6 Љ ϭ 3 Ϫ cos

1
Љϩ Јϩ ϭ
4

2

Љ Ϫ 8 Ј ϩ 20 ϭ 100
Љ ϩ 3 ϭ Ϫ48

2

Ϫ 26

2 3

1
ЉϪ Јϩ
ϭ3ϩ
4

Ϫ2

2

/2

4

Ϫ 3) sin 2

Љ ϩ ϭ 2 sin

ϩ 2 Љ ϩ ϭ ( Ϫ 1) 2
Ϫ Љϭ4 ϩ2

2

ϩ

Ϫ

In Problems 27 – 36 solve the given initial-value problem
...
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...

Briefly discuss your reasoning

(0) ϭ 1,
2

(0) ϭ Ϫ5, Ј(0) ϭ 3,

In Problems 37 – 40 solve the given boundary-value problem
...
[
:
Solve each problem on two intervals, and then find a solution so that and Ј are continuous at ϭ p͞2 (Problem 41)
and at ϭ p (Problem 42)
...
The auxiliary
equation of the associated homogeneous equation is
2
ϩ
ϩ ϭ 0
...

If is a root of the auxiliary equation of multiplicity
one, show that we can find a particular solution of
the form
ϭ
, where
ϭ 1͞(2 ϩ )
...

If is a root of the auxiliary equation of multiplicity
two, show that we can find a particular solution of the
form ϭ 2 , where ϭ 1͞(2 )
...


Solution curve

In Problems 46 and 47 find a particular solution of the given
differential equation
...
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...
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...
1
...
1
...
1)
We saw in Section 4
...
, When it suits our purpose, (1) is also written as ( ) ϭ ( ),
denotes the linear th-order differential, or polynomial, operator
ϩ

Ϫ1

Ϫ1

ϩиииϩ

ϩ

1

(2)

0
...
In this
section there are no special rules; the form of follows almost automatically once we have found
an appropriate linear differential operator that
( ) in (1)
...


When the coefficients , ϭ 0, 1,
...
In other words, if 1 is a
root of the auxiliary equation
ϩ

Ϫ1

Ϫ1

ϩиииϩ

ϩ

1

0

ϭ 0,

then ϭ ( Ϫ 1) ( ), where the polynomial expression ( ) is a linear differential
operator of order Ϫ 1
...
Thus
if a function ϭ ( ) possesses a second derivative, then
(

2

ϩ5

ϩ 6) ϭ (

ϩ 2)(

ϩ 3) ϭ (

ϩ 3)(

ϩ 2)
...


If is a linear differential operator with constant coefficients and is a sufficiently di ferentiable function such that
( ( )) ϭ 0,
then is said to be an
of the function
...
The function ϭ is annihilated by the
differential operator 2 since the first and second derivatives of are 1 and 0,
respectively
...

The differential operator
1,

annihilates each of the functions
2

,


...


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...
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...


(3)

●

As an immediate consequence of (3) and the fact that differentiation can be done
term by term, a polynomial
0

ϩ

ϩ

1

2

2

ϩиииϩ

(4)

Ϫ1

Ϫ1

can be annihilated by finding an operator that annihilates the highest power of
The functions that are annihilated by a linear th-order differential operator
are simply those functions that can be obtained from the general solution of the
homogeneous differential equation ( ) ϭ 0
...
,

To see this, note that the auxiliary equation of the homogeneous equation
the general
( Ϫ a) ϭ 0 is ( Ϫ a) ϭ 0
...


Find a differential operator that annihilates the given function
...


From (5), with a ϭ Ϫ3 and ϭ 1, we see that
(

ϩ 3)

Ϫ3

ϭ 0
...


When a and b, b Ͼ 0 are real numbers, the quadratic formula reveals that
[ 2 Ϫ 2a ϩ (a 2 ϩ b 2)] ϭ 0 has complex roots a ϩ b, a Ϫ b, both of multiplicity From the discussion at the end of Section 4
...

The differential operator [
functions
␣
␣

cos ␤ ,
sin ␤ ,

␣
␣

2

cos ␤ ,
sin ␤ ,

Ϫ 2a ϩ (a 2 ϩ b 2)] annihilates each of the
2 ␣
2 ␣

cos ␤ ,
...
,

Find a differential operator that annihilates 5

Ϫ

Ϫ1 ␣
Ϫ1 ␣

cos 2 Ϫ 9

Ϫ

cos ␤ ,
sin ␤
...
Hence from (7) we conclude that 2 ϩ 2 ϩ 5 will annihilate
each function
...
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...
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...


(8)

For example, 2 ϩ 16 will annihilate any linear combination of sin 4 and cos 4
We are often interested in annihilating the sum of two or more functions
...
This is a direct consequence of Theorem 4
...
2
...
Then the
of differential operators 1 2 annihilates the sum 1 1( ) ϩ 2 2 ( )
...


ϭ
ϭ
ϭ

zero

zero

For example, we know from (3) that 2 annihilates 7 Ϫ and from (8) that
2
ϩ 16 annihilates sin 4 Therefore the product of operators 2 ( 2 ϩ 16) will
annihilate the linear combination 7 Ϫ ϩ 6 sin 4
The differential operator that annihilates a function is not unique
...
For example, ( ϩ 3)( ϩ 1), ( ϩ 3)2, and 3( ϩ 3) all annihilate Ϫ3
...
) As a matter of course, when we seek a differential annihilator for a
function ϭ ( ), we want the operator of
that does the job
...
Suppose that ( ) ϭ ( ) is a linear differential equation with constant
coefficients and that the input ( ) consists of finite sums and products of the functions listed in (3), (5), and (7) —that is, ( ) is a linear combination of functions of
the form
(constant),

,

␣

,

␣

cos ␤ , and

␣

sin ␤ ,

where is a nonnegative integer and a and b are real numbers
...
Applying 1 to both sides of the equation ( ) ϭ ( )
yields 1 ( ) ϭ 1( ( )) ϭ 0
...
We then substitute this assumed form into
( ) ϭ ( ) to find an explicit particular solution
...

Before proceeding, recall that the general solution of a nonhomogeneous
linear differential equation ( ) ϭ ( ) is ϭ ϩ , where
is the complementary function —that is, the general solution of the associated homogeneous
equation ( ) ϭ 0
...


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...
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...


●

Solve Љ ϩ 3 Ј ϩ 2 ϭ 4 2
...

Then, from the auxiliary equation 2 ϩ 3 ϩ 2 ϭ ( ϩ 1)( ϩ 2) ϭ 0 we fin
1 ϭ Ϫ1 and 2 ϭ Ϫ2, and so the complementary function is
ϭ
3

(

2

Ϫ

1

ϩ

Ϫ2

2

Now, since 4 2 is annihilated by the differential operator
ϩ 3 ϩ 2) ϭ 4 3 2 is the same as
3

2

(

3

, we see that

ϩ 3 ϩ 2) ϭ 0
...
Thus its general solution

Ϫ

ϩ

5

Ϫ2


...
We can then argue that a particular solution of (9) should
also satisfy equation (10)
...

For (12) to be a particular solution of (9), it is necessary to find
coefficient
, , and Differentiating (12), we have
ϩ2

,

Љϭ2 ,

ϭ2 ϩ3 ϩ6

ϩ2 ϩ2

Јϭ
and substitution into (9) then gives
Љϩ3 Јϩ2

ϩ2

2

ϭ 4 2
...


ϭ4

1

Ϫ

ϩ

2

Ϫ2

ϩ 0 ϩ 0
...

ϭ 7,

ϭ Ϫ6, and

The general solution of the equation in (9) is ϭ
ϭ

2

ϩ

(13)
ϭ 2
...


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...
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...
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...


●

3

Solve Љ Ϫ 3 Ј ϭ 8

(14)

ϩ 4 sin
...


2

tion Љ Ϫ 3 Ј ϭ 0 is

Now, since ( Ϫ 3) 3 ϭ 0 and ( 2 ϩ 1) sin ϭ 0, we apply the differential operator ( Ϫ 3)( 2 ϩ 1) to both sides of (14):
(

2

Ϫ 3)(

ϩ 1)(

2

(15)

Ϫ 3 ) ϭ 0
...



...


in (14) and simplifying yield

ЉϪ3 Јϭ3

3

ϩ (Ϫ Ϫ 3 ) cos ϩ (3 Ϫ

) sin ϭ 8

Equating coefficients gives 3 ϭ 8, Ϫ Ϫ 3 ϭ 0, and 3 Ϫ
ϭ 6, and ϭ Ϫ2, and consequently,
5
5
8
3

ϭ

3

ϩ

3

ϩ 4 sin
...
We find

ϭ 8,
3

6
2
cos Ϫ sin
...

5
5

ϭ cos Ϫ cos
...
Applying this operator to the differential equation gives
(

2

ϩ 1)2 (

2

ϩ 1) ϭ 0

or

(

2

ϩ 1)3 ϭ 0
...


We substitute
ϭ

cos ϩ

sin ϩ

2

cos ϩ

2

sin

into (16) and simplify:
Љϩ

ϭ4
cos Ϫ 4
ϭ cos Ϫ cos
...
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...
Hence the
4
2
4
general solution of (16) is
1 cos

ϭ

2 sin

ϩ

ϩ

1
1
1
cos Ϫ
sin ϩ
4
2
4

2

sin
...


Ϫ2

The complementary function for the given
ϭ 1 ϩ 2
...


Ϫ2

ϩ 4 ϩ 5 to (17) gives
2

(

2

equation

ϩ4

2

ϩ 5)(

Ϫ2

(18)

ϩ 1) ϭ 0
...


Ϫ2

Determine the form of a particular solution for
2

ЉЈ Ϫ 4 Љ ϩ 4 Ј ϭ 5

2 2

Ϫ6 ϩ4

ϩ3

5


...


3

( Ϫ 2) 3( Ϫ 5) applied to (19) gives
3

(

Ϫ 2)3(

3

Ϫ 5)(
4

(

or

2

ϩ4 ) ϭ0

Ϫ 2)5(

Ϫ 5) ϭ 0
...
Hence
ϭ

1

ϩ

2

ϩ

3

2

ϩ

4

3

ϩ

5

2

ϩ

6

2

ϩ

7

2 2

ϩ

2

8

3 2

ϩ

9

4 2

ϩ

10

5

(20)

2

Because the linear combination 1 ϩ 5
corresponds to the complemenϩ 6
tary function of (19), the remaining terms in (20) give the form of a particular solution of the differential equation:
ϭ

ϩ

2

ϩ

3

ϩ

2 2

ϩ

3 2

ϩ

4 2

ϩ

5


...
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...

()

Find the complementary solution for the homogeneous equation
( ) ϭ 0
...

( ) Find the general solution of the higher-order homogeneous differential
equation 1 ( ) ϭ 0
...
Form a linear
combination of the terms that remain
...

( ) Substitute found in step ( ) into ( ) ϭ ( )
...

( ) With the particular solution found in step ( ), form the general solution
ϭ ϩ of the given differential equation
...
Differential equations in which the input ( ) is a function of this
last kind will be considered in the next section
...
If possible, factor
9 Љ Ϫ 4 ϭ sin

ЉϪ5 ϭ

Љ Ϫ 4 Ј Ϫ 12 ϭ Ϫ 6

ٞϩ4 Љϩ3 Јϭ

2

(4)

ٞϩ4 Јϭ

Ϫ 8 Љ ϩ 16 ϭ (

cos 2

Ϫ

cos Ϫ 3
3

Ϫ2 )

;

ϭ 10

3

Ϫ2

ϩ 64;

2

Ϫ 1;

ϭ4

/2

ϭ

2

ϩ3

Ϫ5

ϭ 2 cos 8 Ϫ 5 sin 8

In Problems 15 – 26 find a linear differential operator that
annihilates the given function
...

4

2

1ϩ7

ϩ8 Јϭ4

(4)

Ϫ2

2 ЉϪ3 ЈϪ2 ϭ1

ٞ ϩ 10 Љ ϩ 25 Ј ϭ
ٞ ϩ 2 Љ Ϫ 13 Ј ϩ 10 ϭ

2

( Ϫ 2)( ϩ 5);

ϩ2

2

Ϫ sin 4
Ϫ

cos 2

2

8 Ϫ sin ϩ 10 cos 5
(2 Ϫ
Ϫ

)2

sin Ϫ

2

cos

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...
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...


●

In Problems 27 – 34 find linearly independent functions that
are annihilated by the given differential operator
...


ٞ ϩ 2 Љ ϩ Ј ϭ 10

Љϩ4 Јϩ4 ϭ2 ϩ6
ٞϩ Љϭ8

ϩ9 ϩ2

ٞϪ3 Љϩ3 ЈϪ ϭ

In Problems 35 – 64 solve the given differential equation by
undetermined coefficients

Љϩ Јϭ3

2

ٞϪ Љϩ ЈϪ ϭ

( Ϫ 5)( Ϫ 7)

Љ Ϫ 9 ϭ 54

Љ ϩ 4 ϭ cos2

ٞ ϩ 8 Љ ϭ Ϫ6

2

ϩ 25

Љ ϩ ϭ 4 cos Ϫ sin

Љ ϩ Ј ϩ ϭ sin

ϩ4

2

Љ ϩ 25 ϭ 20 sin 5

Љϩ Јϭ ,
ϩ4

(0) ϭ 1, Ј(0) ϭ 0
(0) ϭ 1, Ј(0) ϭ 0

Љ Ϫ 5 Ј ϭ Ϫ 2,

(0) ϭ 0, Ј(0) ϭ 2

Љ ϩ 5 Ј Ϫ 6 ϭ 10

2

,

(0) ϭ 1, Ј(0) ϭ 1

Љ ϩ ϭ 8 cos 2 Ϫ 4 sin ,
ϩ 5,

ٞϪ2 Љϩ Јϭ
Љ(0) ϭ Ϫ1
ЉϪ4 Јϩ8 ϭ

3

(4)

,

Ϫ ٞϭ ϩ
ٞ(0) ϭ 0

,

(p>2) ϭ Ϫ1, Ј(p>2) ϭ 0
(0) ϭ 2, Ј(0) ϭ 2,

(0) ϭ 2, Ј(0) ϭ 4
(0) ϭ 0, Ј(0) ϭ 0, Љ(0) ϭ 0,

2 Ϫ

Љϩ2 Јϩ ϭ
ЉϪ2 Јϩ5 ϭ
1
ϭ
Љϩ Јϩ
4

●
●

sin
(sin 3 Ϫ cos 3 )

Suppose is a linear differential operator that factors
but has variable coefficients
...


Basic integration formulas and techniques from calculus
Review Section 2
...
4 and 4
...
4
...
In this section we examine a method for determining a particular solution of
a nonhomogeneous linear DE that has, in theory, no such restrictions on it
...

Before examining this powerful method for higher-order equations we revisit the solution of linear first-order differential equations that have been put into standard form
...
If pressed for time this motivational material could be assigned
for reading
...
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...
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...
3 we saw that the general solution of a linear first-order differential equation 1( ) Ј ϩ 0( ) ϭ ( ) can be
found by first rewriting it in the standard form
(1)

ϩ ( ) ϭ ( )

and assuming that ( ) and ( )are continuous on an common interval
...


The foregoing solution has the same form as that given in Theorem 4
...
6, namely,
ϭ ϩ
...
As a means of motivating a method for solving nonhomogeneous linear equations of higher-order we
propose to rederive the particular solution (3) by a method known as
Suppose that

1

is a known solution of the homogeneous equation (2), that is,
1

ϩ ( )

1

(4)

ϭ 0
...
Variation of parameters consists of finding a particular solution of (1) of the form ϭ 1( ) 1( )
...

Substituting ϭ 1 1 into (1) and using the Product Rule gives

[ 1 1] ϩ
1

1

ϩ

1

1

( )

1 1

ϭ ( )

ϩ ( )

1 1

ϭ ( )

0 because of (4)
4
1

1

ϩ ( )

so

΅ϩ

1
1

1

1

1

ϭ ( )
...



...


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...
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...
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...


●

Next we consider the case of a linear secondorder equation
2(

) Љϩ

1(

) Јϩ

0(

(5)

) ϭ ( ),

although, as we shall see, variation of parameters extends to higher-order equations
...
In (6) we suppose that coefficient functions ( ), ( ), and ( ) are continuous on some common interval
...
3, there is no difficulty in obtaining the complementary
solution ϭ 1 1( ) ϩ 2 2( ), the general solution of the associated homogeneous
equation of (6), when the coefficients are constants
...

Using the Product Rule to differentiate twice, we get
Јϭ

1 Ј
1

ϩ

1 Ј
1

ϩ

2 Ј
2

ϩ

2 Ј
2

Љϭ

Љ
1 1

ϩ 1 1ϩ
Ј Ј

Љ
1 1

ϩ 1 1ϩ
Ј Ј

Љϩ Ј Јϩ
2
2 2

2

2

Љ ϩ Ј Ј
...

1 1
2 2

(8)

Because we seek to determine two unknown functions 1 and 2, reason dictates that
we need two equations
...
This assumption does not
1
2
come out of the blue but is prompted by the first two terms in (8), since if we demand
that 1 Ј ϩ 2 Ј ϭ 0, then (8) reduces to Ј Ј ϩ Ј Ј ϭ ( )
...
By Cramer’s Rule, the solution of the system
2
1 Ј
1

ϩ

2 Ј
2

ϭ0

Ј Јϩ Ј Јϭ ( )
1 1
2 2
can be expressed in terms of determinants:
1

Јϭ
1
where

ϭ

͉

ϭϪ

1

2

Ј
1

Ј
2

͉

,

2

( )

1

ϭ

and

͉

0
( )

2

Јϭ
2
2

Ј
2

͉

,

2

ϭ

ϭ

͉

1

1

Ј
1

( )

,

(9)

͉

0

...
The determinant
is recognized as the Wronskian of 1 and 2
...


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...
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...
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...


●

Usually, it is not a good idea to memorize formulas in lieu of understanding a procedure
...

In this case it is more efficient to simply use the formulas in (9)
...
By dividing by 2, we put the
equation into the standard form Љ ϩ Ј ϩ
ϭ ( ) to determine ( )
...
A particular solution is ϭ 1 1 ϩ 2 2
...


2

Solve Љ Ϫ 4 Ј ϩ 4 ϭ ( ϩ 1)


...
With the identifications 1 ϭ
pute the Wronskian:
2

2

(

2

,

͉

)ϭ

2

ϩ 4 ϭ ( Ϫ 2) 2 ϭ 0 we have
and 2 ϭ 2 , we next com-

2

2

2

2

2

2

ϩ

2

͉

4

ϭ


...
From (10) we obtain
1

ϭ

͉(

0
ϩ 1)

2
2

2

2

2

ϩ

͉ ϭ Ϫ(

4

,

2

ϩ 1)

Ϫ ,

2

ϭ

͉2

2

0
( ϩ 1)

2

2

͉ϭ(

ϩ 1)

4

,

and so from (9)
ЈϭϪ
1
It follows that

1

ϭ Ϫ
and

ϭϪ

4

ϭ Ϫ1
3

΂

4

( ϩ 1)

1
3

ϭ

3

3

Ϫ1
2

Ϫ

2

and

ϩ

΃

2

ϭ

1
2

1

2

2

Јϭ
2

ϭ1
2

ϩ

΂1
2

2

2

ϩ

2

2

4

( ϩ 1)

ϭ

4

ϩ 1
...
Hence

ϩ

΃

2

ϩ

2

1
6

ϭ
3 2

1
6
ϩ

3 2

1
2

ϩ
2 2

1
2

2 2


...

4

Because the roots of the auxiliary equation 2 ϩ 9 ϭ 0 are 1 ϭ 3 and 2 ϭ Ϫ3 , the
complementary function is ϭ 1 cos 3 ϩ 2 sin 3 Using 1 ϭ cos 3 , 2 ϭ sin 3 ,
and ( ) ϭ 1 csc 3 , we obtain
4
(cos 3 , sin 3 ) ϭ

1

ϭ

͉

1
4

0
csc 3

sin 3
3 cos 3

͉

1
ϭϪ ,
4

͉

cos 3
Ϫ3 sin 3
2

ϭ

͉

sin 3
3 cos 3
cos 3
Ϫ3 sin 3

͉

ϭ 3,

1
4

0
csc 3

͉

ϭ

1 cos 3

...
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...
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...
Thus a particular solution is

2

ϭϪ

1
1
cos 3 ϩ
(sin 3 ) ln͉ sin 3 ͉
...

12
36

(11)

Equation (11) represents the general solution of the differential equation on, say,
the interval (0, p͞6)
...
This is because
2
ϭ

ϩ

ϭ

1 1

ϭ(
ϭ

Solve Љ Ϫ

1

1 1

ϩ

2 2

ϩ(

1) 1

ϩ

ϩ(

ϩ

2 2

2

ϩ

1) 1

ϩ

1

1) 2

ϩ

1 1

ϩ(
ϩ

2

1 1

ϩ

1) 2
2 2

2 2
...

The auxiliary equation 2 Ϫ 1 ϭ 0 yields
ϭ 1 ϩ 2 Ϫ
...


Ϫ

,
0

͵

1
2


...


(12)

0

In Example 3 we can integrate on any interval [ 0, ] that does not contain the origin
...
8
...


(13)

is the complementary function for (13), then a

) 1( ) ϩ

2(

) 2( ) ϩ и и и ϩ

( ) ( ),

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...
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...
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...


●

where the Ј, ϭ 1, 2,
...


(14)

The first Ϫ 1 equations in this system, like 1 Ј ϩ 2 Ј ϭ 0 in (8), are assumptions
1
2
that are made to simplify the resulting equation after
ϭ 1( ) 1( ) ϩ и и и ϩ
( ) ( ) is substituted in (13)
...
, ,

where is the Wronskian of 1, 2,
...
When ϭ 2,
we get (9)
...
6
...
The present method is not limited to a function ( ) that is a combination of the four
types listed on page 140
...

Depending on how the antiderivatives of Ј and Ј are found, you might not
1
2
obtain the same as given in the answer section
...
6 both ϭ 1 sin Ϫ 1 cos and ϭ 1 sin Ϫ 2 cos
2
2
4
are valid answers
...
Why?
2

In Problems 1 – 18 solve each differential equation by variation of parameters
...
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...
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...

3
Љ
3
3

●

Љϩ3 Јϩ2 ϭ
ЉϪ2 Јϩ

1
1ϩ

ϭ

Ϫ

ٞ ϩ Ј ϭ tan
ٞ ϩ 4 Ј ϭ sec 2

11 Ϫ

ln

ٞ Ϫ2 ЉϪ Јϩ2 ϭ

ϭ 41

3 ЉϪ6 Јϩ6 ϭ
4 ЉϪ4 Јϩ

ϭ

ٞ Ϫ3 Љϩ2 Јϭ
2

In Problems 19 – 22 solve each differential equation by
variation of parameters, subject to the initial conditions
(0) ϭ 1, Ј(0) ϭ 0
...


In Problems 23 and 24 the indicated functions are known linearly independent solutions of the associated homogeneous
differential equation on (0, )
...

2

1ϩ

In Problems 29 and 30 discuss how the methods of undetermined coefficients and variation of parameters can be
combined to solve the given differential equation
...


/2

4 ЉϪ ϭ

4

2

sec
/2

Љ ϩ Ј ϩ ϭ sec(ln );
ϭ cos(ln ), 2 ϭ sin(ln )

In Problems 25–28 solve the given third-order differential
equation by variation of parameters
...


sin

;

Review the concept of the auxiliary equation in Section 4
...


The same relative ease with which we were able to find explicit solutions of
higher-order linear differential equations with constant coefficients in the preceding sections does
not, in general, carry over to linear equations with variable coefficients
...
However, the type of differential equation that we consider
in this section is an exception to this rule; it is a linear equation with variable coefficients whose
general solution can always be expressed in terms of powers of , sines, cosines, and logarithmic
functions
...


A linear differential equation of the form
Ϫ1

ϩ

Ϫ1

Ϫ1
Ϫ1

ϩиииϩ

1

ϩ

0

ϭ ( ),

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...
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...
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...


●

where the coefficients , Ϫ1,
...

(French, 1789–1857) and
(Swiss, 1707–1783)
...
, 1, 0 of the monomial coefficients
matches the order of differentiation
͞ :
same

same
Ϫ1

–––– ϩ

Ϫ1

Ϫ1

–––––– ϩ
...
3, we start the discussion with a detailed examination of the
forms of the general solutions of the homogeneous second-order equation
2

2

2

ϩ

ϭ 0
...
Also, we can solve the
nonhomogeneous equation 2 Љ ϩ
Ј ϩ ϭ ( ) by variation of parameters,
once we have determined the complementary function
...
Hence to guarantee that the
fundamental results of Theorem 4
...
1 are applicable to the Cauchy-Euler equation, we
focus our attention on finding the general solutions defined on the interval (
)
...
Analogous to what happened when we substituted
into a linear equation with constant coefficients, when we substitute , each term of a Cauchy-Euler
equation becomes a polynomial in times , since
ϭ

(

Ϫ 1)(

Ϫ 2) и и и (

ϩ 1)

Ϫ

Ϫ

For example, when we substitute ϭ
2

(

ϭ

Ϫ 1)(

Ϫ 2) и и и (

Ϫ

ϩ 1)
...


is a solution of the
ϩ

ϭ 0
...
In the last
case the roots appear as a conjugate pair
...
Then 1 ϭ
the general solution is

1

and

2

Let 1 and 2 denote the real roots of (1) such
ϭ 2 form a fundamental set of solutions
...


2
2

Ϫ2

Ϫ 4 ϭ 0
...
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...
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...
3
...
Now (
Ϫ1
ϩ 2 4
...
When the roots of the
2
quadratic equation
ϩ ( Ϫ ) ϩ ϭ 0 are equal, the discriminant of the coefficients is necessarily zero
...
2
...
Thus

1

Ϫ /

ؒ

Ϫ2

Ϫ /

ؒ

( Ϫ )/

ϭ

1

1

;

Ϫ( / )ln

; Ϫ2

1

ϭ

ln

Ϫ /

ϭ

Ϫ /

ϭ ( Ϫ )/

ln
...


(4)

2
2

ϩ8

ϭ 0
...
Since
the general solution is ϭ 1 Ϫ1/2 ϩ 2 Ϫ1/2 ln
For higher-order equations, if
1

,

1

ln ,

1

(ln )2,
...
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...
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...
Correspondingly, the general solution of the differential equation must then contain a linear combination of these solutions
...


But when the roots of the auxiliary equation are complex, as in the case of equations
with constant coefficients, we wish to write the solution in terms of real functions
only
...

ϭ cos(b ln ) Ϫ sin(b ln )
...
From the fact that ϭ
of the constants, we see, in turn, for
ϭ

1

1

or

y

1

␣

(

ϭ2

␤

␣

Ϫ␤

ϩ

)

and

x

0

ϭ

␣

cos( ␤ ln )

2
2

and
a

ϭ

␣

ϭ2

(

␤
␣

Ϫ

2

ϭ

␣

)

sin( ␤ ln )

sin(b ln )) ϭ b

and

Ϫ␤

2aϪ1

0, b Ͼ 0 on

sin( ␤ ln )

constitute a fundamental set of real solutions of the differential equation
...
Since ( a cos(b ln ),
the interval (0, ), we conclude that
1

b

␣

[ 1 cos( ␤ ln ) ϩ

2

sin( ␤ ln )]
...

2

The Ј term is missing in the given Cauchy-Euler equation; nevertheless, the substitution ϭ yields

10

4

5
x

2

Љ ϩ 17 ϭ

(4 (

50

75

100

solution for 0 Ͻ Յ100
Solution curve of IVP
in Example 3

(4

2

Ϫ4

ϩ 17) ϭ 0

when 4 2 Ϫ 4 ϩ 17 ϭ 0
...
With the identifications ␣ ϭ 2 and b ϭ 2 we see from
(5) that the general solution of the differential equation is
ϭ

25

Ϫ 1) ϩ 17) ϭ

1/2

[ 1 cos(2 ln ) ϩ

2 sin(2

ln )]
...
Hence the solution
of the initial-value problem is ϭ Ϫ 1/2 cos(2 ln )
...
7
...
The particular
solution is seen to be oscillatory and unbounded as :
...
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...
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...
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...


Solve

3

3

ϩ5

3

2

2

2

ϩ7

ϩ 8 ϭ 0
...


In this case we see that ϭ
will be a solution of the differential equation
Hence the general solution is
for
1 ϭ Ϫ2,
2 ϭ 2 , and
3 ϭ Ϫ2
ϭ 1 Ϫ2 ϩ 2 cos(2 ln ) ϩ 3 sin(2 ln )
...
5 and 4
...
Consequently, in our
next example the method of variation of parameters is employed
...


Since the equation is nonhomogeneous, we first solve the associated
homogeneous equation
...
Now before using variation of parameters to find a particular solution
ϭ 1 1 ϩ 2 2, recall that the formulas Ј ϭ 1> and Ј ϭ 2> , where 1,
1
2
are the determinants defined on page 158, were derived under the assump2, and
tion that the differential equation has been put into the standard form Љ ϩ ( ) Ј ϩ
( ) ϭ ( )
...
Now with
3

2

2 5
ϭϪ
2 3

3

Јϩ

3

2

2

͉ ϭ Ϫ2

1
5

and

ϭ ,
,

Јϭ
2

2
2

ϭ

ϭ

͉1

2 3
ϭ
2 3

3

, and
0
2

2

͉ϭ2

3

,


...
Hence
by parts twice
...



...
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...
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...
For example, when the roots of the auxiliary
equations for Љ ϩ Ј ϩ ϭ 0 and 2 Љ ϩ
Ј ϩ ϭ 0 are distinct and real,
the respective general solutions are
ϭ

1

1

ϩ

and

2

2

ϭ

1

1

ϩ

2

2

,

Ͼ 0
...
The idea is to solve the new differential equation
in terms of the variable , using the methods of the previous sections, and, once the
general solution is obtained, resubstitute ϭ ln
...


Solve

2

Ј ϩ ϭ ln

ЉϪ

With the substitution ϭ
ϭ
2
2

ϭ
ϭ

or ϭ ln , it follows that

1

ϭ

; Chain Rule

΂ ΃ ϩ ΂Ϫ 1 ΃

1

; Product Rule and Chain Rule

2

1

΂

2

΃ ϩ ΂Ϫ 1 ΃ ϭ 1 ΂

1
2

2

2

2

2

Ϫ

΃
...


Since this last equation has constant coefficients, its auxiliary equation is
2
Ϫ 2 ϩ 1 ϭ 0, or ( Ϫ 1) 2 ϭ 0
...

By undetermined coefficients we try a particular solution of the form ϭ ϩ
This assumption leads to Ϫ2 ϩ ϩ ϭ , so ϭ 2 and ϭ 1
...


2

By resubstituting ϭ and ϭ ln we see that the general solution of the original
differential equation on the interval (0, ) is ϭ 1 ϩ 2 ln ϩ 2 ϩ ln
In the preceding discussion we have solved Cauchy-Euler
equations for Ͼ 0
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

See Problems 37 and 38 in Exercises 4
...

A second-order equation of the form
2

2
0)

( Ϫ

2

ϩ ( Ϫ

0)

ϭ0

ϩ

(6)

is also a Cauchy-Euler equation
...

We can solve (6) as we did (1), namely, seeking solutions of ϭ ( Ϫ 0) and
using
2

ϭ

( Ϫ

0)

Ϫ1

and

2

(

ϭ

Ϫ 1)( Ϫ

0)

Ϫ2


...

See Problems 39–42 in Exercises 4
...


2

2

ЉϪ2 ϭ0

4

Љϩ Јϭ0
Љϩ

2

ЉϪ3 ЈϪ2 ϭ0
2

2

Љ Ϫ 3 Ј ϩ 4 ϭ 0,

2

Љ ϩ 25 Ј ϩ ϭ 0

Љϩ5 Јϩ3 ϭ0

2

Јϩ4 ϭ0

Љϩ3 ЈϪ4 ϭ0
2

4

Ј ϩ ϭ 0,

Љϩ Јϭ ,

Љϩ4 ЈϪ ϭ0

Љϩ5 Јϩ4 ϭ0

2

Љϩ8 Јϩ6 ϭ0

Љϩ6 Јϩ ϭ0

2

Љ Ϫ 7 Ј ϩ 41 ϭ 0

3

ٞϩ

2

3

Љϩ ϭ0

ЉϪ3 Јϭ0

2

25

2

Љϩ

2

In Problems 1 – 18 solve the given differential equation
...
Solve the original equation
by solving the new equation using the procedures in
Sections 4
...
5
...

ЉϪ4 Јϭ
2

2
2
2

Љϩ

Љ ϩ 10 Ј ϩ 8 ϭ
Љ Ϫ 4 Ј ϩ 6 ϭ ln

2

Љ Ϫ 3 Ј ϩ 13 ϭ 4 ϩ 3
ٞϪ3

4

Љϩ5 Јϩ ϭ

ЉϪ

Љ Ϫ 9 Ј ϩ 25 ϭ 0

2

ЈϪ ϭ0

2
2

ٞϪ6 ϭ0

Љ ϩ 9 Ј Ϫ 20 ϭ 0

3

3

2

2

2

Ј Ϫ ϭ ln

ЉϪ2 Јϩ2 ϭ

Љ ϩ 3 Ј ϭ 0,

2

Љ Ϫ 5 Ј ϩ 8 ϭ 0,

Љ ϩ 6 Ј Ϫ 6 ϭ 3 ϩ ln

3

Љϩ

ЈϪ

ϭ

4

1
ϩ1

In Problems 25 – 30 solve the given initial-value problem
...

2

2

In Problems 37 and 38 use the substitution ϭ Ϫ to solve
the given initial-value problem on the interval (Ϫϱ, 0)
...


(Ϫ2) ϭ 8, Ј(Ϫ2) ϭ 0
ϭ( Ϫ

0)

to solve the given

( ϩ 3)2 Љ Ϫ 8( Ϫ 1) Ј ϩ 14 ϭ 0
( Ϫ 1)2 Љ Ϫ ( Ϫ 1) Ј ϩ 5 ϭ 0

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...
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...
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...


●

In Problems 41 and 42 use the substitution ϭ
solve the given differential equation
...
7
...


In Problems 47– 50 solve the given differential equation by
using a CAS to find the (approximate) roots of the auxiliary
equation
...


●
●

ٞ Ϫ 10
...
5 Ј ϩ 1
...
Use a CAS as an aid in computing roots of
the auxiliary equation and the determinants given in
(15) of Section 4
...


See the
at the end of Section 4
...
1 and Section 4
...
6

,

, and


...
In the mathematical analysis of physical systems it is
( ) of (1) subject to either initial conditions or
often desirable to express the
or
( )
...

To see how this is done, we start by examining solutions of initial-value problems in which the
DE (1) has been put into the standard form
Љϩ ( ) Јϩ

( ) ϭ ( )

(2)

by dividing the equation by the lead coefficient 2( )
...


We will see as the discussion unfolds that the
solution ( ) of the second order initial-value problem
Љϩ ( ) Јϩ

( ) ϭ ( ),

( 0) ϭ

0,

Ј( 0) ϭ

1

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...
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...
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...


(3)

●

can be expressed as the superposition of two solutions:
( )ϭ

( )ϩ

( ),

(4)

where ( ) is the solution of the associated homogeneous DE with nonhomogeneous
initial conditions
ϭ

Љϩ ( ) Јϩ

᭤

( ) ϭ 0,

( 0) ϭ

0,

Ј( 0) ϭ

1

(5)

and ( ) is the solution of the nonhomogeneous DE with homogeneous (that is,
zero) initial conditions
Љϩ ( ) Јϩ

( ) ϭ ( ),

( 0) ϭ 0, Ј( 0) ϭ 0
...
3 to find the general solution
of the homogeneous DE and then use the given initial conditions to determine the
two constants in that solution
...

Because of the zero initial conditions, the solution of (6) could describe a physical
system that is initially at rest and so is sometimes called a

...
Recall from (3) of Section 4
...


(7)

) in (7) are defined by (9) of Section 4
...


(8)

The linear independence of 1( ) and 2( ) on the interval guarantees that the
Wronskian ϭ ( 1( ), 2( )) 0 for all in
...


(10)

0

The function ( , ) in (10),
( , )ϭ

᭤

1(

) 2( ) Ϫ 1( ) 2( )
()

(11)

is called the
for the differential equation (2)
...
Therefore all linear second-order differential equations
(2) with the same left-hand side but with different forcing functions have the same
the Green’s function
...


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...
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...
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...


●

Use (10) and (11) to find a particular solution of Љ Ϫ

ϭ ( )
...
It follows from (11) that the Green’s
1
function is
Ϫ

( , )ϭ

Ϫ

Ϫ
Ϫ2

Ϫ

Ϫ
2

ϭ

Ϫ( Ϫ )

(12)

ϭ sinh( Ϫ )
...


0

Find the general solution of following nonhomogeneous differential equations
...
Moreover, as pointed out in the paragraph preceding Example 1,
the Green’s function for both differential equations is (12)
...
Thus the general solu-

͵

tion

ϭ

0

of the given DE on any interval [ 0, ] not containing the origin is

ϩ

ϭ

1

ϩ

2

Ϫ

ϩ

͵

sinh( Ϫ )

(14)


...
6
...
The general solution ϭ ϩ is then
ϭ

1

ϩ

2

Ϫ

ϩ

͵

sinh( Ϫ )

2


...
One way of solving the problem when ( ) 0 has already been illustrated in Sections 4
...
6, that is, apply the initial conditions
( 0) ϭ 0, Ј( 0) ϭ 0 to the general solution of the nonhomogeneous DE
...


The function

( ) defined in (10) is the solution of the initial-value problem (6)
...
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...
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...
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...
Next, because a definite integral has the property ͵ ϭ 0 we have
( 0) ϭ

͵

0

( 0, ) ( )

ϭ 0
...


ϭ 0
...
8
...

Identifying 0 ϭ 0 and ( ) ϭ

2

( )ϭ

, we see from (15) that the solution of the IVP is

͵

sinh( Ϫ )

2


...


Solve the initial-value problem
Љϩ4 ϭ ,

(0) ϭ 0, Ј(0) ϭ 0
...


* This formula, usually discussed in advanced calculus, is given by

͵

( )

( , )
( )

ϭ ( , ( )) Ј( ) Ϫ ( , ( )) Ј( ) ϩ

͵

( )
( )

Ѩ
Ѩ

( , )
...
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...
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...
From (11), with (cos2 , sin2 ) ϭ 2, we find
2
᭤
Ϫ

ϭ

1(

) ϭ cos 2 and

cos2 sin2 Ϫ cos2 sin2
ϭ 1 sin2( Ϫ )
...


0

If we wish to evaluate the integral, we first write
( ) ϭ 1 sin2
2

͵

Ϫ 1 cos2
2

cos2

0

͵

sin2

0

and then use integration by parts:
( ) ϭ 1 sin2
2

[1
2

sin2 ϩ 1 cos2
4

or

( )ϭ

1
4

]0 Ϫ 1 cos2 [Ϫ1
2
2

cos2 ϩ 1 sin2
4

]0

Ϫ 1 sin 2
...
8
...
It is simply the function already given in (4)
...


Because ( ) is a linear combination of the fundamental solutions, it
follows from (10) of Section 4
...
Moreover, since satisfies the initial-conditions in (5) and satisfies the
initial conditions in (6), we have,
( 0) ϭ

( 0) ϩ

( 0) ϭ

0

ϩ0ϭ

0

Ј( 0) ϭ Ј ( 0) ϩ Ј ( 0) ϭ

1

ϩ0ϭ

1
...
See pages 200–202
...
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...
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...
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...


We solve two initial-value problems
...
By applying the initial conditions to the general solution ( ) ϭ 1 cos2 ϩ 2 sin2 of the homogeneous DE,
we find that 1 ϭ 1 and 2 ϭ Ϫ1
...

Next we solve Љ ϩ 4 ϭ sin2 , (0) ϭ 0, Ј(0) ϭ 0
...
With ( ) ϭ sin2 we see from (10) that
2
the solution of this second problem is ( ) ϭ 1͵0 sin 2( Ϫ )sin2
...
8
...


(19)

0

If desired, we can integrate the definite integral in (19) by using the trigonometric identity
sin sin
with

ϭ 2( Ϫ ) and

ϭ 1[cos( Ϫ ) Ϫ cos ( ϩ )]
2

ϭ2:
( )ϭ

͵
͵

1
2

ϭ1
4

sin2( Ϫ )sin2

0

[

ϭ 1 Ϫ1 sin(2 Ϫ 4 ) Ϫ cos2
4
4
ϭ

(20)

[cos(2 Ϫ 4 ) Ϫ cos2 ]

0

1
8 sin2

Ϫ

1
4

]0

cos2
...

8
4

or

(21)

Note that the physical significance indicated in (18) is lost in (21) after combining
like terms in the two parts of the solution ( ) ϭ ( ) ϩ ( )
...
For example, if the problem in Example 5 is changed to
Љϩ4 ϭ ,

(0) ϭ 1, Ј(0) ϭ Ϫ2,

we simply replace sin2 in the integral in (19) by and the solution is then
( )ϭ

( )ϩ

( )

ϭ cos 2 Ϫ sin2 ϩ 1
2

͵

sin2( Ϫ )

; see Example 4

0

ϭ 1 ϩ cos2 Ϫ 9 sin2
...


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...
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...
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...


●

Solve the initial-value problem
Љ ϩ 4 ϭ ( ),

(0) ϭ 1, Ј(0) ϭ Ϫ2,

where the forcing function is piecewise defined

Ά

0,
Ͻ0
( ) ϭ sin 2 , 0 Յ Յ 2␲
0,
Ͼ 2␲
...


0

Because is defined in three pieces, we consider three cases in the evaluation of the
definite integral
...

2
Hence

( ) is
( )ϭ

Ά

0,
1
1
8 sin2 Ϫ4 cos 2 ,
Ϫ1 ␲ cos2 ,
2

Ͻ0
0 Յ Յ 2␲
Ͼ 2␲
...


Putting all the pieces together we get

Ά

cos 2 Ϫ sin 2 ,
Ͻ0
( ) ϭ (1 Ϫ 1 ) cos 2 Ϫ 7 sin 2 , 0 Յ Յ 2␲
4
8
(1 Ϫ 1␲)cos2 Ϫ sin2 ,
Ͼ 2␲
...
8
...

We next examine how a boundary value problem (BVP) can be solved using a
different kind of Green’s function
...
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...
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...
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...
Conditions such as
( ) ϭ 0, ( ) ϭ 0;
( ) ϭ 0, Ј( ) ϭ 0;
Ј( ) ϭ 0, Ј( ) ϭ 0
...
Specificall , our goal is to find a integral
solution ( ) that is analogous to (10) for nonhomogeneous boundary-value problems
of the form
Љ ϩ ( ) Ј ϩ ( ) ϭ ( ),
(24)
1 ( ) ϩ 1 Ј( ) ϭ 0
( ) ϩ 2 Ј( ) ϭ 0
...
This latter assumption is sufficient to guarantee
that a unique solution of (24) exists and is given by an integral ( ) ϭ ͵ ( , ) ( ) ,
where ( , ) is a Green’s function
...

Suppose 1( ) and 2( ) are linearly independent
solutions on [ , ] of the associated homogeneous form of the DE in (24) and that
is a number in the interval [ , ]
...

()

(25)

The reason for integrating 1( ) and Ј( ) over different intervals will become clear
Ј
2
shortly
...

()

(26)

The right-hand side of (26) can be written compactly as a single integral
( )ϭ

͵

( , ) () ,

(27)

where the function ( , ) is

( , )ϭ

Ά

1(

) 2( )
,
()
1( ) 2( )
,
()

Յ Յ
(28)
Յ Յ
...
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...
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...
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...
It can be proved that ( , ) is a continuous function of on the
interval [ , ]
...

2

(29)
2(

) Ј( ) ϩ
2

2(

) Ј( )
2
(30)

Before proceeding, observe in (25) that 1( ) ϭ 0 and 2( ) ϭ 0
...
From (29) and (30) we hav
0

1

( )ϩ

1

Ј( ) ϭ
ϭ

Likewise,

1(

1[ 1( ) 1( ) ϩ
1(

)[

1 1(

)ϩ

0

2

2( ) 2( )] ϩ
1 Ј(
1

) ϭ 0 implies that whenever

2

2

ϭ

) Ј( )]
2

2(

) satisfies (23) so does ( ):
0

2

2[ 1( ) 1( ) ϩ
2(

2(

)] ϭ 0
...


2 2
2 2
(1111)1111*
0 from (22)

The next theorem summarizes these results
...
Then the
function ( ) defined in (27) is a solution of the boundary-value problem (24)

Ј
ϭ
ϭ
p
ϭp

Solve the boundary-value problem
ϭ

᭤

ϭ
ϭ
ϭ

ϭ

Љ ϩ 4 ϭ 3,

Ј(0) ϭ 0,

(p>2) ϭ 0
...
The Wronskian is ( 1, 2) ϭ 2, and so from (28) we see that the
Green’s function for the boundary-value problem is
( , )ϭ

Ά

1
2 cos

2 sin 2 ,

1
2 cos

2 sin 2 ,

0Յ Յ
Յ Յ p>2
...
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...
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...
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...
8
...

4
4

Don’t infer from the preceding example that the demand that 1( ) satisfy (22)
and 2( ) satisfy (23) uniquely determines these functions
...


Solve the boundary-value problem
2

Љ Ϫ 3 Ј ϩ 3 ϭ 24 5,

(1) ϭ 0,

(2) ϭ 0
...
From
the auxiliary equation ( Ϫ 1) Ϫ 3 ϩ 3 ϭ ( Ϫ 1)( Ϫ 3) ϭ 0 the general solution of the associated homogeneous equation is ϭ 1 ϩ 2 3
...
By choosing 2 ϭ Ϫ1 we get
Ϫ 3
...
The choice 2 ϭ Ϫ1 now gives 1 ϭ 4 and so
3

...

1Ϫ3 2 4Ϫ3 2

͉

͉

Hence the Green’s function for the boundary-value problem is

Ά

( Ϫ 3)(4 Ϫ 3)
, 1Յ Յ
63
( , )ϭ
( Ϫ 3)(4 Ϫ 3)
, Յ Յ 2
...

From this equation we see that ( ) ϭ 24 3 and so

͵

2

( ) ϭ 24

3

( , )

1

ϭ 4(4 Ϫ

3

͵

)

1

᭤

( ) in (27) becomes

( Ϫ 3)

ϩ 4( Ϫ

͵

2

3

)

(4 Ϫ 3)
...


We have barely scratched the surface of the elegant, albeit complicated, theory
of Green’s functions
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 1–6 proceed as in Example 1 to find a particular
solution ( ) of the given differential equation in the
integral form (10)
...

2

( )ϭ

Љ Ϫ 2 Ј ϩ 2 ϭ cos

Љ Ϫ Ј ϭ 1, (0) ϭ 0, Ј(0) ϭ 0

ЉϪ4 ϭ

where

2

ϩ sin

2

ϭ ( ), (0) ϭ 8, Ј(0) ϭ 2,

where

2

In Problems 13–18 proceed as in Example 3 to find a solution of the given initial-value problem
...

ЉϪ4 ϭ

ЉϪ

ЉϪ

ЉϪ2 Јϩ2 ϭ ( )

In Problems 7–12 proceed as in Example 2 to find the
general solution of the given differential equation
...
Do not evaluate the integral
that defines ( )
...

In Problem 36 find a solution of the BVP when ( ) ϭ
...

Љϩ

ϭ 1, (0) ϭ 0, (1) ϭ 0

Љ ϩ 9 ϭ 1, (0) ϭ 0, Ј(p) ϭ 0
ЉϪ2 Јϩ2 ϭ
ЉϪ Јϭ

2

, (0) ϭ 0, (p>2) ϭ 0

, (0) ϭ 0, (1) ϭ 0

2

Љϩ

2

ЉϪ4 Јϩ6 ϭ

Ј ϭ 1, (

) ϭ 0, (1) ϭ 0

Ϫ1

4

, (1) Ϫ Ј(1) ϭ 0, (3) ϭ 0

Љ Ϫ 2 Ј ϩ 2 ϭ ln , (1) ϭ 1, Ј(1) ϭ 0

2

In Problems 35 and 36,
use (27) and (28) to find a solution of the boundary-value problem
...


Јϩ

ϭ

2

, (1) ϭ 4, Ј(1) ϭ 3

Suppose the solution of the boundary-value problem
Љϩ

Јϩ

ϭ ( ), ( ) ϭ 0, ( ) ϭ 0,

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...
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...
Editorial review has
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...


●

Ͻ , is given by ( ) ϭ ͵ ( , ) ( ) where 1( ) and
2( ) are solutions of the associated homogeneous
differential equation chosen in the construction of ( , ) so
that 1( ) ϭ 0 and 2( ) ϭ 0
...
Reread the assumptions fol1
lowing (24)
...


ϭ ( ), ( ) ϭ , ( ) ϭ

Јϩ

is given by
( )ϭ

( )ϩ

●

1(

)

1(

)ϩ

2(

)

2(

)
...
3, 4
...
5), and 4
...


Simultaneous ordinary differential equations involve two or more equations that
contain derivatives of two or more dependent variables—the unknown functions—with respect to a
single independent variable
...
We
shall see that the analogue of
an algebraic equation by a constant is
on an ODE
with some combination of derivatives
...
Recall from Section 4
...
,
(

ϩ

( Ϫ1)

ϩиииϩ

Јϩ

1

0

ϭ ( ),

are constants, can be written as
Ϫ1

( Ϫ1)

ϩиииϩ

1

ϩ

0)

ϭ ( )
...
Now, for example, to rewrite the system
Љϩ2 Јϩ Љϭ

ϩ 3 ϩ sin

Ј ϩ Ј ϭ Ϫ4 ϩ 2 ϩ

Ϫ

in terms of the operator , we first bring all terms involving the dependent variables
to one side and group the same variables:
Љ ϩ 2 Ј Ϫ ϩ Љ Ϫ 3 ϭ sin
ЈϪ4 ϩ ЈϪ2 ϭ Ϫ

is the same as

(

2

ϩ 2 Ϫ 1) ϩ (
( Ϫ 4) ϩ (

2

Ϫ 3) ϭ sin
Ϫ 2) ϭ Ϫ
...
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...
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...
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...


or, equivalently,

(1)

Operating on the first equation in (1) by while multiplying the second by Ϫ3 and
then adding eliminates from the system and gives 2 Ϫ 6 ϭ 0
...


Multiplying the first equation in (1) by 2 while operating on the second by and
then subtracting gives the differential equation for , 2 Ϫ 6 ϭ 0
...


Now (2) and (3) do not satisfy the system (1) for every choice of 1, 2, 3, and
because the system itself puts a constraint on the number of parameters in a solu4
tion that can be chosen arbitrarily
...
These two equations enable us to write 3
as a multiple of 1 and 4 as a multiple of 2 :
΂Ϫ16

3

1

Ϫ 3 3΃

ϭϪ

16
3
16

Ϫ16

ϩ ΂16

2

and

1

Ϫ 3 4΃

16
3
4

ϭ

16
3

ϭ 0
...


Hence we conclude that a solution of the system must be
()ϭ

1

Ϫ16

ϩ

2

,

()ϭϪ

1

Ϫ16

ϩ

16
3

(4)

2

16


...


Solve
(

ϩ(
Ϫ 3) Ϫ

ϩ 2) ϭ 0
2 ϭ 0
...
It follows that the differential equation
for is
[(
Since
2
ϩ

Ϫ 3)(

ϩ 2) ϩ 2 ] ϭ 0

or

(

2

ϩ

Ϫ 6) ϭ 0
...


is
(6)

Ϫ 6) ϭ 0, from which we fin
Ϫ3


...
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...
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...
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...
Substituting (6) and (7) into the first equation of (5) give
(4

ϩ 2 3)

1

2

ϩ (Ϫ

2

Ϫ 3 4)

ϭ 0
...

3


...

2

᭤

4

1

and

2,

the

It sometimes pays to keep one’s eyes open when solving systems
...
You should verify that substituting
( ) into ϭ 1 ( Ϫ 3 ) yields ϭ Ϫ1 3 2 Ϫ 3 4 Ϫ3
...


ЈϪ4 ϩ Љϭ 2
Јϩ
ϩ Ј ϭ 0
...


(9)

Then, by eliminating , we obtain
[(

ϩ 1)

2

Ϫ(

or

(

ϩ 1) 2 Ϫ (

Ϫ 4) ] ϭ (
3

ϩ4 ) ϭ

2

Ϫ 4)0

ϩ2
...


The last equality implies that 12 ϭ 1, 8 ϭ 2, and 6 ϩ 4 ϭ 0; hence
1
ϭ 12, ϭ 1, and ϭ Ϫ1
...


2

Ϫ

1

...

8

(11)

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Now 4 and 5 can be expressed in terms of 2 and 3 by substituting (10)
and (11) into either equation of (8)
...
Solving for 4 and 5 in terms of 2
and 3 gives 4 ϭ Ϫ1 (4 2 ϩ 2 3 ) and 5 ϭ 1 (2 2 Ϫ 4 3 )
...

8

Ϫ

In (3) of Section 3
...
3
...
At that time we were not able to solve
the system
...


2
1
Operating on the first equation by ϩ 25, multiplying the second equation by 50,
2
adding, and then simplifying gives (625 ϩ 100 ϩ 3) 1 ϭ 0
...
We can now obtain 2 ( ) by
2
ϭ 50( ϩ 25) 1
...


In the original discussion on page 108 we assumed that the initial conditions were
1 (0) ϭ 25 and
2 (0) ϭ 0
...
Solving these equations simultaneously gives
25
1 ϭ 2 ϭ 2
...


The graphs of both of these equations are given in Figure 4
...
1
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 1 –20 solve the given system of differential
equations by systematic elimination
...


Ϫ

ϩ 5) Ϫ
Ϫ2 ϩ (

2

Ϫ2

(1) ϭ 0, (1) ϭ 1

A projectile shot from a gun has
weight ϭ
and velocity tangent to its path of
motion
...

See Figure 4
...
2
...
[
: Use Newton’s
second law of motion in the and directions
...
See Figure 4
...
3
...
[
: is a multiple of velocity,
say, b ]

Ϫ Ϫ 1) Ϫ (2 ϩ 1) ϭ 1
( Ϫ 1) ϩ
ϭ Ϫ1
Ϫ5 ϩ

ϭ

Ϫ

ϭ5

ϩ

ϩ

ϭ

2

Ϫ

2

ϩ

ϩ

ϩ

ϭ0

( Ϫ 1) ϩ ( 2 ϩ 1) ϭ 1
( 2 Ϫ 1) ϩ ( ϩ 1) ϭ 2
2

Ϫ 2(
ϩ

ϭ
ϭ
ϭ

2

Forces in Problem 24

ϩ ) ϭ sin
ϭ0
ϩ
( Ϫ 1) ϩ
ϩ
ϩ2 ϩ

ϭ6

ϭ
ϭ0
ϭ

Examine and discuss the following system:
(

Ϫ2
ϭ 2
ϩ 1) Ϫ 2( ϩ 1) ϭ 1
...
9
...
Then use a rootfinding application to determine when tank contains
more salt than tank

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...
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...
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...


●

Reread Problem 8 of Exercises 3
...
In that problem
you were asked to show that the system of differential equations
1

ϭϪ

1
50
2
3
ϭ
75
2

ϭ

1
50

1

1

Ϫ

2

2
75
1
Ϫ
25

2

3

is a model for the amounts of salt in the connected
mixing tanks , , and shown in Figure 3
...
7
...


●
●
●

Use a CAS to graph 1( ), 2( ), and 3( ) in the
same coordinate plane (as in Figure 4
...
1) on the
interval [0, 200]
...
Use a root-findin
application of a CAS to determine the time when
the amount of salt in each tank is less than or equal
to 0
...
When will the amounts of salt 1( ),
2( ), and 3( ) be simultaneously less than or equal
to 0
...
2 and 2
...
2
A review of Taylor series from calculus is also recommended
...
Two of the solution methods
considered in this section employ a change of variable to reduce a nonlinear second-order DE to a
first-order DE
...
2

There are several significan differences between linear
and nonlinear differential equations
...
1 that homogeneous linear
equations of order two or higher have the property that a linear combination of solutions is also a solution (Theorem 4
...
2)
...
See Problems 1 and 18 in Exercises 4
...
We can fin
general solutions of linear first-orde DEs and higher-order equations with constant
coefficients Even when we can solve a nonlinear first-orde differential equation in
the form of a one-parameter family, this family does not, as a rule, represent a general solution
...
But the major difference between linear
and nonlinear equations of order two or higher lies in the realm of solvability
...
On the other hand, nonlinear higher-order differential equations virtually defy solution by analytical methods
...
As was pointed out at the end of
Section 1
...

Let us make it clear at the outset that nonlinear higher-order differential equations
are important—dare we say even more important than linear equations?—because as
we fine-tune the mathematical model of, say, a physical system, we also increase the
likelihood that this higher-resolution model will be nonlinear
...


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...
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...


●

Nonlinear second-order differential equations ( , Ј, Љ) ϭ 0,
where the dependent variable is missing, and ( , Ј, Љ) ϭ 0, where the independent variable is missing, can sometimes be solved by using first-order methods
...

The next example illustrates the substitution
technique for an equation of the form ( , Ј, Љ) ϭ 0
...
If we can solve this last equation for , we can f nd
by integration
...


Solve Љ ϭ 2 ( Ј)2
...
After substituting, the second-order
equation reduces to a first-order equation with separable variables; the independent
variable is and the dependent variable is :
ϭ2

͵

2

or

Ϫ2

Ϫ

2

ϭ
Ϫ1

ϭ

͵

ϭ2

ϩ

2
1
...
The reason should be
1
obvious in the next few steps
...


Next we show how to solve an equation
that has the form ( , Ј, Љ) ϭ 0
...


In this case the first-order equation that we must now solve i

΂,
Solve

,

΃ ϭ 0
...


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...
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...
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...


●

With the aid of ϭ Ј, the Chain Rule shown above, and separation of
variables, the given differential equation becomes

΂ ΃ϭ

2

or


...
We now resubstitute ϭ ͞ , separate variables once again, integrate, and relabel constants a second time:

͵

ϭ

͵

2

ln͉ ͉ ϭ

or

2

ϩ

or

3

ϭ

4

2


...


Let us assume that a solution of the initial-value problem
Љϭ

ϩ

Ϫ

2

,

(0) ϭ Ϫ1,

Ј(0) ϭ 1

(1)

exists
...
(2)

Note that the values of the first and second terms in the series (2) are known
since those values are the specified initial conditions (0) ϭ Ϫ1, Ј(0) ϭ 1
...
We can then fin
expressions for the higher derivatives ٞ, (4),
...
Now using (0) ϭ Ϫ1 and Ј(0) ϭ 1, we find from (3) that ٞ(0) ϭ 4
...
With
the additional information that ٞ(0) ϭ 4, we then see from (5) that (5)(0) ϭ 24
...


Numerical methods, such as Euler’s method or the
Runge-Kutta method, are developed solely for first-order differential equations and then
are extended to systems of first-order equations
...
In brief, here is how it is done for a second-order initial-value problem: First, solve

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...
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...


●

for Љ—that is, put the DE into normal form Љ ϭ ( , , Ј)—and then let Ј ϭ
example, if we substitute Ј ϭ in

For

2
2

ϭ ( , , Ј),

( 0) ϭ

0,

Ј( 0 ) ϭ

0,

(6)

then Љ ϭ Ј and Ј( 0 ) ϭ ( 0 ), so the initial-value problem (6) becomes

Ά ЈЈ ϭ
ϭ

:
:

( 0) ϭ

( , , )
0 , ( 0) ϭ

0
...


require*

y

Following the foregoing procedure, we find that the second-order initial-value problem in Example 3 is equivalent to

Taylor
polynomial

ϭ
ϭ
x
solution curve
generated by a
numerical solver

Comparison of two
approximate solutions in Example 1
y

x
10

20

Numerical solution
curve for the IVP in (1)

ϩ

Ϫ

2

with initial conditions (0) ϭ Ϫ1, (0) ϭ 1
...
10
...
For comparison the graph of the
fifth-degree Taylor polynomial 5( ) ϭ Ϫ1 ϩ Ϫ 2 ϩ 2 3 Ϫ 1 4 ϩ 1 5 is shown
3
3
5
in red
...

The blue numerical solution curve in Figure 4
...
1
raises some questions of a qualitative nature: Is the solution of the original initial-value
problem oscillatory as : ? The graph generated by a numerical solver on the larger
interval shown in Figure 4
...
2 would seem to
that the answer is yes
...
Also, what is happening to the solution curve in Figure 4
...
2 when is
near Ϫ1? What is the behavior of solutions of the differential equation as : Ϫϱ?
Are solutions bounded as : ? Questions such as these are not easily answered, in
general, for nonlinear second-order differential equations
...
1, are the kind that have no explicit
dependence on the independent variable
...
For an in-depth
treatment of the topic of stability of autonomous second-order differential equations
and autonomous systems of differential equations, refer to Chapter 10 in

*

Some numerical solvers require only that a second-order differential equation be expressed in normal
form Љ ϭ ( , , Ј)
...


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...
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...


●

In Problems 1 and 2 verify that 1 and 2 are solutions of the
given differential equation but that ϭ 1 1 ϩ 2 2 is, in
general, not a solution
...

[1 ϩ ( Ј) 2]3 / 2

Find ϭ ( ) for which k ϭ 1
...
]

2

ϭ

In calculus the curvature of a curve that is defined by a
function ϭ ( ) is defined a

: For simplicity,

In Problems 3 – 8 solve the given differential equation by
using the substitution ϭ Ј
...

2 Ј Љ ϭ 1, (0) ϭ 2, Ј(0) ϭ 1
Љ ϩ ( Ј)2 ϭ 0, (1) ϭ 4, Ј(1) ϭ 2
Consider the initial-value problem
Љϩ

Ј ϭ 0,

(0) ϭ 1, Ј(0) ϭ Ϫ1
...

Find an explicit solution of the IVP
...

Find an interval of definition for the solution in
part (b)
...

2
2
2

In Problem 1 we saw that cos and were solutions of
the nonlinear equation ( Љ) 2 Ϫ 2 ϭ 0
...
Without attempting to solve the
differential equation, discuss how these explicit solutions
can be found by using knowledge about linear equations
...


Discuss how the method of reduction of order considered in this section can be applied to the third-order
differential equation ٞ ϭ 11 ϩ ( Љ)2
...


Discuss how to find an alternative two-parameter family of solutions for the nonlinear differential equation
Љ ϭ 2 ( Ј) 2 in Example 1
...
]
1

Find two solutions of the initial-value problem
( Љ)2 ϩ ( Ј)2 ϭ 1,

Use a numerical solver to graph the solution curves
...
Solve this equation (see Section 2
...

Љ ϭ Ј ϩ ( Ј) 3

Љ ϭ Ј ϩ ( Ј) 2

In Problems 15– 18 proceed as in Example 3 and obtain the
first six nonzero terms of a Taylor series solution, centered
at 0, of the given initial-value problem
...

Љϭ ϩ

2

,

(0) ϭ 1, Ј(0) ϭ 1
(0) ϭ 2, Ј(0) ϭ 3

Љϩ

2

ϭ 1,

Љϭ

2

ϩ

Љϭ

,

2

Ϫ 2 Ј,

(0) ϭ 1, Ј(0) ϭ 1

(0) ϭ 0, Ј(0) ϭ Ϫ1

2
2

ϭ Ϫ 2
...
Show that the velocity of
the body at time is given by 2 ϭ 2 2 (1͞ Ϫ 1͞ 0 )
...


Use a numerical solver to graphically investigate the solutions of the equation subject to (0) ϭ 0, Ј(0) ϭ 1,

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...
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...


●

Ն 0
...
Investigate the equation
1

in the same manner
...


2
2

ϩ

ϩ sin ϭ 0

Answer Problems 1 – 10 without referring back to the text
...


1(
1(

The only solution of the initial-value problem
Љ ϩ 2 ϭ 0, (0) ϭ 0, Ј(0) ϭ 0 is __________
...
If ϭ 1 Ϫ ϩ 6 2 ϩ 3 is a solution of a homogeneous fourth-order linear differential equation with constant coefficients, then the roots of the auxiliary equation
are _______
...

ϭ 2 is particular solution of
ϭ _______
...

If 1 ϭ and 2 ϭ Ϫ are solutions of homogeneous
linear
differential
equation,
then
necessarily
ϭ Ϫ5 Ϫ ϩ 10 is also a solution of the DE
...

1( ) ϭ
Then give an interval over which the set consisting of
1 and 2 is linearly dependent
...

2
, (0, )
1( ) ϭ ln , 2 ( ) ϭ ln
ϩ1
, 2( ) ϭ
, ϭ 1, 2,
...
Write down the general solution of the
corresponding homogeneous linear DE if it is
an equation with constant coefficients
a Cauchy-Euler equation
...
Choose the input functions ( ) for which the method of undetermined coeffi
cients is applicable and the input functions for which the
method of variation of parameters is applicable
...
__________


...


) ϭ cos

1(

For the method of undetermined coefficients, the
assumed form of the particular solution
for
Љ Ϫ ϭ 1 ϩ is __________
...

ЉϪ2 ЈϪ2 ϭ0
2 Љϩ2 Јϩ3 ϭ0
ٞ ϩ 10 Љ ϩ 25 Ј ϭ 0
2 ٞ ϩ 9 Љ ϩ 12 Ј ϩ 5 ϭ 0
3 ٞ ϩ 10 Љ ϩ 15 Ј ϩ 4 ϭ 0
2

(4)

ϩ3 ٞϩ2 Љϩ6 ЈϪ4 ϭ0

ЉϪ3 Јϩ5 ϭ4
ЉϪ2 Јϩ ϭ

3

Ϫ2

2

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...
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...


●

In Problems 35– 40 solve the given differential equation subject to the indicated conditions
...
Do not determine the coefficients i
...


2

ϩ 2 ) Ј ϩ ( ϩ 2) ϭ

Find a member of the family of solutions of
Љ ϩ Ј ϩ 1 ϭ 0 whose graph is tangent to the
-axis at ϭ 1
...

In Problems 43– 46 use systematic elimination to solve the
given system
...


Verify that 1 ϭ is one solution of the associated
homogeneous equation
...
2 leads to a
second solution 2 of the homogeneous equation as well
as a particular solution of the nonhomogeneous equation
...


ϩ 64 ٞ ϩ 59 Љ Ϫ 23 Ј Ϫ 12 ϭ 0
...

Solve the DE in part (a) subject to the initial conditions (0) ϭ Ϫ1, Ј(0) ϭ 2, Љ(0) ϭ 5, ٞ(0) ϭ 0
...


Consider the differential equation
ЉϪ(

(4)

12

Write the general solution of the fourth-order DE
(4)
Ϫ 2 Љ ϩ ϭ 0 entirely in terms of hyperbolic
functions
...
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...
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...
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...
1 Linear Models: Initial-Value Problems
5
...
1 Spring/Mass Systems: Free Undamped Motion
5
...
2 Spring/Mass Systems: Free Damped Motion
5
...
3 Spring/Mass Systems: Driven Motion
5
...
4 Series Circuit Analogue
5
...
3 Nonlinear Models
Chapter 5 in Review

We have seen that a single differential equation can serve as a mathematical model
for diverse physical systems
...
1
...
Forms of this linear second-order equation appear in the
analysis of problems in many different areas of science and engineering
...
1
we deal exclusively with initial-value problems, whereas in Section 5
...
In Section 5
...
Section 5
...


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...
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●

●
●
●

Sections 4
...
3, and 4
...
3
Problems 27–36 in Exercises 4
...


Recall that the function is the
or
of the system
...


Suppose that a flexible spring is suspended vertically from a rigid
support and then a mass is attached to its free end
...
By Hooke’s law the spring itself exerts a restoring force opposite to the direction of elongation and proportional to the amount of
elongation Simply stated, ϭ , where is a constant of proportionality called the
The spring is essentially characterized by the number For example,
if a mass weighing 10 pounds stretches a spring 1 foot, then 10 ϭ ΂ 1΃ implies
2
2
ϭ 20 lb/ft
...

5

unstretched
equilibrium
position
0

motion

Spring/mass system

0

After a mass is attached to a spring, it stretches
the spring by an amount and attains a position of equilibrium at which its
weight
is balanced by the restoring force
Recall that weight is defined by
ϭ
, where mass is measured in slugs, kilograms, or grams and ϭ 32 ft /s2,
9
...
As indicated in Figure 5
...
1(b), the condition
of equilibrium is
ϭ or
Ϫ ϭ 0
...

Assuming that there are no retarding forces acting on the system and assuming that
the mass vibrates free of other external forces —
— we can equate
Newton’s second law with the net, or resultant, force of the restoring force and the
weight:
2

––– ϭ Ϫ ( ϩ ) ϩ
2

0
0

Direction below the
equilibrium position is positive
...


(1)

zero

The negative sign in (1) indicates that the restoring force of the spring acts opposite
to the direction of motion
...
See Figure 5
...
2
...
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...
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...
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...
For example, if 0 Ͼ 0, 1 Ͻ 0, the mass starts from a point
the
equilibrium position with an imparted
velocity
...
For example, if 0 Ͻ 0, 1 ϭ 0, the mass is released
from
from a point ͉ 0 ͉ units
the equilibrium position
...
3 we find the general solution of (2) to b
()ϭ

1 cos

␻ ϩ

2

sin ␻
...
The number represents the time
(measured in seconds) it takes the mass to execute one cycle of motion
...
From a graphical viewpoint ϭ 2p͞v seconds is the
length of the time interval between two successive maxima (or minima) of ( )
...
We refer to either case as an
of
the mass
...
For example, if ( ) ϭ 2 cos 3p Ϫ 4 sin 3p , then the period
is ϭ 2p͞3p ϭ 2͞3 s, and the frequency is ϭ 3͞2 cycles/s
...
The number ␻ ϭ 1 > (measured in radians per second) is called the
of the system
...

Finally, when the initial conditions are used to determine the constants 1 and 2 in (3),
we say that the resulting particular solution or response is the

(

)

A mass weighing 2 pounds stretches a spring 6 inches
...

3
Determine the equation of motion
...
ϭ 1 ft; 8 in
...

2
3
In addition, we must convert the units of weight given in pounds into units of mass
...
Also, from Hooke’s law, 2 ϭ ΂ 1΃
From ϭ ͞ we have
2
implies that the spring constant is ϭ 4 lb/ft
...


The initial displacement and initial velocity are (0) ϭ 2, Ј(0) ϭ Ϫ4, where the neg3
3
ative sign in the last condition is a consequence of the fact that the mass is given an
initial velocity in the negative, or upward, direction
...
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...
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...


Applying the initial conditions to ( ) and Ј( ) gives
equation of motion is
()ϭ

1

(4)
ϭ 2 and
3

2

ϭ Ϫ1
...

3
6

(5)

When 1 0 and 2 0, the actual
of free vibrations is not obvious from inspection of equation (3)
...
Hence it is often con3
venient to convert a solution of form (3) to the simpler form

where

ϭ 2

()ϭ
2
1

ϩ

2
2

sin(␻ ϩ ),

and f is a

(6)

defined b

·

ϭ

cos

1

ϭ

sin

2

tan

ϭ

1


...
1
...


ϭ

then (8) becomes

A relationship between
1 Ͼ 0, 2 Ͼ 0 and phase angle f

1

1
2
1

2
2

ϩ

2

cos ␻ ϩ

ϭ

1

,

cos

sin ␻ ϭ

1

ϭ

1

cos ␻ ϩ

2

2
2
1

ϩ

2
2

ϭ

2

(8)

,

sin ␻ ϭ ( )
...
Computation of the amplitude is straightforward,
ϭ 2 2 2 ϩ Ϫ1 2 ϭ 217 Ϸ 0
...
With 1 ϭ 2 and 2 ϭ Ϫ1 we find
3
6
tan f ϭ Ϫ4, and a calculator then gives tan Ϫ1(Ϫ4) ϭ Ϫ1
...
This is
the
phase angle, since tanϪ1(Ϫ4) is located in the
and therefore contradicts the fact that sin f Ͼ 0 and cos f Ͻ 0 because 1 Ͼ 0 and 2 Ͻ 0
...
326) ϭ 1
...

Thus (5) is the same as

()

( )

()ϭ
The period of this function is

117
sin(8 ϩ 1
...

6

(9)

ϭ 2p͞8 ϭ p͞4 s
...
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...
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...
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...
In this case the radian measured angle f is defined in slightly
1
2
different manner than in (7):
sin

·

2

ϭ

cos

tan

ϭ

1

ϭ

2


...

3
6
4
Because sin f Ͻ 0 and cos f Ͼ 0 the angle f lies in the fourth quadrant and so
rounded to three decimal places f ϭ tanϪ1(Ϫ1) ϭ Ϫ0
...
From (6Ј) we obtain
4
a second alternative form of solution (5):
()ϭ

217
cos(8 Ϫ (Ϫ0
...
245)
...
1
...
Reading left to right, the
first five positions (marked with black dots) correspond to the initial position of the
mass below the equilibrium position ϭ 2 , the mass passing through the equilibrium
3
position for the first time heading upward ( ϭ 0), the mass at its extreme displacement
above the equilibrium position ( ϭ Ϫ 117͞6), the mass at the equilibrium position
for the second time heading downward ( ϭ 0), and the mass at its extreme displacement below the equilibrium position ( ϭ 117͞6)
...
1
...
Note, however,
that in Figure 5
...
4(b) the positive direction in the -plane is the usual upward

(

negative

)

17
6

0
positive

0

0
17
6

2
3

(0, 2 )
3

amplitude

positive

17
6

0
negative
4

period

Simple harmonic motion

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...
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...
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...


●

direction and so is opposite to the positive direction indicated in Figure 5
...
4(a)
...
1
...
1
...

Form (6) is very useful because it is easy to find values of time for which
the graph of ( ) crosses the positive -axis (the line ϭ 0)
...

In the model discussed above we
assumed an ideal world — a world in which the physical characteristics of the spring
do not change over time
...
In one model for the
the spring constant in (1) is
replaced by the decreasing function ( ) ϭ Ϫa , Ͼ 0, a Ͼ 0
...
Nevertheless, we can obtain two linearly independent solutions using the methods in Chapter 6
...
1, Example 4 in
Section 6
...
4
...
The resulting model,
Љϩ
ϭ 0, is a form of
Like the equation for
an aging spring, Airy’s equation can be solved by the methods of Chapter 6
...
1, Example 5 in Section 6
...
4
...
Unless the mass is suspended in a perfect vacuum, there will be at least a resisting force due to the surrounding medium
...
1
...

In the study of mechanics, damping forces
acting on a body are considered to be proportional to a power of the instantaneous
velocity
...

Dividing (10) by the mass , we find that the differential equation of
is
2
b
ϩ
ϩ
ϭ0
2
2

or
Damping devices

where

2

ϩ2

2 ϭ

␤

,

ϩ ␻ 2 ϭ 0,

(11)

␻2 ϭ

(12)


...
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...
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...
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...


We can now distinguish three possible cases depending on the algebraic sign of
Ϫl
l2 Ϫ v 2
...

L ؊V Ͼ
In this situation the system is said to be
because the damping coefficient b is large when compared to the spring constant
The corresponding solution of (11) is ( ) ϭ 1 1 ϩ 2 2 or
()ϭ
Motion of an
overdamped system

Ϫ

1 2Ϫ␻2

(1

ϩ

Ϫ1 2 Ϫ ␻ 2

2

)
...
Figure 5
...
6 shows
two possible graphs of ( )
...
The general
solution of (11) is ( ) ϭ 1 1 ϩ 2 1 or
()ϭ

Motion of a critically
damped system

undamped
underdamped

(

Ϫ

1

ϩ

)
...
1
...
Notice that the motion is
quite similar to that of an overdamped system
...

ϭ Ϫ ϩ 1␻ 2 Ϫ

ϭ Ϫ Ϫ 1␻ 2 Ϫ

L ؊V Ͻ
In this case the system is said to be
since the damping coefficient is small in comparison to the spring constant
...



...
1
...


It is readily verified that the solution of the initial-value proble
2
2

is

ϩ5

ϩ 4 ϭ 0,
()ϭ

5
3

Ϫ

(0) ϭ 1,
Ϫ

2
3

Ϫ4


...
The mass is initially released from a position 1 unit
the equilibrium
position with a
velocity of 1 ft/s
...
Differentiating (16) gives Ј( ) ϭ Ϫ5 Ϫ ϩ 8 Ϫ4 , so Ј( ) ϭ 0 implies that
3
3

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...
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...


●

ϭ 8 or ϭ 1 ln 8 ϭ 0
...
It follows from the first derivative test, as well as our
5
3
5
physical intuition, that (0
...
069 ft is actually a maximum
...
069 feet below the equilibrium position
...
This cannot happen in this
instance because the equation ( ) ϭ 0, or 3 ϭ 2, has the physically irrelevant solu5
tion ϭ 1 ln 2 ϭ Ϫ0
...

3
5
The graph of ( ), along with some other pertinent data, is given in
Figure 5
...
9
...
Assuming that a damping force
numerically equal to 2 times the instantaneous velocity acts on the system, determine
the equation of motion if the mass is initially released from the equilibrium position
with an upward velocity of 3 ft/s
...
The differential equation of motion is then
4
1
4

2

2

ϭ Ϫ4 Ϫ 2

2

or

ϩ8

2

ϩ 16 ϭ 0
...



...
Thus the equation of motion is

1
4

0
...


1

ϭ0
(19)

To graph ( ), we proceed as in Example 3
...
The corresponding extreme displacement is
4
1
1 Ϫ1
ϭ Ϫ0
...
As shown in Figure 5
...
10, we interpret this value
4 ϭ Ϫ3 4
to mean that the mass reaches a maximum height of 0
...


()

()

A mass weighing 16 pounds is attached to a 5-foot-long spring
...
2 feet
...

The elongation of the spring after the mass is attached is 8
...
2 ft,
so it follows from Hooke’s law that 16 ϭ (3
...
In addition,
ϭ 16 ϭ 1 slug, so the differential equation is given by
32
2
1
2

2

2
2

or

ϭ Ϫ5 Ϫ

2

ϩ2

ϩ 10 ϭ 0
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


(21)

●

Finally, the initial conditions (0) ϭ Ϫ2 and Ј(0) ϭ 0 yield
so the equation of motion is
()ϭ

Ϫ

ϭ Ϫ2 and

1

΂Ϫ2 cos 3 Ϫ 2 sin 3 ΃
...

The coefficient Ϫl is sometimes called the
Because (23) is not a periodic function, the number 2␲ ͞1␻2 Ϫ 2 is called the
and 1␻2 Ϫ 2 ͞2␲ is the
The quasi period is the
time interval between two successive maxima of ( )
...
391
...


2

()ϭ

2110
3

Ϫ

sin(3 ϩ 4
...


Suppose we now take into consideration an external force ( ) acting on a vibrating mass on a spring
...
See Figure 5
...
11
...


(24)

ϩ ␻ 2 ϭ ( ),

(25)

Ϫ␤

gives
2

Oscillatory vertical
motion of the support

2

ϩ2

where ( ) ϭ ( )͞ and, as in the preceding section, 2l ϭ b͞ , v 2 ϭ ͞ To solve
the latter nonhomogeneous equation, we can use either the method of undetermined
coefficients or variation of parameters

Interpret and solve the initial-value problem
1
5

2
2

ϩ 1
...


(26)

We can interpret the problem to represent a vibrational system consisting of a mass ( ϭ 1 slug or kilogram) attached to a spring ( ϭ 2 lb/ft or N/m)
...
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...
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...
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...
The motion is damped (b ϭ 1
...
Intuitively, we would expect that even
with damping, the system would remain in motion until such time as the forcing
function was “turned off,” in which case the amplitudes would diminish
...

We first multiply the di ferential equation in (26) by 5 and solve
2
2

ϩ 10 ϭ 0

by the usual methods
...
Using the method of undetermined coefficients,
we assume a particular solution of the form ( ) ϭ cos 4 ϩ sin 4 Differentiating
( ) and substituting into the DE gives

x
steady state
xp (t)

1

ϩ6

Љ ϩ 6 Ј ϩ 10
t

ϭ (Ϫ6 ϩ 24 ) cos 4 ϩ (Ϫ24 Ϫ 6 ) sin 4 ϭ 25 cos 4
...


Ϫ24 Ϫ 6 ϭ 0

It follows that

transient

_1

()ϭ
/2

x(t)=transient
+ steady state

()ϭ

t

_1
/2

Graph of solution in
(28) of Example 6
x

(

1

cos ϩ

2

sin ) Ϫ

25
50
cos 4 ϩ
sin 4
...
By differentiating
51
the expression and then setting ϭ 0, we also find that 2 ϭ Ϫ86
...


(28)

When is a periodic function, such as
( ) ϭ 0 sin g or ( ) ϭ 0 cos g , the general solution of (25) for l Ͼ 0 is the sum
of a nonperiodic function ( ) and a periodic function ( )
...
Thus for large values of time, the displacements of the mass are closely approximated by the particular solution ( )
...
In the
particular solution (28), Ϫ3 38 cos Ϫ 86 sin is a transient term, and ( ) ϭ
51
51
25
Ϫ102 cos 4 ϩ 50 sin 4 is a steady-state term
...
1
...
1
...


(

)

The solution of the initial-value problem
2
2

t

x1=_3

where

1

ϩ2

ϩ 2 ϭ 4 cos ϩ 2 sin ,

1

Ϫ 2)

Ϫ

1,

sin ϩ 2 sin
...
1
...

The graphs show that the influence of the transient term is negligible for about
Ͼ 3p͞2
...
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...
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...
Also,
we shall see that a periodic impressed force with a frequency near or the same as the
frequency of free undamped vibrations can cause a severe problem in any oscillatory
mechanical system
...


The complementary function is ( ) ϭ 1 cos v ϩ
particular solution, we assume ( ) ϭ cos g ϩ sin g so that

Љ ϩ ␻2

(29)

2

sin v To obtain a

ϭ (␻ 2 Ϫ ␥ 2) cos ␥ ϩ (␻ 2 Ϫ ␥ 2) sin ␥ ϭ

Equating coefficients immediately gives

ϭ 0 and

0 ͞(v

ϭ

2

0

sin ␥
...
Therefore

0
sin ␥
...
Thus the solution is

()ϭ

()ϭ

1

cos ␻ ϩ

0

␻ (␻ 2 Ϫ ␥ 2)

2

sin ␻ ϩ

(Ϫ␥ sin ␻ ϩ ␻ sin ␥ ),

␻
...
This limiting process is analogous to “tuning in” the frequency of the driving
force (g ͞2p) to the frequency of free vibrations (v͞2p)
...
For g ϭ v we define the solution to b

( ) ϭ lim

␥ :␻

0

Ϫ␥ sin ␻ ϩ ␻ sin ␥
ϭ
␻ (␻ 2 Ϫ ␥ 2)

␥

(Ϫ␥ sin ␻ ϩ ␻ sin ␥ )

␥ :␻

(␻ 3 Ϫ ␻␥ 2)
␥
Ϫsin ␻ ϩ ␻ cos ␥
ϭ 0 lim
␥ :␻
Ϫ2␻␥

ϭ
ϭ

Pure resonance

0 lim

(31)

Ϫsin ␻ ϩ ␻ cos ␻
Ϫ2␻ 2

0

0

2␻ 2

sin ␻ Ϫ

0

2␻

cos ␻
...
The phenomenon that we have just described is known
as
The graph given in Figure 5
...
14 shows typical motion in this case
...
Alternatively, equation (31) follows
by solving the initial-value problem
2
2

ϩ ␻2 ϭ

0

sin ␻ ,

(0) ϭ 0,

Ј(0) ϭ 0

directly by conventional methods
...
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...
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...
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...
Large oscillations of the mass
would eventually force the spring beyond its elastic limit
...
1
...
Although it is true that
pure resonance cannot occur when the smallest amount of damping is taken into consideration, large and equally destructive amplitudes of vibration (although bounded
as : ϱ) can occur
...
1
...


ϩ

(32)

If ( ) denotes current in the
shown in Figure 5
...
15,
then the voltage drops across the inductor, resistor, and capacitor are as shown in
Figure 1
...
4
...


But the charge ( ) on the capacitor is related to the current ( ) by ϭ
(33) becomes the linear second-order differential equation
2
2

ϩ

ϩ

1

(33)
͞ , so

ϭ ( )
...

If ( ) ϭ 0, the
of the circuit are said to be
Because
2
the auxiliary equation for (34) is
ϩ
ϩ 1͞ ϭ 0, there will be three forms of
the solution with
0, depending on the value of the discriminant 2 Ϫ 4 ͞ We
say that the circuit is
2

if
and

if

Ϫ 4 / Ͼ 0,

2

Ϫ 4 / ϭ 0,

2

if

Ϫ 4 / Ͻ 0
...
In the underdamped case when (0) ϭ 0 , the charge on the
capacitor oscillates as it decays; in other words, the capacitor is charging and discharging as : ϱ
...
25 henry (h),
ϭ 10 ohms (⍀), ϭ 0
...

Since 1͞ ϭ 1000, equation (34) becomes
1
Љ ϩ 10 Ј ϩ 1000 ϭ 0
4

or

Љ ϩ 40 Ј ϩ 4000 ϭ 0
...
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...
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...
Applying the initial conditions, we find 1 ϭ 0 and 2 ϭ 1 0
...


Using (23), we can write the foregoing solution as
()ϭ

Ϫ20

3

sin(60 ϩ 1
...


When there is an impressed voltage ( ) on the circuit, the electrical vibrations
are said to be
In the case when
0, the complementary function ( ) of
(34) is called a
If ( ) is periodic or a constant, then the particular solution ( ) of (34) is a

Find the steady-state solution ( ) and the
circuit when the impressed voltage is ( ) ϭ
The steady-state solution

in an
0

-series

sin g

( ) is a particular solution of the differential

equation
2
2

ϩ

ϩ

1

ϭ

0

sin ␥
...


in terms of some new symbols
...


), so the steady-state charge is

0

␥

1
...


Now the steady-state current is given by ( ) ϭ Ј ( ):
The quantities ϭ g Ϫ 1͞ g and ϭ 1 2 ϩ 2 defined in Example 10 are
called the
and
respectively, of the circuit
...

()ϭ

0

΂

sin ␥ Ϫ

΃

cos ␥
...
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...
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...
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...
What is the period of simple
harmonic motion?
A 20-kilogram mass is attached to a spring
...
Initially, the mass is
released from rest from a point 3 inches above the equilibrium position
...

Determine the equation of motion if the mass in
Problem 3 is initially released from the equilibrium
position with a downward velocity of 2 ft/s
...

The mass is initially released from rest from a point
6 inches below the equilibrium position
...

What is the velocity of the mass when ϭ 3pͲ16 s?
In which direction is the mass heading at this instant?
At what times does the mass pass through the equilibrium position?
A force of 400 newtons stretches a spring 2 meters
...
Find the
equation of motion
...
A mass of 20 kilograms is attached to the second spring, and both masses
are initially released from the equilibrium position with
an upward velocity of 10 m/s
...

Determine the amplitude and period of motion if the
mass is initially released from a point 1 foot above the
equilibrium position with an upward velocity of 2 ft/s
...

When set in motion, the spring/mass system exhibits
simple harmonic motion
...

2
( ) Express the equation of motion in the form given
in (6)
...

A mass weighing 10 pounds stretches a spring 1 foot
...
6 slugs, which is initially released from a point 1 foot
3
above the equilibrium position with a downward velocity of 5 ft/s
...

( ) Express the equation of motion in the form given
in (6Ј)
( ) Use one of the solutions obtained in parts (a) and (b)
to determine the times the mass attains a displacement below the equilibrium position numerically
equal to 1 the amplitude of motion
...
32 foot
...

Find the equation of motion
...
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...
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...
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...
The mass is initially released
from a point 1 foot above the equilibrium position with
an upward velocity of 13 ft/s
...

Under some circumstances when two parallel springs,
with constants 1 and 2, support a single mass, the
of the system is given by
ϭ 4 1 2 ͞( 1 ϩ 2 )
...
The springs are attached to a common rigid
support and then to a metal plate
...
1
...
Determine the
effective spring constant of this system
...


Graph for Problem 17

Graph for Problem 18
1

2

20 lb

Double-spring system in
Problem 13

A certain mass stretches one spring 1 foot and another
3
spring 1 foot
...
1
...
The first mass is set aside, a mass weighing 8 pounds is attached to the double-spring arrangement, and the system is set in motion
...


Graph for Problem 19

A model of a spring/mass system is 4 Љ ϩ Ϫ0
...
By
inspection of the differential equation only, discuss the behavior of the system over a long period of time
...

By inspection of the differential equation only, discuss
the behavior of the system over a long period of time
...

Use the graph to determine
whether the initial displacement is above or below the
equilibrium position and
whether the mass is initially released from rest, heading
downward, or heading upward
...
The medium offers a damping force
that is numerically equal to the instantaneous velocity
...
Determine the time at which the mass passes
through the equilibrium position
...
What is the position of the mass at this
instant?

Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it
...
Find
the equation of motion if the mass is initially released
from the equilibrium position with a downward velocity
of 5 ft/s
...

What is the position of the mass at this instant?
A 1-kilogram mass is attached to a spring whose constant
is 16 N/m, and the entire system is then submerged in a
liquid that imparts a damping force numerically equal to
10 times the instantaneous velocity
...

In parts (a) and (b) of Problem 23 determine whether the
mass passes through the equilibrium position
...
What is the
position of the mass at this instant?
A force of 2 pounds stretches a spring 1 foot
...
2 pounds is attached to the spring, and the
system is then immersed in a medium that offers a
damping force that is numerically equal to 0
...

Find the equation of motion if the mass is initially
released from rest from a point 1 foot above the
equilibrium position
...

Find the first time at which the mass passes through
the equilibrium position heading upward
...
This mass is removed
and replaced with another mass that weighs 8 pounds
...

Find the equation of motion if the mass is initially
released from a point 1 foot below the equilibrium
2
position with a downward velocity of 1 ft/s
...

Find the times at which the mass passes through the
equilibrium position heading downward
...

A mass weighing 10 pounds stretches a spring 2 feet
...
Determine the values of the damping constant b so that the subsequent motion is
overdamped,
critically damped, and
underdamped
...

The subsequent motion takes place in medium that offers
a damping force numerically equal to b (b Ͼ 0) times
the instantaneous velocity
...

3

A mass weighing 16 pounds stretches a spring 8 feet
...
Find the equation of motion if the mass is
driven by an external force equal to ( ) ϭ 10 cos 3
A mass of 1 slug is attached to a spring whose constant
is 5 lb/ft
...

Find the equation of motion if the mass is driven by an
external force equal to ( ) ϭ 12 cos 2 ϩ 3 sin 2
Graph the transient and steady-state solutions on the
same coordinate axes
...

A mass of 1 slug, when attached to a spring, stretches it
2 feet and then comes to rest in the equilibrium position
...
Find the equation of motion if
the surrounding medium offers a damping force that is
numerically equal to 8 times the instantaneous velocity
...

When a mass of 2 kilograms is attached to a spring
whose constant is 32 N/m, it comes to rest in the equilibrium position
...
Find the
equation of motion in the absence of damping
...
What is the
amplitude of vibrations after a very long time?

Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

A mass is attached to the end of a spring whose constant is After the mass reaches equilibrium, its support
begins to oscillate vertically about a horizontal line
according to a formula ( )
...
1
...

Determine the differential equation of motion if
the entire system moves through a medium offering a damping force that is numerically equal to
b( ͞ )
...


Evaluate lim
␥ ␻
:

0

␻2 Ϫ ␥ 2

Compare the result obtained in part (b) of Problem 39
with the solution obtained using variation of parameters
when the external force is 0 cos v
Show that ( ) given in part (a) of Problem 39 can
be written in the form
()ϭ

Ϫ2 0
1
1
sin (␥ Ϫ ␻) sin (␥ ϩ ␻)
...
After the mass reaches equilibrium, its support oscillates according to the formula
( ) ϭ sin 8 , where represents displacement from its
original position
...
1
...

In the absence of damping, determine the equation
of motion if the mass starts from rest from the equilibrium position
...

In Problems 37 and 38 solve the given initial-value problem
...


Beats phenomenon in Problem 41

2

2

0

2␧␥

When ␧ is small, the frequency g ͞2p of the
impressed force is close to the frequency v ͞2p of
free vibrations
...
1
...
Oscillations of this
kind are called
and are due to the fact that
the frequency of sin ␧ is quite small in comparison to the frequency of sin g The dashed curves,
or envelope of the graph of ( ), are obtained from
the graphs of Ϯ( 0 ͞2␧g) sin ␧ Use a graphing
utility with various values of 0 , ␧, and g to verify
the graph in Figure 5
...
22
...


(0) ϭ 0,

Ј(0) ϭ 0

(cos ␥ Ϫ cos ␻ )
...

Show that the general solution of
2
2

ϩ2

ϩ ␻2 ϭ

0 sin

␥

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

is
()ϭ

1(␻ 2 Ϫ ␥ 2)2 ϩ 4 2␥ 2
Ϫl

sin΂2v2 Ϫ l2 ϩ f΃

2
2
where ϭ 1 1 ϩ 2 and the phase angles f
and u are, respectively, defined by sin f ϭ 1 ͞ ,
cos f ϭ 2 ͞ and

ϩ

0

sin ␪ ϭ
cos ␪ ϭ

sin(␥ ϩ ␪ ),

Ϫ2 ␥
,
1(␻ 2 Ϫ ␥ 2) 2 ϩ 4 2␥ 2
1(␻ 2

␻2 Ϫ ␥ 2

...
Inspection shows that ( ) is transient, and hence for large values of time, the solution
is approximated by ( ) ϭ (g) sin(g ϩ u), where
1(␻ 2

0

Although the amplitude (g) of ( ) is bounded as
: ϱ, show that the maximum oscillations will
occur at the value ␥1 ϭ 1␻ 2 Ϫ 2 2
...

( )ϭ

When

0

ϭ 2,

Ϫ

␥ 2) 2

ϩ4

1(4 Ϫ ␥ 2 )2 ϩ ␤ 2␥ 2

ϭ 1, and ϭ 4,

(␥) ϭ

2␥ 2

2


...


Construct a table of the values of g1 and (g1) corresponding to the damping coefficients b ϭ 2, b ϭ 1,
␤ ϭ 3, ␤ ϭ 1, and ␤ ϭ 1
...
Use the same coordinate axes
...
What is
g1 approaching as ␤ : 0? What is happening to the
resonance curve as ␤ : 0?
Consider a driven undamped spring/mass system
described by the initial-value problem
2
2

ϩ ␻2 ϭ

0 sin

␥,

(0) ϭ 0,

Ј(0) ϭ 0
...

For ϭ 3, discuss why there are two frequencies
g1 ͞2p and g 2 ͞2p at which the system is in pure
resonance
...
Use a numerical solver
to obtain the graph of the solution of the initial-value
problem for ϭ 2 and g ϭ g1 in part (a)
...


Find the charge on the capacitor in an
-series circuit
at ϭ 0
...
05 h, ϭ 2 ⍀, ϭ 0
...
Determine the
first time at which the charge on the capacitor is equal to
zero
...
Is the charge on the capacitor ever equal to zero?
In Problems 47 and 48 find the charge on the capacitor and
the current in the given
-series circuit
...

ϭ 5 h, ϭ 10 ⍀,
3
(0) ϭ 0 A

ϭ

ϭ 1 h,
ϭ 100 ⍀,
(0) ϭ 0 C, (0) ϭ 2 A

1
30

f, ( ) ϭ 300 V, (0) ϭ 0 C,
ϭ 0
...
25 f, and ( ) ϭ 50 cos V
...

Use Problem 50 to show that the steady-state current
in an
-series circuit when ϭ 1 h, ϭ 20 ⍀,
2
ϭ 0
...
160 sin(60 Ϫ 0
...

Find the steady-state current in an
-series
ϭ 1 h,
circuit when
ϭ 20 ⍀,
ϭ 0
...

Find the charge on the capacitor in an
-series circuit
when ϭ 1 h, ϭ 10 ⍀,
ϭ 0
...
What is the charge on the
capacitor after a long time?

Show that if , , , and 0 are constant, then the
amplitude of the steady-state current in Example 10 is a
maximum when ␥ ϭ 1> 1
...

Find the charge on the capacitor and the current in
an
series circuit when ϭ 0
...
1 f, ( ) ϭ
100 sin g V, (0) ϭ 0 C, and (0) ϭ 0 A
...

In Problem 57 find the current when the circuit is in
resonance
...
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...
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...
1 (page 117)
Problems 37–40 in Exercises 4
...
4

The preceding section was devoted to systems in which a second-order mathematical model was accompanied by initial conditions — that is, side conditions that are specified on
the unknown function and its first derivative at a single point
...


axis of symmetry

deflection curve

Many structures are constructed by using girders or
beams, and these beams deflect or distort under their own weight or under the influenc
of some external force
...

To begin, let us assume that a beam of length is homogeneous and has uniform
cross sections along its length
...
2
...
If a load is applied to the beam in a vertical plane containing the axis of symmetry, the beam, as shown in Figure 5
...
1(b),
undergoes a distortion, and the curve connecting the centroids of all cross sections is
called the
or
The deflection curve approximates the
shape of the beam
...
In the
theory of elasticity it is shown that the bending moment ( ) at a point along the
beam is related to the load per unit length ( ) by the equation

Deflection of
homogeneous beam

2

In addition, the bending moment
tic curve

( )
...
The product is called the
of the beam
...
When the
deflection ( ) is small, the slope Ј Ϸ 0, and so [1 ϩ ( Ј)2]3/2 Ϸ 1
...
The second derivative of this last expression is
2

2
2

ϭ

4

2

Љϭ

Using the given result in (1) to replace 2 ͞
satisfies the fourth-order di ferential equation

2

4


...


(4)

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Boundary conditions associated with equation (4) depend on how the ends of the
beam are supported
...
A diving board, an outstretched arm, an airplane wing, and a balcony are common examples of such beams, but even trees, flagpoles, skyscrapers,
and the George Washington Monument can act as cantilever beams because they
are embedded at one end and are subject to the bending force of the wind
...


3
The function ( ) ϭ
͞ ϭ
͞ 3 is called the shear force
...
Table 5
...
1 summarizes
the boundary conditions that are associated with (4)
...
2
...


0

Beams with various
end conditions

Ends of the Beam
embedded
free
simply supported
or hinged

Boundary Conditions
ϭ 0,
Љ ϭ 0,

Јϭ0
ٞϭ0

ϭ 0,

A beam of length is embedded at both ends
...


Љϭ0

From (4) we see that the deflection ( ) satisfie
4
0
...

Thus the boundary conditions are
(0) ϭ 0,

Ј(0) ϭ 0,

Ј( ) ϭ 0
...
Either way, we
find the general solution of the equation ϭ ϩ to be
( )ϭ

1

ϩ

ϩ

2

3

2

ϩ

4

3

ϩ

Now the conditions (0) ϭ 0 and Ј(0) ϭ 0 give, in turn,

0

24
1

2

2

3

ϩ

3

4

ϩ3

4

ϩ
2

0

4

0

6

3

3

2

2

ϭ 0, whereas the

ϩ

4

3

ϩ

0

24

ϭ0
ϭ 0
...


ϭ 0 and

remaining conditions ( ) ϭ 0 and Ј( ) ϭ 0 applied to ( ) ϭ
yield the simultaneous equations
3

4

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...
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...
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...


4

●

Solving this system gives

0
...
By choosing
24
deflection curve in Figure 5
...
3

or ( ) ϭ

͞12

4ϭϪ 0

0

0

Thus the deflection is

4

24

ϭ 24 , and

ϭ 1, we obtain the

Many applied problems demand that we
solve a two-point boundary-value problem (BVP) involving a linear differential
equation that contains a parameter l
...


Deflection curve fo
BVP in Example 1

Solve the boundary-value problem
ϭ 0,

Љϩ

( ) ϭ 0
...

For l ϭ 0 the solution of Љ ϭ 0 is ϭ 1 ϩ 2
...
Hence for l ϭ 0
the only solution of the boundary-value problem is the trivial solution ϭ 0
...
With this notation the roots of the auxiliary equation 2 Ϫ a 2 ϭ 0 are 1 ϭ a
and 2 ϭ Ϫa
...
Thus ϭ 2 sinh a The second condition,
( ) ϭ 0, demands that 2 sinh a ϭ 0
...
Again the only solution of the BVP is the trivial solution ϭ 0
...
Because the
auxiliary equation 2 ϩ a 2 ϭ 0 has complex roots 1 ϭ a and 2 ϭ Ϫ a, the
general solution of Љ ϩ a 2 ϭ 0 is ϭ 1 cos a ϩ 2 sin a As before, (0) ϭ 0
yields 1 ϭ 0, and so ϭ 2 sin a Now the last condition ( ) ϭ 0, or
2 sin

␣ ϭ 0,

is satisfied by choosing 2 ϭ 0
...
If we require
sin a ϭ 0 is satisfied whenever a is an integer multiple of p
...


Therefore for any real nonzero 2 , ϭ 2 sin( p ͞ ) is a solution of the problem for
each Because the differential equation is homogeneous, any constant multiple of a
solution is also a solution, so we may, if desired, simply take 2 ϭ 1
...
,
2

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...
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...


●

y
1

the corresponding function in the sequence
1

2

3
1

x

–1

4

5

Graphs of
eigenfunctions ϭ sin( p > ),
for ϭ 1, 2, 3, 4, 5

ϭ sin

␲

,

2

ϭ sin

2␲

is a nontrivial solution of the problem
ϭ 1, 2, 3,
...


,

ϭ sin

3

3␲

,
...
for which the boundary-value
problem in Example 2 possesses nontrivial solutions are known as
The
nontrivial solutions that depend on these values of l , ϭ 2 sin( p ͞ ) or simply
ϭ sin( p ͞ ), are called
The graphs of the eigenfunctions for
ϭ 1, 2, 3, 4, 5 are shown in Figure 5
...
4
...


It follows from Example 2 and the preceding disucussion that the boundary-value
problem
0

Љ ϩ 5 ϭ 0, (0) ϭ 0, ( ) ϭ 0
possesses only the trivial solution ϭ 0 because 5 is

an eigenvalue
...

Consider a long, slender vertical column of uniform cross section and length
...
2
...
By comparing bending moments at any point along the column, we obtain
2

2
2

Elastic column
buckling under a compressive force

or

ϭϪ

2

ϩ

ϭ 0,

(5)

where is Young’s modulus of elasticity and is the moment of inertia of a cross
section about a vertical line through its centroid
...


subjected to a

The boundary-value problem to be solved is
2
2

ϩ

ϭ 0,

(0) ϭ 0,

( ) ϭ 0
...
This solution has
a simple intuitive interpretation: If the load
is not great enough, there is no
deflection
...
From Case III of that discussion we see
that the deflections are ( ) ϭ 2 sin( p ͞ ) corresponding to the eigenvalues
l ϭ ͞ ϭ 2p 2 ͞ 2, ϭ 1, 2, 3,
...
These different forces are called

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...
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...


●

The deflection corresponding to the smallest critical load
called the
is 1( ) ϭ 2 sin(p ͞ ) and is known as the

1

ϭ p2

͞ 2,

The deflection curves in Example 4 corresponding to ϭ 1, ϭ 2, and ϭ 3
are shown in Figure 5
...
6
...
2
...
If restraints are put on
the column at ϭ ͞3 and at ϭ 2 ͞3, then the column will not buckle until the
critical load 3 ϭ 9p 2 ͞ 2 is applied, and the deflection curve will be as shown in
Figure 5
...
6(c)
...
2
...
In Section 5
...
It is apparent when the model for the deflection of a thin
column in (5) is written as 2 ͞ 2 ϩ ( ͞ ) ϭ 0 that it is the same as (6)
...
The physical situation is analogous to when two people hold a jump rope and twirl it in a synchronous
manner
...
2
...
2
...

Suppose a string of length with constant linear density r (mass per unit length)
is stretched along the -axis and fixed at ϭ 0 and ϭ Suppose the string is then
rotated about that axis at a constant angular speed v
...
If the magnitude of the tension ,
acting tangential to the string, is constant along the string, then the desired differential equation can be obtained by equating two different formulations of the net force
acting on the string on the interval [ , ϩ ⌬ ]
...
2
...


(7)

When angles u1 and u 2 (measured in radians) are small, we have sin u 2 Ϸ tan u 2 and
sin u1 Ϸ tan u1
...


Thus (7) becomes

1

Ϸ

Rotating string and
forces acting on it

[ Ј( ϩ ⌬ ) Ϫ Ј( )]
...

With ⌬ small we take ϭ Thus the net vertical force is also approximated by
Ϸ Ϫ( ␳ ⌬ ) ␻ 2,

(9)

where the minus sign comes from the fact that the acceleration points in the direction
opposite to the positive -direction
...


(10)

For ⌬ close to zero the difference quotient in (10) is approximately the second
derivative 2 ͞ 2
...


(11)

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...
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...


●

Since the string is anchored at its ends ϭ 0 and ϭ , we expect that the solution ( )
of equation (11) should also satisfy the boundary conditions (0) ϭ 0 and ( ) ϭ 0
...
See Problems 34–38 in Exercises 5
...

( ) Boundary conditions applied to a general solution of a linear differential
equation can lead to a homogeneous algebraic system of linear equations in
which the unknowns are the coefficients in the general solution
...
But a homogeneous system of linear
equations in unknowns has a nontrivial solution if and only if the determinant of the coefficients equals zero
...
2
...
The beam is of length , and 0 is a
constant
...

The beam is simply supported at both ends, and
( ) ϭ 0, 0 Ͻ Ͻ
Use a graphing utility to graph the deflection curve
when 0 ϭ 24 and ϭ 1
...

Use a graphing utility to graph the deflection curve
when 0 ϭ 48 and ϭ 1
...

Use a root-finding application of a CAS (or a
graphic calculator) to approximate the point in the
graph in part (b) at which the maximum deflectio
occurs
...

Use a root-finding application of a CAS (or a
graphic calculator) to approximate the point in the

graph in part (b) at which the maximum deflectio
occurs
...

How does the maximum deflection of a beam that is
half as long compare with the value in part (a)?
Find the maximum deflection of the simply supported beam in Problem 2
...
When the origin is taken at its free end,
as shown in Figure 5
...
8, the deflection ( ) of the beam
can be shown to satisfy the differential equation
Љϭ

Ϫ

( )
...


0

Deflection of cantilever beam in Problem 7

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...
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...


●

When a compressive instead of a tensile force is applied
at the free end of the beam in Problem 7, the differential
equation of the deflection i
ЉϭϪ
Solve this equation if
(0) ϭ 0, Ј( ) ϭ 0
...

2

Ϫ
( )ϭ

0

, 0 Ͻ Ͻ , and

0

In Problems 9 – 18 find the eigenvalues and eigenfunctions
for the given boundary-value problem
...


(0) ϭ 0,

Ј(0) ϭ 0,

(5) ϭ 0

Ј(1) ϭ 0

2

Љϩ

Ј ϩ l ϭ 0,

(1) ϭ 0,

2

Љϩ

Ј ϩ l ϭ 0,

Ј(

Ϫ1

) ϭ 0,

(

p

)ϭ0

(4)

(0) ϭ 0,

Љ(0) ϭ 0,

(1) ϭ 0,

(4)

Ј(0) ϭ 0,

ٞ(0) ϭ 0,

(p) ϭ 0,

Ϫ l ϭ 0,
Љ(p) ϭ 0

Consider Figure 5
...
6
...

The critical loads of thin columns depend on the end
conditions of the column
...
Suppose that a thin
vertical homogeneous column is embedded at its base
( ϭ 0) and free at its top ( ϭ ) and that a constant
axial load is applied to its free end
...
2
...
In either case the differential equation for the deflection ( ) is
2
2

ϩ

ϭ ␦
...
Consider only
the case l ϭ a 4, a Ͼ 0
...
In
general, the differential equation governing the deflectio
of the column is given by
2

΂

΃

2
2

2

ϩ

2

ϭ 0
...
Show that
the solution of this fourth-order differential equation
subject to the boundary conditions (0) ϭ 0, Љ(0) ϭ 0,
( ) ϭ 0, Љ( ) ϭ 0 is equivalent to the analysis in
Example 4
...

( ) Use a graphing utility to graph the first buckling
mode
...


For constant and r, define the critical speeds of angular rotation v as the values of v for which the boundaryvalue problem has nontrivial solutions
...


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...
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...


●

When the magnitude of tension is not constant, then a
model for the deflection curve or shape ( ) assumed by
a rotating string is given by

΄

( )


...
25, show that
the critical speeds of angular rotation are
␻ ϭ 1 2(4 2␲2 ϩ 1)> ␳ and the corresponding
2
deflections ar
2

2

2

Ϫ1/2

sin( p ln ),

ϭ 1, 2, 3,
...
Choose 2 ϭ 1
...
See
Figure 5
...
10
...
Solve for ( )
...

ln( > )

are constants
...


Suppose that 1 Ͻ Ͻ and that ( ) ϭ

( )ϭ

where

The model
Љϩ ϭ0
for simple harmonic motion, discussed in Section 5
...

Consider a free undamped spring/mass system for
which the spring constant is, say, ϭ 10 lb/ft
...

How many times will each mass
pass through the
equilibrium position in the time interval 0 Ͻ Ͻ 1?
Assume that the model for the
spring/mass system in Problem 29 is replaced by
Љϩ2 Јϩ

ϭ 0
...
With the same initial conditions and spring
constant as in Problem 29, investigate whether a mass
can be found that will pass through the equilibrium
position at ϭ 1 second
...


1

0

Љ ϩ 16 ϭ 0,

(0) ϭ

0,

(p͞2) ϭ

Љ ϩ 16 ϭ 0,

(0) ϭ 1, ( ) ϭ 1

1

Consider the boundary-value problem
Concentric spheres in Problem 27

The temperature ( ) in the
circular ring shown in Figure 5
...
11 is determined from
the boundary-value problem
2
2

ϩ

ϭ 0,

( )ϭ

0,

( )ϭ

1,

Љϩ

ϭ 0,

(Ϫ␲) ϭ (␲),

Ј(Ϫ␲) ϭ Ј(␲)
...

Find the eigenvalues and eigenfunctions of the
problem
...
Verify your geometric interpretation of
the boundary conditions given in part (a)
...
are the consecutive positive roots of
the equation tan a ϭ Ϫa
...
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...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...
Use a CAS to
approximate the first four eigenvalues l1, l2, l3, and l4
...
Explain why the negative roots of the
equation can be ignored
...


ϭ 0,

Љϩ
(4)

(0) ϭ 0,

(1) Ϫ 1 Ј(1) ϭ 0
2

Ϫ l ϭ 0, (0) ϭ 0, Ј(0) ϭ 0, (1) ϭ 0, Ј(1) ϭ 0
: Consider only l ϭ a 4, a Ͼ 0
...


●

Section 4
...
We are able to solve some of these models using the substitution method (leading to
reduction of the order of the DE) introduced on page 186
...
1 has the form
2
2

ϩ ( ) ϭ 0,

(1)

where ( ) ϭ
Because denotes the displacement of the mass from its equilibrium
position, ( ) ϭ is Hooke’s law — that is, the force exerted by the spring that tends
to restore the mass to the equilibrium position
...
Depending on how it is constructed and the material that is used, a spring
can range from “mushy,” or soft, to “stiff,” or hard, so its restorative force may vary
from something below to something above that given by the linear law
...
A spring whose
mathematical model incorporates a nonlinear restorative force, such as
2
2

ϩ

3

2

ϭ0

or

2

ϩ

ϩ

1

3

ϭ 0,

(2)

is called a
In addition, we examined mathematical models in which
damping imparted to the motion was proportional to the instantaneous velocity ͞
and the restoring force of a spring was given by the linear function ( ) ϭ
But these
were simply assumptions; in more realistic situations damping could be proportional to
some power of the instantaneous velocity ͞ The nonlinear differential equation
2
2

ϩ

͉ ͉

ϩ

ϭ0

(3)

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...
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...
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...


●

is one model of a free spring/mass system in which the damping force is proportional
to the square of the velocity
...
The point is that nonlinear characteristics of a physical system lead
to a mathematical model that is nonlinear
...


When the displacements are small, the values of
are negligible for sufficiently large
...
For the force at Ͼ 0,
( )ϭ
hard
spring

linear spring

0

ϩ

ϩ

1

2

2

ϩ

3

3

,

and for the force at Ϫ Ͻ 0,
(Ϫ ) ϭ

soft spring

0

Ϫ

1

ϩ

2

2

Ϫ

3

3

to have the same magnitude but act in the opposite direction, we must have
(Ϫ ) ϭ Ϫ ( )
...
Had we used only the first two terms in the
series, the same argument yields the linear function ( ) ϭ 1 A restoring force with
mixed powers, such as ( ) ϭ 1 ϩ 2 2, and the corresponding vibrations are said
to be

...

Hard and soft springs

Let us take a closer look at the equation in (1) in
the case in which the restoring force is given by ( ) ϭ
ϩ 1 3, Ͼ 0
...
Graphs of three types of
restoring forces are illustrated in Figure 5
...
1
...


x
x(0)=2,
x'(0)=_3

t

x(0)=2,
x'(0)=0

The differential equations
2
2

x

ϩ

3

ϭ0

(4)

2

ϩ

Ϫ

3

ϭ0

(5)

2

x(0)=2,
x'(0)=0

and

t
x(0)=2,
x'(0)=_3

Numerical solution
curves

ϩ

are special cases of the second equation in (2) and are models of a hard spring and
a soft spring, respectively
...
3
...
3
...
The
curves shown in red are solutions that satisfy the initial conditions (0) ϭ 2,
Ј(0) ϭ Ϫ3; the two curves in blue are solutions that satisfy (0) ϭ 2, Ј(0) ϭ 0
...
But we must be careful about drawing conclusions based on a
couple of numerical solution curves
...


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...
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...
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...


●

Any object that swings back and forth is called a
The
is a special case of the physical pendulum and consists of a rod of length to which a mass is attached at one end
...
The displacement angle u of the pendulum,
measured from the vertical as shown in Figure 5
...
3, is considered positive when
measured to the right of
and negative to the left of
Now recall the arc of a
circle of radius is related to the central angle u by the formula ϭ u
...

ϭ 2ϭ
2

sin

cos

Simple pendulum

From Newton’s second law we then have
2

ϭ

␪

ϭ

2


...
3
...
In direction this force is Ϫ sin u because it points
to the left for u Ͼ 0 and to the right for u Ͻ 0
...


ϩ

2

(6)

␪ (0)= , ␪ Ј(0)=2

Because of the presence of sin u, the model in (6) is nonlinear
...
For example, the
Maclaurin series for sin u is given by

t

␲

␪3 ␪5
...


Observe that this last equation is the same as the second nonlinear equation in (2) with
ϭ 1, ϭ ͞ , and 1 ϭ Ϫ ͞6 However, if we assume that the displacements u are
small enough to justify using the replacement sin u Ϸ u, then (6) becomes
2

␪
2

␪
␪Ј

ϭ
ϭ

␪
␪Ј

ϭ
ϭ

In Example 2,
oscillating pendulum in (b); whirling
pendulum in (c)

ϩ

␪ ϭ 0
...
3
...
1 that is a model for the free undamped vibrations of a linear spring/mass system
...
2
...
Because the general solution of (7) is u( ) ϭ
cos v ϩ 2 sin v , this linearization suggests that for initial conditions amenable
1
to small oscillations the motion of the pendulum described by (6) will be periodic
...
3
...
The blue
curve depicts the solution of (6) that satisfies the initial conditions
␪ (0) ϭ 1, ␪ Ј(0) ϭ 1, whereas the red curve is the solution of (6) that satisfie
2
2

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...
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...
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...


●

␪ (0) ϭ 1, uЈ(0) ϭ 2
...
3
...
The red curve shows that u increases without bound as time increases — the
pendulum, starting from the same initial displacement, is given an initial velocity of
magnitude great enough to send it over the top; in other words, the pendulum is
about its pivot as shown in Figure 5
...
4(c)
...

The first-order differential equation
͞ ϭ ͞ 1 is
equation (16) of Section 1
...
This differential equation, established with the aid
of Figure 1
...
8 on page 26, serves as a mathematical model for the shape of a flex
ible cable suspended between two vertical supports when the cable is carrying a
vertical load
...
2 we solved this simple DE under the assumption that
the vertical load carried by the cables of a suspension bridge was the weight of a
horizontal roadbed distributed evenly along the -axis
...
We are now in a position to determine the shape of a uniform flexible cable hanging only under its own weight, such as a wire strung between
two telephone posts
...
3
...

⌻1

and

2

(8)

is given by

΂ ΃

2

1ϩ

B

,

(9)

it follows from the fundamental theorem of calculus that the derivative of (9) is
ϭ

΂ ΃
...

2

1ϩ

(11)

In the example that follows we solve (11) and show that the curve assumed by
the suspended cable is a
Before proceeding, observe that the nonlinear
second-order differential equation (11) is one of those equations having the form
( , Ј, Љ) ϭ 0 discussed in Section 4
...
Recall that we have a chance of solving an
equation of this type by reducing the order of the equation by means of the substitution ϭ Ј
...
3
...

␳
If we substitute ϭ Ј, then the equation in (11) becomes
ϭ
11 ϩ 2
...


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...
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...
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...


●

Now, Ј(0) ϭ 0 is equivalent to (0) ϭ 0
...
Finally, by integrating both sides of
ϭ sinh

␳

,

we get

ϭ

1

1

␳

cosh

␳

ϩ

1

1

ϭ 0, so

2
...

Thus we see that the shape of the hanging wire is given by
ϭ ( 1> ␳) cosh( ␳ > 1) ϩ Ϫ 1> ␳
...

In (12) of Section 1
...


or simply

ϭϪ

2

ϭϪ ,

where represents the distance from the surface of the Earth to the object and the
positive direction is considered to be upward
...
3
...
If the positive direction is upward and air resistance is ignored, then the
differential equation of motion after fuel burnout is
2

2
2

ϭϪ

or

2

2

ϭϪ

2

,

(12)

where is a constant of proportionality, is the distance from the center of the
Earth to the rocket, is the mass of the Earth, and is the mass of the rocket
...


(13)

See Problem 14 in Exercises 5
...

Notice in the preceding discussion that we described the motion
of the rocket after it has burned all its fuel, when presumably its mass is constant
...
We saw in (17) of Exercises 1
...


(14)

is constant, then (14) yields the more familiar form ϭ
͞ ϭ
, where
is acceleration
...


If

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

5 lb
upward
force
()

Chain pulled upward
by a constant force in Example 4

A uniform 10-foot-long chain is coiled loosely on the ground
...
The chain weighs
1 pound per foot
...
3
...

Let us suppose that ϭ ( ) denotes the height of the end of the chain in
the air at time , ϭ ͞ , and the positive direction is upward
...


Thus from (14) we have
Product Rule

( )

––– –––
32
Because ϭ

ϭ5Ϫ

––– ϩ ––– ϭ 160 Ϫ 32
...


(16)

The nonlinear second-order differential equation (16) has the form ( , Ј, Љ) ϭ 0,
which is the second of the two forms considered in Section 4
...
To solve (16), we revert back to (15) and use ϭ Ј
along with the Chain Rule
...


(17)

On inspection (17) might appear intractable, since it cannot be characterized as any
of the first-order equations that were solved in Chapter 2
...
From ( Ϫ )͞ ϭ 1͞ we see from (13) of Section 2
...
When (18) is multiplied by m( ) ϭ , the
resulting equation is exact (verify)
...
4, we obtain
1
2

2 2

ϩ

32
3

3

Ϫ 80

2

ϭ

1
...
This last condition applied to (19) yields 1 ϭ 0
...

160 Ϫ
3
B

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

8
7
6
5
4
3
2
1

x

The last equation can be solved by separation of variables
...


(20)

This time the initial condition (0) ϭ 0 implies that 2 ϭ Ϫ3110͞8
...
5

1
...
5

()ϭ

t

Example 4

In addition to Problems 24 and 25, all
or portions of Problems 1 – 6, 8 –13, 15, 20, and 21 could
serve as Computer Lab Assignments
...
For each equation use a numerical
solver to plot the solution curves that satisfy the given initial
conditions
...

2

ϩ

3

ϭ 0,

(0) ϭ 1, Ј(0) ϭ 1;
2
2

ϩ 4 Ϫ 16

3

2
2

ϩ2 Ϫ

2

2

ϩ

0
...
Use a numerical solver to estimate the
smallest value of ͉ 1͉ at which the motion of the mass
is nonperiodic
...
Use
a numerical solver to estimate an interval Յ 0 Յ for
which the motion is oscillatory
...


(0) ϭ Ϫ 12, Ј(0) ϭ Ϫ1
...
Predict
the behavior of each system as : ϱ
...

2

ϩ

ϩ

ϩ

3

ϭ 0,

(0) ϭ Ϫ3, Ј(0) ϭ 4;
2

(0) ϭ

(21)

Consider the model of an undamped nonlinear
spring/mass system given by Љ ϩ 8 Ϫ 6 3 ϩ 5 ϭ 0
...

2

The graph of (21) given in Figure 5
...
7 should not, on physical grounds, be taken at
face value
...
3
...
Use a numerical solver to investigate the behavior of the system for
values of 1 Ͼ 0 ranging from 1 ϭ 0
...

State your conclusions
...

Consider the initial-value problem
Љϩ ϩ

1

3

ϭ cos 3 ,
2

Find values for
oscillatory
...


Ͻ 0 for which the system is

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Consider the model of the free damped nonlinear
pendulum given by
2

␪
2

␪

ϩ2

Determine how long it takes for the chain to fall
completely to the ground
...


Use a numerical solver to investigate whether the motion
in the two cases l2 Ϫ v 2 Ͼ 0 and l2 Ϫ v 2 Ͻ 0 corresponds, respectively, to the overdamped and underdamped
cases discussed in Section 5
...
For
l2 Ϫ v2 Ͼ 0, use lϭ2, vϭ1, u(0)ϭ1, and uЈ(0) ϭ 2
...

3

Use the substitution ϭ ͞ to solve (13) for in
terms of Assuming that the velocity of the rocket
at burnout is ϭ 0 and Ϸ at that instant, show
that the approximate value of the constant of
integration is ϭ Ϫ ϩ 1 02
...

[
: Take : ϱ and assume Ͼ 0 for all time ]
The result in part (b) holds for any body in the Solar
System
...

Find the escape velocity from the Moon if the
acceleration of gravity is 0
...


In a naval exercise a ship 1 is pursued
by a submarine 2 as shown in Figure 5
...
8
...
The submarine 2 keeps ship 1 in visual contact, indicated by the
straight dashed line in the figure while traveling at a constant speed 2 along a curve Assume that ship 2 starts
at the point ( , 0), Ͼ 0, at ϭ 0 and that is tangent to
Determine a mathematical model that describes the
curve
Find an explicit solution of the differential equation
...

Determine whether the paths of 1 and 2 will ever
intersect by considering the cases Ͼ 1, Ͻ 1, and ϭ 1
...
3
...

Why would you expect ( ) to be a periodic solution?
A uniform chain of length , measured in feet, is held
vertically so that the lower end just touches the floo
...
The upper end that is held is
released from rest at ϭ 0 and the chain falls straight
down
...

in terms of

Pursuit curve in Problem 17

In another naval exercise a destroyer
pursues a submerged submarine 2
...
The captain of the destroyer
1 assumes that the submarine will take immediate evasive action and conjectures that its likely new course is
the straight line indicated in Figure 5
...
9
...

Explain why the captain waits until 1 reaches (3, 0)
before ordering a course change to
Using polar coordinates, find an equation ϭ (u)
for the curve
Let denote the time, measured from the initial
detection, at which the destroyer intercepts the submarine
...
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...
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...
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...

Invented in 1742 by the English engineer Benjamin
Robins, the ballistic pendulum is simply a plane pendulum consisting of a rod of negligible mass to which a
block of wood of mass
is attached
...
In (7) of this section, we saw that in the
case of small oscillations, the angular displacement u( )
of a plane pendulum shown in Figure 5
...
3 is given by
the linear DE uЉ ϩ ( > )u ϭ 0, where u Ͼ 0 corresponds to motion to the right of vertical
...
3
...

Intuitively, the horizontal velocity of the combined mass (wood plus bullet) after impact is only a
fraction of the velocity of the bullet, that is,
ϩ

΃


...


Solve the initial-value problem
2

u
2

ϩ

ϭ

΂

ϩ

΃ 22


...


and

␪ max

Ballistic pendulum in Problem 19

As shown in Figure 5
...
11, a plane
flying horizontally at a constant speed 0 drops a relief
supply pack to a person on the ground
...
Under the assumption
that the horizontal and vertical components of the air
resistance are proportional to ( > )2 and ( > )2, respectively, and if the position of the supply pack is
given by ( ) ϭ ( ) ϩ ( ) , then its velocity is
( ) ϭ ( > ) ϩ ( > )
...
By differentiating the last
formula with respect to time , it follows that the
angular velocity v of the mass and its linear velocity
are related by ϭ v
...


ϩ

(3, 0)

΂

ϩ

Use Figure 5
...
10 to express cos umax in terms of
and
...

Finally, show that
is given (approximately) by

2

ϭ

΂

u ϭ 0, u(0) ϭ 0, uЈ(0) ϭ v0
...


2

2

2

0

2

Solve both of the foregoing initial-value problems
by means of the substitutions ϭ > , ϭ > ,
and separation of variables
...
2
...
Assume that the
constant of proportionality for air resistance is
ϭ 0
...

Use a root-finding application of a CAS or a graphic

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

calculator to determine the horizontal distance the
pack travels, measured from its point of release to
the point where it hits the ground
...


Consider the initial-value problem
2

␪

ϩ sin ␪ ϭ 0, ␪ (0) ϭ

2

supply
pack

Airplane drop in Problem 20

Discuss why the damping term in equation (3) is
written as

͉ ͉

΂ ΃
...

Then use a graphing utility to plot the graphs of
ϭ and ϭ sin on the same coordinate axes
for 0 Յ Յ p͞2
...
Since we cannot solve the
differential equation, we can find no explicit solution of
this problem
...
3
...

In this problem and the next we examine several ways
to proceed
...
Then plot solution curves of the
initial-value problems for several values of u 0 for
which u 0 Ͼ u 1
...
Use a numerical solver to generate a
numerical solution curve for the nonlinear model
(6) subject to the initial conditions u(0) ϭ 1,
uЈ(0) ϭ 2
...
165 for the acceleration of gravity on the Moon
...
Which pendulum has the
greater amplitude of motion?

␲
,
12

1
␪ Ј(0) ϭ Ϫ
...
10 to find the first four nonzero terms of a
Taylor series solution u( ) centered at 0 for the nonlinear initial-value problem
...

Use the first three terms of the Taylor series in
part (b) to approximate 1
...

In this part of the problem you are led through
the commands in
that enable you to
approximate the root 1
...
(
)
Precisely reproduce and then, in turn, execute each
line in the given sequence of commands
...


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...
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...
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...


●

Consider a pendulum that is released from rest from an
initial displacement of u 0 radians
...
The period of
oscillations predicted by this model is given by the
familiar formula ϭ 2␲ ͞1 / ϭ 2␲ 1 /
...
In other words, the linear model predicts that the
time it would take the pendulum to swing from an initial displacement of, say, u 0 ϭ p͞2 (ϭ 90°) to Ϫp͞2
and back again would be exactly the same as the time
it would take to cycle from, say, u 0 ϭ p ͞360 (ϭ 0
...
This is intuitively unreasonable; the actual
period must depend on u 0
...

Let us compare this last number with the period

Answer Problems 1–8 without referring back to the text
...

If a mass weighing 10 pounds stretches a spring
2
...

The period of simple harmonic motion of mass weighing 8 pounds attached to a spring whose constant is
6
...

The differential equation of a spring/mass system is
Љ ϩ 16 ϭ 0
...

Pure resonance cannot take place in the presence of a
damping force
...

A mass on a spring whose motion is critically damped
can possibly pass through the equilibrium position
twice
...

If

simple harmonic motion is described by
ϭ (12>2)sin(2 ϩ f), the phase angle f is
when the initial conditions are (0) ϭ Ϫ1 and Ј(0) ϭ 1
...
, and ϭ cos ,
respectively
...


predicted by the nonlinear model when u 0 ϭ p͞4
...
As in Problem 25, if 1 denotes
the first time the pendulum reaches the position
in
Figure 5
...
3, then the period of the nonlinear pendulum is
4 1
...

Experiment with small step sizes and advance the time,
starting at ϭ 0 and ending at ϭ 2
...
Use the value 1 to determine
the true value of the period of the nonlinear pendulum
...


A solution of the BVP when l ϭ 8 is
because

...

A free undamped spring/mass system oscillates with a
period of 3 seconds
...
What
was the weight of the original mass on the spring?
A mass weighing 12 pounds stretches a spring 2 feet
...

Find the equation of motion
...
With one
end held fixed, a mass weighing 8 pounds is attached
to the other end
...
Initially, the mass is displaced
4 inches above the equilibrium position and released
from rest
...


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...
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...
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...


●

A mass weighing 32 pounds stretches a spring 6 inches
...
Determine the values of b Ͼ 0 for which
the spring/mass system will exhibit oscillatory motion
...
If a mass is suspended from the
spring, determine the values of for which the subsequent free motion is nonoscillatory
...
Determine the maximum vertical
displacement of the mass
...

A periodic force equal to ( ) ϭ cos g ϩ sin g is
impressed on the system starting at ϭ 0
...

A mass weighing 4 pounds is suspended from a spring
whose constant is 3 lb/ft
...
Beginning at ϭ 0, an
external force equal to ( ) ϭ Ϫ is impressed on the
system
...

Two springs are attached in series as shown in
Figure 5
...
1
...

A mass weighing
pounds stretches a spring
1
1
2 foot and stretches a different spring 4 foot
...
R
...
Assume
that the motion is free and that there is no damping
force present
...

Show that the maximum speed of the mass is
2
3 23 ϩ 1
...
Initially, the charge
and
current are zero
...

Determine the current ( )
...

Show that the current ( ) in an
-series circuit
2
1
satisfies
ϩ
ϩ
ϭ Ј( ), where Ј( )
2
denotes the derivative of ( )
...
If (0) ϭ 0 and (0) ϭ 0 ,
what is Ј(0)?
Consider the boundary-value problem
Љϩ

ϭ 0,

(0) ϭ (2␲),

Ј(0) ϭ Ј(2␲)
...

A bead is constrained to slide along a frictionless rod of
length The rod is rotating in a vertical plane with a
constant angular velocity v about a pivot fixed at the
midpoint of the rod, but the design of the pivot allows
the bead to move along the entire length of the rod
...
R
...
To
apply Newton’s second law of motion to this rotating
frame of reference, it is necessary to use the fact that the
net force acting on the bead is the sum of the real forces
(in this case, the force due to gravity) and the inertial
forces (coriolis, transverse, and centrifugal)
...


Solve the foregoing DE subject to the initial
conditions (0) ϭ 0 , Ј(0) ϭ 0
...
What is the minimum length of the rod for which it can accommodate simple harmonic motion of the bead?
For initial conditions other than those obtained in
part (b), the bead must eventually fly off the rod
...

Suppose v ϭ 1 rad/s
...
1,
and 17
...
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...
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...
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...
For each
pair of initial conditions in part (d), use a rootfinding application to find the total time that the
bead stays on the rod
...
R
...
If the spring constants are 1 and 2, determine a differential equation for the displacement ( ) of
the freely sliding mass
...
In Figure 5
...
3(a) the mass is shown at the equilibrium position
ϭ 0, that is, the spring is neither
stretched nor compressed
...
R
...

Determine a differential equation for the displacement
( ) of the freely sliding mass
...
1
...
If the retarding force of
,
kinetic friction has the constant magnitude ϭ
where
is the weight of the mass, and acts opposite
to the direction of motion, then it is known as
By using the
sgn( Ј) ϭ

ΆϪ1,
1,

Ј Ͻ 0 (motion to left)
Ј Ͼ 0 (motion to right)

determine a piecewise-defined differential equation for
the displacement ( ) of the damped sliding mass
...

Find the displacement ( ) of the mass if it is released from rest from a point 5
...
5, Ј(0) ϭ 0
...

Give a time interval [0, 1] over which this solution
is defined
...
Using initial conditions at 1, find ( ) and give
a time interval [ 1, 2] over which this solution is
defined
...

Using initial conditions at 2, find ( ) and give a
time interval [ 2, 3] over which this solution is
defined
...

Graph the displacement ( ) on the interval [0, 3]
...
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...
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...
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...
1
6
...
3
6
...
The only
exception was the Cauchy-Euler equation in Section 4
...
In applications, higherorder linear equations with variable coefficients are just as important as, if not mor
than, differential equations with constant coefficients
...
7,
even a simple linear second-order equation with variable coefficients such a
Љ ϩ ϭ 0 does not possess solutions that are elementary functions
...
In Sections 6
...
4 we shall see that the functions that are solutions of this
equation are defined by infinite series
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

●

●

Infinite series of constants, -series, harmonic series, alternating harmonic series, geometric
series, tests for convergence especially the ratio test
Power series, Taylor series, Maclaurin series (See any calculus text)

In Section 4
...
By finding the roots of the auxiliary equation, we
could write a general solution of the DE as a linear combination of the elementary functions
a
, a , a cos b , and a sin b
...
7,
most linear higher-order DEs with variable coefficients cannot be solved in terms of elementary
functions
...
4)
...
2 we consider linear second-order DEs with variable coefficients that
possess solutions in the form of a power series, and so it is appropriate that we begin this chapter
with a review of that topic
...


Such a series also said to be a
For example, the power
series ͚ϱϭ0 ( ϩ 1) is centered at ϭ Ϫ1
...
For example,
ϱ

͚2
ϭ0

ϭ1ϩ2 ϩ4

2

ϩ
...

The following bulleted list summarizes some important
facts about power series ͚ϱϭ0 ( Ϫ )
...
If the limit does not exist at , then the series
:ϱ
is said to be

...
The center of the interval of convergence is the center of the series
...
If the series
converges only at its center , then ϭ 0
...
Recall, the absolute-value inequality ͉ Ϫ ͉ Ͻ is
equivalent to the simultaneous inequality Ϫ Ͻ Ͻ ϩ
...

•
Within its interval of convergence a power series
In other words, if is in the interval of convergence
and is not an endpoint of the interval, then the series of absolute values
͚ϱϭ0͉ ( Ϫ ) ͉ converges
...
1
...


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...
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...
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...


●

•

Convergence of power series can often be determined by the
0 for all in ͚ϱϭ0 ( Ϫ ) , and that
Suppose
lim
:ϱ

͉

ϩ1(

Ϫ )
( Ϫ )

ϩ1

͉ϭ ͉

Ϫ ͉ lim
:ϱ

͉ ͉ϭ
ϩ1


...
The ratio test is always inconclusive at an
endpoint

...

2

The ratio test gives

͉

͉

( Ϫ 3) ϩ1
lim 2 ϩ1 ( ϩ 1)
ϭ ͉ Ϫ 3 ͉ lim
:ϱ
:ϱ
( Ϫ 3)
2

ϩ1 1
ϭ ͉ Ϫ 3 ͉
...

2
This last inequality defines the
interval of convergence
...
At the left endpoint ϭ 1 of the open
interval of convergence, the series of constants ͚ϱϭ1 ((Ϫ1) > ) is convergent by
the alternating series test
...
The interval of convergence of the series is [1, 5), and the
radius of convergence is ϭ 2
...
If the radius of convergence is Ͼ 0 or ϭ ϱ,
then is continuous, differentiable, and integrable on the intervals
( Ϫ , ϩ ) or (Ϫϱ, ϱ), respectively
...
Convergence at an
endpoint may be either lost by differentiation or gained through integration
...


Ϫ1
is a power series in , then the first two derivatives are Ј ϭ ͚ϱϭ0
and
ϱ
Ϫ2
Љ ϭ ͚ ϭ0 ( Ϫ 1)

...
We omit these zero
terms and write
ϱ

Јϭ

͚
ϭ1

Ϫ1

ϭ

1

ϩ2

2

ϩ3

3

2

ϩ4

4

3

(1)

ϱ

Љϭ

͚
ϭ2

( Ϫ 1)

Ϫ2

ϭ2

2

ϩ6

3

ϩ 12

ϩ
...


Be sure you understand the two results given in (1); especially note where
the index of summation starts in each series
...

•
If ͚ϱϭ0 ( Ϫ ) ϭ 0, Ͼ 0, for all numbers in
some open interval, then ϭ 0 for all
...
In calculus it is seen that infinitel

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...
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...
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...


●

differentiable functions such as , sin , cos ,
be represented by Taylor series
ϱ

͚
ϭ0

ln(1 ϩ ), and so on, can

( )

Љ( )
( )
Ј( )
( Ϫ ) ϭ ( )ϩ
( Ϫ )ϩ
( Ϫ )2 ϩ
...


You might remember some of the following Maclaurin series representations
...
ϭ

͚ (2
ϭ0

1

ϱ

ϱ

ϩ
...
ϭ

ϱ

(Ϫϱ, ϱ)

(Ϫ1)
ϩ 1)!

ϱ

ϩ
...
ϭ

7

4

ϩ

3

sinh ϭ

7!
7

Ϫ

1
!

ϩ
...
ϭ

6!

5

ϩ

2

cosh ϭ 1 ϩ

6

Ϫ

5

ϩ

ϩ
...
For example, if we wish to find the Maclaurin series representatio
2
of, say, we need only replace in the Maclaurin series for :
2

2

ϭ1ϩ

1!

4

ϩ

2!

6

ϩ

3!

ϩ
...


Similarly, to obtain a Taylor series representation of ln centered at
we replace by Ϫ 1 in the Maclaurin series for ln(1 ϩ ):
ln ϭ ln(1 ϩ ( Ϫ 1)) ϭ ( Ϫ 1) Ϫ

( Ϫ 1)2 ( Ϫ 1)3 ( Ϫ 1)4
...

2

᭤

The interval of convergence for the power series representation of is the
same as that of , that is, (Ϫϱ, ϱ)
...

•
Power series can be combined through the
operations of addition, multiplication, and division
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Find a power series representation of

sin
...
΃
2
6
24
1 1
1 1
1
1
1
ϩ (1) ϩ ΂Ϫ ϩ ΃ ϩ ΂Ϫ ϩ ΃ ϩ ΂
Ϫ ϩ
6 2
6 6
120 12 24΃
2

sin ϭ 1 ϩ
ϭ (1)

ϭ

ϩ

and sin :

3

4

ϩ

ϩ

ϩ

2

2

3

5

3

3

ϩ

3

ϩ
...


ϩ
...
Problems involving multiplication or division of
power series can be done with minimal fuss using a computer algebra system
...

As the next example illustrates, combining two or more summations as a single summation often requires a reindexing, that is, a shift in the index of summation
...

In order to add the two series given in summation notation, it is necessary that both indices of summation start with the same number and that the powers
of in each series be “in phase,” in other words, if one series starts with a multiple
of, say, to the first power, then we want the other series to start with the same power
...
By writing the first term of the first series outside of the summation notation,
series starts
with
for ϭ 3
ϱ

͚
ϭ2

( Ϫ 1)

ϱ

Ϫ2

ϩ͚

ϩ1

ϭ0

ϭ2и1

2

0

ϱ

series starts
with
for ϭ 0
ϱ

ϩ ͚ ( Ϫ 1)

Ϫ2

ϭ3

ϩ͚

ϩ1

(3)

ϭ0

we see that both series on the right side start with the same power of , namely, 1
...

For ϭ 3 in ϭ Ϫ 2 we get ϭ 1, and for ϭ 0 in ϭ ϩ 1 we get ϭ 1, and
so the right-hand side of (3) becomes
same

2

ϱ

2

ϩ ͚ ( ϩ 2)( ϩ 1)
ϭ1

same

ϱ

ϩ2

ϩ͚

Ϫ1


...
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...
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...
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...
In both cases takes
on the same successive values ϭ 1, 2, 3,
...
for ϭ Ϫ 1 and ϭ 0, 1, 2,
...
We are now in a
position to add the series in (4) term-by-term:
ϱ

͚ (
ϭ2

ϱ

Ϫ 1)

Ϫ2

ϱ

ϩ͚

ϩ1

ϭ2 2ϩ

ϭ0

͚ [(
ϭ1

ϩ 2)( ϩ 1)

ϩ2 ϩ

Ϫ1]


...

The point of this section is to remind you of the salient facts about
power series so that you are comfortable using power series in the next section to fin
solutions of linear second-order DEs
...
2
...
Also suspend, for the sake of illustration, the fact that you already
know how to solve the given equation by the integrating-factor method in Section 2
...


ϱ

Find a power series solution ϭ

͚
ϭ0

of the differential equation Ј ϩ

ϭ 0
...

( ) First calculate the derivative of the assumed solution:
ϱ

Јϭ

͚
ϭ1

Ϫ1

; see the first line in (1)

( ) Then substitute and Ј into the given DE:
ϱ

Јϩ

ϭ

ϱ

͚
ϭ1

Ϫ1

͚
ϭ0

ϩ


...
When the indices of summation have the
same starting point and the powers of agree, combine the summations:
ϱ

Јϩ

ϭ

ϱ

͚
ϭ1

Ϫ1

ϩ

͚
ϭ0

ϭ Ϫ1

ϭ

ϱ

ϭ

͚
ϭ0

ϭ

ϱ

͚[
ϭ0

ϩ1(

ϩ 1)

ϩ

͚
ϭ0

ϱ

ϩ1(

ϩ 1) ϩ

]
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

( ) By letting take on successive integer values starting with ϭ 0, we fin
1

2

3

4

1
ϭϪ
1
1
ϭϪ
2
1
ϭϪ
3
1
ϭϪ
4

0

1

2

2

ϭϪ

0

1
1
ϭ Ϫ (Ϫ 0) ϭ 0
2
2
1 1
1
ϭϪ
0 ϭ Ϫ
3 2
3ؒ2

΂ ΃
1
1
1
ϭ Ϫ ΂Ϫ
ϭ
4 3ؒ2 ΃ 4ؒ3ؒ2
0

0

0

and so on, where 0 is arbitrary
...


ϩ

0

1
4ؒ3ؒ2

1
4ؒ3ؒ2

4

4

Ϫ
...


It should be fairly obvious that the pattern of the coefficients in part ( ) is
ϭ 0(Ϫ1) > !, ϭ 0, 1, 2,
...


(8)

From the first power series representation in (2) the solution in (8) is recognized
as ϭ 0 Ϫ
...
3, you would have found that
ϭ Ϫ is a solution of Ј ϩ ϭ 0 on the interval (Ϫϱ, ϱ)
...


In Problems 1–10 find the interval and radius of convergence
for the given power series
...
Write your answer in
summation notation
...
Write your answer in summation notation
...
]
: ϭ 2[1 ϩ ( Ϫ 2)>2]]

In Problems 19 and 20 the given function is analytic at
ϭ 0
...

sin cos

Ϫ

cos

In Problems 21 and 22 the given function is analytic at
ϭ 0
...

sec

tan

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●

In Problems 31–34 verify by direct substitution that the
given power series is a solution of the indicated differential
equation
...
]

In Problems 23 and 24 use a substitution to shift the summation index so that the general term of given power series
involves
...


ϱ

ϩ

ϭ1

ϱ

͚
ϭ2

(1 ϩ

order differential equation
...


͚
ϭ1

(Ϫ1)
!

ϭ

ϩ2

ϱ

Ϫ2

ϩ 3͚

In Problem 21, what do you think is the interval of convergence for the Maclaurin series of sec ?

ϭ1

Power series, analytic at a point, shifting the index of summation in Section 6
...
In this section we turn to the more important
problem of finding power series solutions of linear second-order equations
...
We begin with the definition of an
ordinary point
...


(2)

We have the following definition

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●

A point ϭ 0 is said to be an
of the differential of the
differential equation (1) if both coefficients ( ) and ( ) in the standard
form (2) are analytic at 0
...


A homogeneous linear second-order differential equation with constant coefficients
such as
ϭ0

Љϩ

and

Љ ϩ 3 Ј ϩ 2 ϭ 0,

can have no singular points
...

Every finite value of is an ordinary point of the differential equation
Ј ϩ (sin ) ϭ 0
...
1 that both and sin are analytic at this point
...
2
...


0

( ) The differential equation
Љϩ

Ј ϩ (ln ) ϭ 0

is already in standard form
...


Now ( ) ϭ is analytic at every real number, and ( ) ϭ ln is analytic at every
real number
...
We
conclude that ϭ 0 is a singular point of the DE
...


ϭ 0,
ϭ 0
...
A polynomial function is analytic at any value of , and a rational function is
analytic except at points where its denominator is zero
...
It is possible for a
ODE to have, say, a singular point at infinit
...
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...
It follows, then, that

(1)
(1)

2( 0)

2( 0)

ϭ 0
...
All other values of are ordinary points
...


ϭ 0
...
The equation
(

2

ϩ 1) Љ ϩ

ϭ0

ЈϪ

2

has singular points at the solutions of ϩ 1 ϭ 0—namely, ϭ
of , real or complex, are ordinary points
...
All other values

We state the following theorem about the existence of power series solutions
without proof
...


A power series solution converges at least on some interval defined by
͉ Ϫ 0 Ͻ , where is the distance from 0 to the closest singular point
...
2
...


value or

Find the minimum radius of convergence of a power series solution of the secondorder differential equation
(

2

Ϫ 2 ϩ 5) Љ ϩ

about the ordinary point ϭ 0,

ЈϪ

ϭ0

about the ordinary point ϭ Ϫ1
...


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...
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...


●

Because ϭ 0 is an ordinary point of the DE, Theorem 6
...
1 guarantees that we
can find two power series solutions centered at 0
...
See Figure 6
...
1
...
2
...
Each of
power series converges at least for ͉ ϩ 1 ͉ Ͻ 212 since the distance from each of
the singular points to Ϫ1 (the point (Ϫ1, 0)) is ϭ 18 ϭ 212
...


In the examples that follow as well as in the problems of Exercises 6
...
If it is necessary to find a power series solutions of an ODE about an
ordinary point 0 0, we can simply make the change of variable ϭ Ϫ 0 in the
equation (this translates ϭ 0 to ϭ 0), find solutions of the new equation of the
form ϭ ͚ϱϭ0
, and then resubstitute ϭ Ϫ 0
...
4
...
1 here,
in brief, is the idea
...
1, and then equate the all coefficients to
the right-hand side of the equation to determine the coefficients
...
1, that all coefficients of must be equated to zero
...
2
...
We will see in Example 5 how the single
assumption that ϭ ͚ϱϭ0
ϭ 0 ϩ 1 ϩ 2 2 ϩ
...
The general solution of the differential equation is
ϭ 1 1( ) ϩ 2 2( ); indeed, it can be shown that 1 ϭ 0 and 2 ϭ 1
...


Solve Љ ϩ

Since there are no singular points, Theorem 6
...
1 guarantees two
power series solutions centered at 0 that converge for ͉ ͉ Ͻ ϱ
...
1) into the differential equation give
ϱ

Љϩ

ϭ

͚
ϭ2

ϱ

( Ϫ 1)

Ϫ2

ϩ

͚
ϭ0

ϱ

ϭ

͚
ϭ2

ϱ

( Ϫ 1)

Ϫ2

ϩ͚


...
From the result given in (5) of Section 6
...


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...
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...


(4)

●

At this point we invoke the identity property
...


(5)

ϭ 0 obviously dictates that 2 ϭ 0
...


(6)

This relation generates consecutive coefficients of the assumed solution one at a time
as we let take on the successive integers indicated in (6):
ϭ 1,

3

ϭϪ

ϭ 2,

4

ϭϪ

ϭ 3,

5

ϭϪ

ϭ 4,

6

0

2ؒ3
1

3ؒ4
2

4ؒ5

ϭ0

;

ϭ 5,

3

5ؒ6

5

ϭ 6,

8

ϭϪ

ϭ 7,

9

ϭϪ

ϭ 8,

10

ϭϪ

ϭ 9,

11

ϭϪ

7ؒ8

ϭ

is zero

8

is zero

1

1
ϭϪ
8ؒ9
2ؒ3ؒ5ؒ6ؒ8ؒ9
6

0

1
ϭϪ
9 ؒ 10
3 ؒ 4 ؒ 6 ؒ 7 ؒ 9 ؒ 10
7

8

5

0

ϭ0

10 ؒ 11

is zero

;

1
2ؒ3ؒ5ؒ6
1
4
ϭ
7 ϭ Ϫ
6ؒ7 3ؒ4ؒ6ؒ7
ϭϪ

2

1

ϭ0

;

and so on
...


After grouping the terms containing 0 and the terms containing 1 , we obtain
ϭ 0 1( ) ϩ 1 2 ( ), where
ϱ
1
1
(Ϫ1)
6
9
3
ϩ
Ϫ
ϩ и и и ϭ 1 ϩ͚
2ؒ3ؒ5ؒ6
2ؒ3ؒ5ؒ6ؒ8ؒ9
ϭ1 2 ؒ 3 и и и (3 Ϫ 1)(3 )
ϩ

1
3ؒ4ؒ6ؒ7

7

Ϫ

1
3 ؒ 4 ؒ 6 ؒ 7 ؒ 9 ؒ 10

10

ϱ

ϩиииϭ

ϩ

(Ϫ1)

͚ 3 ؒ 4 и и и (3 )(3
ϭ1

3 ϩ1

ϩ 1)


...
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...
As was mentioned prior to this example, the linear
combination ϭ 0 1( ) ϩ 1 2 ( ) actually represents the general solution of the
differential equation
...
2
...
This fact can also be verified by the ratio test
The differential equation in Example 5 is called
and is named
after the English mathematician and astronomer
(1801–1892)
...
Other common
forms of Airy’s equation are Љ Ϫ ϭ 0 and Љ ϩ ␣ 2 ϭ 0
...
4 for an application of the last equation
...


As we have already seen on page 240, the given differential equation has
singular points at ϭ Ϯ , and so a power series solution centered at 0 will converge at
least for ͉ ͉ Ͻ 1, where 1 is the distance in the complex plane from 0 to either or Ϫ
The assumption ϭ ͚ϱϭ0
and its first two derivatives lead t
(

ϱ

2

ϱ

ϩ 1) ͚ ( Ϫ 1)

Ϫ2

ϭ2

ϱ

0

Ϫ

ϭ0
ϱ

ϩ6

ϱ

ϩ͚

Ϫ2

ϭ2

0

0

Ϫ͚

ϩ ͚ ( Ϫ 1)

ϭ2

2

Ϫ1

ϱ

ϭ ͚ ( Ϫ 1)
ϭ2

ϱ

͚
ϭ1

ϩ

Ϫ͚

ϭ1

ϭ0

ϱ

ϩ

3

Ϫ

1

ϩ ͚ ( Ϫ 1)

1

ϭ2

ϭ
ϱ

ϱ

ϩ ͚ ( Ϫ 1)

Ϫ2

ϭ4

ϩ͚

2

Ϫ

0

ϩ6

Ϫ͚

ϭ2

ϭ Ϫ2

ϭ2

ϱ

ϭ2

ϭ

ϭ

ϱ

3

ϩ ͚ [ ( Ϫ 1) ϩ ( ϩ 2)( ϩ 1)

ϩ2

ϩ

Ϫ ]

ϭ2

ϭ2

2

Ϫ

0

ϩ6

ϱ

3

ϩ ͚ [( ϩ 1)( Ϫ 1) ϩ ( ϩ 2)( ϩ 1)

ϩ2]

ϭ2

From this identity we conclude that 2
( ϩ 1)( Ϫ 1)
Thus

1
2

2

ϭ

3

2

Ϫ

0

ϭ 0, 6

3

ϭ 0, and

ϩ ( ϩ 2)( ϩ 1)

ϩ2

ϭ 0
...


Substituting ϭ 2, 3, 4,
...
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...


●

5

ϭϪ

2
5

3

ϭ0

6

ϭϪ

3
6

4

ϭ

7

ϭϪ

4
7

5

ϭ0

8

ϭϪ

5
8

6

ϭϪ

9

ϭϪ

6
9

7

ϭ 0,

10

ϭϪ

7
10

3
2ؒ4ؒ6

ϭ

8

;

3

0

;

is zero

1ؒ3
23 3!

ϭ
5

is zero

3ؒ5
2ؒ4ؒ6ؒ8
;

7

0

0

ϭϪ

1ؒ3ؒ5
24 4!

0

is zero

3ؒ5ؒ7
2 ؒ 4 ؒ 6 ؒ 8 ؒ 10

0

ϭ

1ؒ3ؒ5ؒ7
25 5!

0,

and so on
...


The solutions are the polynomial
1(

5

)ϭ1ϩ

1
2

2

2(

1 ؒ 3 ؒ 5 и и и ΂2 Ϫ 3΃
2 !

ϱ

ϩ

) ϭ and the power series

͚ (Ϫ1) Ϫ1
ϭ2

If we seek a power series solution ϭ ͚ϱϭ0

2

,

͉ ͉ Ͻ 1
...


It follows from these two results that all coefficients , for
3, are expressed in
terms of
0, 1 ϭ 0; this
0 and 1
...
Next, if
we choose 0 ϭ 0, 1 0, then coefficients for the other solution are expressed
in terms of 1
...
gives
0,

0

1
2

2

ϭ

3

ϭ
ϭ
ϭ

3

0ϭ

2

5

ϭ0

1

4

1

0,
1
2

2

ϭ

ϩ 0
ϭ 0 ϭ 0
2ؒ3
2ؒ3
6

3

ϭ

4

ϭ
ϭ

ϭ0

2

5

0

1

ϩ 1
0
ϭ
ϭ 0
3ؒ4
2 ؒ 3 ؒ 4 24

1

3

0

΄

΅

ϩ 2
1 1
ϭ 0
ϩ
ϭ 0
4ؒ5
4ؒ5 6 2
30

0

ϩ 0
ϭ 1 ϭ 1
2ؒ3
2ؒ3
6
ϩ 1
ϭ 1 ϭ 1
3ؒ4
3 ؒ 4 12
ϩ 2
1
ϭ
ϭ 1
4ؒ5
4 ؒ 5 ؒ 6 120

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...


●

and so on
...


Each series converges for all finite values of
The next example illustrates how to find a
power series solution about the ordinary point 0 ϭ 0 of a differential equation when
its coefficients are not polynomials
...


Solve Љ ϩ (cos ) ϭ 0
...
Using the Maclaurin series for cos
given in (2) of Section 6
...
1 we fin

΂

ϱ

Љ ϩ (cos ) ϭ ͚ ( Ϫ 1)

Ϫ2

ϭ2

ϭ2

2

ϩ6

3

ϭ2

2

ϩ

ϩ (6

0

ϩ 12

3

2

ϩ 1Ϫ

4

2

ϩ 20

1)

ϩ

2!

4

ϩ
3

5

΂

ϩ 12

4!

΃͚

6

Ϫ

ϱ

ϩиии

6!

ϭ0

΂

2

ϩиииϩ 1Ϫ

4

ϩ

2

Ϫ

1
2

΃

0

2!
2

΃

4

ϩ

4!

ϩиии (

΂

ϩ 20

5

ϩ

3

0

Ϫ

ϩ

1
2

ϩ

1

΃

1

3

2

2

ϩ

3

3

ϩ и и и)

ϩ и и и ϭ 0
...
This gives 2 ϭ Ϫ1 0 , 3 ϭ Ϫ1 1 , 4 ϭ 12 0 , 5 ϭ 30 1,
...


Because the differential equation has no finite singular points, both power series converge for ͉ ͉ Ͻ ϱ
...
We can always resort to graphing the
͚ϱϭ0
terms in the sequence of partial sums of the series — in other words, the graphs of the

...
We can also obtain an approximate or numerical solution curve by using a solver as we did in Section 4
...

For example, if you carefully scrutinize the series solutions of Airy’s equation in

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...
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...


●

y1

Example 5, you should see that
value problems

3

1(

) and

2(

) are, in turn, the solutions of the initial-

2

Љϩ

ϭ 0,

(0) ϭ 1,

Ј(0) ϭ 0,

1

Љϩ

ϭ 0,

(0) ϭ 0,

Ј(0) ϭ 1
...
Now if your numerical solver requires
a system of equations, the substitution Ј ϭ in Љ ϩ ϭ 0 gives Љ ϭ Ј ϭ Ϫ ,
and so a system of two first-order equations equivalent to Airy’s equation is

y2

1

Јϭ

_1
_2
_3
2

4

6

8

10

(12)

ЈϭϪ
...
The graphs of 1( )
and 2 ( ) shown in Figure 6
...
2 were obtained with the aid of a numerical solver
...
1 (page 197) in the form
Љϩ
ϭ 0 as a model of a spring whose “spring constant” ( ) ϭ increases
with time
...
Even though we can generate as
many terms as desired in a series solution ϭ ͚ϱϭ0
either through the use
of a recurrence relation or, as in Example 8, by multiplication, it might not be
possible to deduce any general term for the coefficients
...

( ) A point 0 is an ordinary point of a
linear second-order
DE Љ ϩ ( ) Ј ϩ ( ) ϭ ( ) if ( ), ( ), and ( ) are analytic at 0
...
2
...
See Problem 26 in
Exercises 6
...


In Problems 1 and 2 without actually solving the given
differential equation, find the minimum radius of convergence
of power series solutions about the ordinary point ϭ 0
...

(

2

(

2

Ϫ 25) Љ ϩ 2 Ј ϩ
Ϫ 2 ϩ 10) Љ ϩ

ЉϪ2 Јϩ

ЈϪ4 ϭ0

Љϩ2 Јϭ0

ϭ0

ЉϪ

ϭ0

In Problems 3–6 find two power series solutions of the given
differential equation about the ordinary point ϭ 0
...
3
...

Љϩ ϭ0
ЉϪ ϭ0
ЉϪ Јϭ0

In Problems 7–18 find two power series solutions of the
given differential equation about the ordinary point ϭ 0
...
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...

( Ϫ 1) Љ Ϫ

Јϩ

ϭ 0, (0) ϭ Ϫ2, Ј(0) ϭ 6

( ϩ 1) Љ Ϫ (2 Ϫ ) Ј ϩ

ϭ 0, (0) ϭ 2, Ј(0) ϭ Ϫ1
Find two power series solutions for Љ ϩ Ј ϩ ϭ 0
and express the solutions 1 ( ) and 2 ( ) in terms of
summation notation
...
Repeat using the
1 ( )
...


Љ Ϫ 2 Ј ϩ 8 ϭ 0, (0) ϭ 3, Ј(0) ϭ 0
(

2

ϩ 1) Љ ϩ 2 Ј ϭ 0, (0) ϭ 0, Ј(0) ϭ 1

In Problems 23 and 24 use the procedure in Example 8 to
find two power series solutions of the given differential
equation about the ordinary point ϭ 0
...
Use
the initial-conditions 1(0) ϭ 1, 1 (0) ϭ 0, and
Ј
Ј
2 (0) ϭ 0, 2 (0) ϭ 1
...
Express
this series as an elementary function
...
2 to find a second solution of the equation
...


Without actually solving the differential equation
(cos ) Љ ϩ Ј ϩ 5 ϭ 0, find the minimum radius of
convergence of power series solutions about the ordinary point ϭ 0
...


Find one more nonzero term for each of the solutions 1 ( ) and 2 ( ) in Example 8
...

Use a CAS to graph the partial sums ( ) for the
solution ( ) in part (b)
...

Compare the graphs obtained in part (c) with the
curve obtained using a numerical solver for the
initial-value problem in part (b)
...

Is ϭ 0 an ordinary or a singular point of the differential equation Љ ϩ (sin ) ϭ 0? Defend your answer
with sound mathematics
...


●
●

Section 4
...
2
...
That is all they have in common
...
2 that there was no problem in finding two distinct power series solutions centered
at that point
...
When only one solution is found, we can use the formula given in (5) of Section 4
...
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...
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...
The classification again depends on
the functions and in the standard form
Љϩ ( ) Јϩ

( ) ϭ 0
...
A singular point that is not regular is said to be an
of the equation
...
3
...

As in Section 6
...
We have already seen that if 2( 0) ϭ 0, then ϭ 0 is a
singular point of (1), since at least one of the rational functions ( ) ϭ 1( ) 2( )
and ( ) ϭ 0( ) 2( ) in the standard form (2) fails to be analytic at that point
...
This means that after 1( ) 2 ( )
and 0 ( ) 2 ( ) are reduced to lowest terms, the factor Ϫ 0 must remain, to some
positive integer power, in one or both denominators
...
We are led to the conclusion that multiplying ( ) by Ϫ 0 and ( ) by ( Ϫ 0) 2 has the effect (through
cancellation) that Ϫ 0 no longer appears in either denominator
...

Moreover, observe that if ϭ 0 is a regular singular point and we multiply (2) by
( Ϫ 0) 2, then the original DE can be put into the form
( Ϫ
where and

2
0)

Љϩ( Ϫ

are analytic at ϭ

0)

( ) Ј ϩ ( ) ϭ 0,

(3)

0
...


After dividing the equation by ( 2 Ϫ 4) 2 ϭ ( Ϫ 2) 2 ( ϩ 2) 2 and reducing the
coefficients to lowest terms, we find th
( )ϭ

3
( Ϫ 2)( ϩ 2)2

We now test ( ) and

and

( )ϭ

5

...


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...
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...
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...


●

For ϭ 2 to be a regular singular point, the factor Ϫ 2 can appear at most to the
first power in the denominator of ( ) and at most to the second power in the denominator of ( )
...
Alternatively, we are led
to the same conclusion by noting that both rational functions
3
5
( ) ϭ ( Ϫ 2) ( ) ϭ
and
( ) ϭ ( Ϫ 2)2 ( ) ϭ
( ϩ 2)2
( ϩ 2)2
are analytic at ϭ 2
...
This also follows from the fact that
3
( ) ϭ ( ϩ 2) ( ) ϭ
( Ϫ 2)( ϩ 2)
is not analytic at ϭ Ϫ2
...

( ϩ 2)2
( ϩ 2)2
As another example, we can see that ϭ 0 is an irregular singular point
of 3 Љ Ϫ 2 Ј ϩ 8 ϭ 0 by inspection of the denominators of ( ) ϭ Ϫ2 2
and ( ) ϭ 8 3
...
A singular point can be a complex
number
...

Any second-order Cauchy-Euler equation 2 Љ ϩ
Ј ϩ ϭ 0, where
, , and are real constants, has a regular singular point at ϭ 0
...
The fact that we would not obtain the second solution is not surprising because ln (and consequently 2 ϭ 2 ln ) is not analytic
at ϭ 0 — that is, 2 does not possess a Taylor series expansion centered at ϭ 0
...


If ϭ 0 is a regular singular point of the differential equation (1), then there
exists at least one solution of the form
ϱ

ϭ( Ϫ

0)

͚
ϭ0

ϱ

( Ϫ

0)

ϭ͚

( Ϫ

0)

ϩ

,

(4)

ϭ0

where the number is a constant to be determined
...
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...
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...
3
...
This means
that in contrast to Theorem 6
...
1, Theorem 6
...
1 gives us no assurance that
series
solutions of the type indicated in (4) can be found
...
However, we have an additional task in this procedure: Before determining the coefficients, we must find the unknown exponent If is found to be a number that is not
a nonnegative integer, then the corresponding solution ϭ͚ϱϭ0 ( Ϫ 0 ) ϩ is not
a power series
...


Because ϭ 0 is a regular singular point of the differential equation
3

ϭ 0,

Љϩ ЈϪ
ϭ ͚ϱϭ0

we try to find a solution of the form
ϱ

ϩ

(5)

...


ϭ 0, we must then have
(6)

(3 Ϫ 2) ϭ 0
and

]

ϭ0

which implies that
and

΅

ϱ

͚(
ϭ1

ϭ Ϫ1

ϭ

ϩ

ϭ0

ϱ

(3 Ϫ 2)

Ϫ͚

ϱ

ϭ ͚ ( ϩ )(3 ϩ 3 Ϫ 2)
ϭ

ϱ

ϩ Ϫ1

,

ϭ 0, 1, 2,
...


(8)

,

ϭ 0, 1, 2,
...
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...
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...


ϭ

0

!1 ؒ 4 ؒ 7 и и и (3 Ϫ 2)


...
If we omit this term, the series
solutions are
1(

)ϭ

2/ 3

΄1 ϩ ͚
ϱ

ϭ1

2(

)ϭ

0

΄

ϱ

1ϩ͚

ϭ1

΅

1
!5 ؒ 8 ؒ 11и и и (3 ϩ 2)
1
!1 ؒ 4 ؒ 7 и и и (3 Ϫ 2)

(10)

΅
...
Also, it should be apparent from the form of these
solutions that neither series is a constant multiple of the other, and therefore 1 ( ) and
2 ( ) are linearly independent on the entire -axis
...
On any interval that does not
contain the origin, such as (0, ϱ), this linear combination represents the general solution of the differential equation
...
In general, after substituting ϭ ͚ϱϭ0
into the given differential equation and simplifying, the indicial equation is a quadratic equation in
that results from equating the
We
solve for the two values of and substitute these values into a recurrence relation
such as (7)
...
3
...

It is possible to obtain the indicial equation in advance of substituting
ϩ
ϭ ͚ϱϭ0
into the differential equation
...
3
...
By multiplying
(2) by 2, we get the form given in (3):
2

Љϩ [

( )] Ј ϩ [

2

( )] ϭ 0
...
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...
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...
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...

ϩ

ϱ

2 Љ ϩ (1 ϩ ) Ј ϩ ϭ 2 ͚ ( ϩ )( ϩ Ϫ 1)
ϱ

ϩ͚( ϩ )

ϩ͚

ϩ

ϭ0

ϩ

ϭ0

ϱ

ϭ ͚ ( ϩ )(2 ϩ 2 Ϫ 1)

ϱ

ϩ Ϫ1

ϭ0

(2 Ϫ 1)

[

ϩ Ϫ1

ϭ0

ϱ

ϩ͚( ϩ )

[

gives
ϱ

ϩ Ϫ1

ϭ0

ϭ

0

are as defined in (12)
...
3

Substituting ϭ ͚ϱϭ0

ϭ

ϩ

0

ϩ ͚ ( ϩ ϩ 1)

ϱ

0

ϩ ͚ ( ϩ )(2 ϩ 2 Ϫ 1)

Ϫ1

ϩ

ϭ0

ϭ1

ϩ ͚ ( ϩ ϩ 1)
ϭ0

ϭ Ϫ1

(2 Ϫ 1)

ϱ

0

Ϫ1

ϩ

͚ [(
ϭ0

ϩ ϩ 1)(2 ϩ 2 ϩ 1)

( ϩ ϩ 1)(2 ϩ 2 ϩ 1)

and

ϭ

ϩ1

ϩ1

2

ϭ

Ϫ
,
2( ϩ 1)

ϭ

3

4

1

ϭ 0,
ϭ 1 and
2

ϭ 0, 1, 2,
...


(17)

ϭ 0, (16) becomes
ϩ1

1

]

(15)

ϩ ( ϩ ϩ 1)

ϭ 0, 1, 2,
...


(18)

From (18) we fin
1

2

3

4

Ϫ 0
1
Ϫ 1
ϭ
ϭ 0
3
1ؒ3
ϭ

Ϫ 2
Ϫ 0
ϭ
5
1ؒ3ؒ5
Ϫ 3
0
ϭ
ϭ
7
1ؒ3ؒ5ؒ7
ϭ

и
и
и

и
и
и
ϭ

(Ϫ1)
2 !

0


...

1 ؒ 3 ؒ 5 ؒ 7 и и и (2 Ϫ 1)

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...
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...
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...


●

Thus for the indicial root

ϭ 1 we obtain the solution
2

1

΄1 ϩ ͚ (Ϫ1)! ΅ ϭ ͚ (Ϫ1)!
2
2
ϱ

ϱ

ϭ1

1(

ϭ0

1/2

)ϭ

ϩ1/2

,

where we have again omitted 0
...
For 2 ϭ 0 a
second solution is
(Ϫ1)
1 ؒ 3 ؒ 5 ؒ 7 и и и (2 Ϫ 1)
ϭ1
ϱ

2(

) ϭ 1ϩ͚

On the interval (0, ϱ) the general solution is ϭ

Solve

1 1(

͉ ͉ Ͻ ϱ
...


Љ ϩ ϭ 0
...
You should verify that the two recurrence relations
corresponding to the indicial roots 1 ϭ 1 and 2 ϭ 0 yield exactly the same set of
coefficients
...


For the sake of discussion let us again suppose that ϭ 0 is a
regular singular point of equation (1) and that the indicial roots 1 and 2 of the
singularity are real
...
In the first two cases the
symbol 1 denotes the largest of two distinct roots, that is, 1 Ͼ 2
...

If 1 and 2 are distinct and the difference 1 Ϫ 2 is not a positive integer, then there exist two linearly independent solutions of equation (1) of the form
ϱ

1( ) ϭ ͚

ϱ

ϩ

,

1

2( ) ϭ ͚

0,

0

ϭ0

ϩ

2

,

0

0
...

Next we assume that the difference of the roots is , where
integer
...


is a positive

If 1 and 2 are distinct and the difference 1 Ϫ 2 is a positive integer,
then there exist two linearly independent solutions of equation (1) of the form
ϱ

1(

) ϭ͚

ϩ

,

1

0

(19)

0,

ϭ0
ϱ

2(

)ϭ

1(

) ln ϩ ͚

ϩ

,

2

0

0,

(20)

ϭ0

where

is a constant that could be zero
...
The situation is analogous to the solution of a Cauchy-Euler
equation when the roots of the auxiliary equation are equal
...
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...
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...
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...


(22)

ϭ0
ϱ

2(

)ϭ

) ln ϩ ͚

1(

ϩ

1

ϭ1

When the difference 1 Ϫ 2 is a positive integer
(Case II), we
or
be able to find two solutions having the
ϩ
form ϭ ͚ϱϭ0

...
We just may be lucky enough
ϩ1
to find two solutions that involve only powers of , that is, 1( ) ϭ ͚ϱϭ0
ϱ
ϩ2
(equation (19)) and 2( ) ϭ ͚ ϭ0
(equation (20) with ϭ 0)
...
3
...
In this situation equation (20), with
0, indicates what
the second solution looks like
...
One
way to obtain the second solution with the logarithmic term is to use the fact that
2(

)ϭ

1(

)

͵

Ϫ͵ ( )
2
1(

(23)

)

is also a solution of Љ ϩ ( ) Ј ϩ ( ) ϭ 0 whenever
We illustrate how to use (23) in the next example
...


Љ ϩ ϭ 0
...
Those with the time,
energy, and patience can carry out the drudgery of squaring a series, long division,
and integration of the quotient by hand
...
We give the results:
2(

)ϭ

ϭ

1(

1(

)

)

͵
͵

Ϫ0

ϭ

[ 1( )]2

΄

͵΄

2

1

3

Ϫ
1

ϩ

1(

5
12

)

4

͵
Ϫ

΄

Ϫ

7
72

5

1(

)

ϭ

1(

1
7
19
) Ϫ ϩ ln ϩ
ϩ
12
144

ϭ

1(

) ln ϩ

ϩ

ϩ

΄

1(

΄

2

ϩ

1
12

3

΅

ϩиии

΅

2

1
144

4

΅

2

ϩиии

; after long division

΅

ϩиии

1
7
19
ϩ
) Ϫ ϩ
12
144

Ϫ

; after squaring

7
19
ϩ
ϩиии
12 72

ϭ

2

1
2

2

; after integrating

΅

ϩиии ,

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...
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...
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...


●

or

2(

)ϭ

1(

΄

) ln ϩ Ϫ1 Ϫ

1
1
ϩ
2
2

2

΅

ϩиии
...


)
...
But on a practical
level, when it comes to actually solving a differential equation using the method
of Frobenius, it is advisable to work with the form of the DE given in (1)
...
See Problems 31 and 32 in Exercises 6
...
We shall not, however, investigate this case
...


In Problems 1 –10 determine the singular points of the given
differential equation
...

3

Љϩ4

2

Јϩ3 ϭ0

In Problems 13 and 14, ϭ 0 is a regular singular point of the
given differential equation
...
Without solving, discuss the number of series solutions
you would expect to find using the method of Frobenius

( ϩ 3) 2 Љ Ϫ ϭ 0
2

(

2

Ϫ 9) 2 Љ ϩ ( ϩ 3) Ј ϩ 2 ϭ 0

ЉϪ
(

3
2

(

(

3

2

Ϫ 25) ϭ 0

ϩ Ϫ 6) Љ ϩ ( ϩ 3) Ј ϩ ( Ϫ 2) ϭ 0
ϩ 1) 2 Љ ϩ ϭ 0
2

Ϫ 25)( Ϫ 2) Љ ϩ 3 ( Ϫ 2) Ј ϩ 7( ϩ 5) ϭ 0

Ϫ2

2

2

ϩ 3 ) Љ ϩ ( Ϫ 3) Ј Ϫ ( ϩ 1) ϭ 0

In Problems 11 and 12 put the given differential equation
into form (3) for each regular singular point of the equation
...

2

Ϫ 1) Љ ϩ 5( ϩ 1) Ј ϩ (

Љ ϩ ( ϩ 3) Ј ϩ 7

ЈϪ1 ϭ0
3

In Problems 15 –24, ϭ 0 is a regular singular point of
the given differential equation
...
Use the method
of Frobenius to obtain two linearly independent series
solutions about ϭ 0
...

ЉϪ Јϩ2 ϭ0

2

Љϩ5 Јϩ

4

1
2

2
3
2

(

)

2

ϩ

2

2

2

(5
3

Љ ϩ Ј ϩ 10 ϭ 0

1
ϭ0
( Ϫ 1) 3

ϩ4 ) ЉϪ2 Јϩ6 ϭ0

2

(

(

Јϩ

( Ϫ 5) 2 Љ ϩ 4 Ј ϩ (

2

3

1

Љϩ

2

ϭ0

2

Ϫ ) ϭ0
2

Љϩ
2

ЉϪ

ϭ0

Јϩ

ϭ0

Јϩ(

2

ϩ 1) ϭ 0

Љ ϩ (2 Ϫ ) Ј Ϫ ϭ 0
ЉϪ

(

Ϫ2
9

)

ϭ0

Љ Ϫ (3 ϩ 2 ) Ј ϩ ϭ 0

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...
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...
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...


●

2

Љϩ

Јϩ
2

( 2 Ϫ 4)
9

In this problem let us assume that the column is of
length , is hinged at both ends, has circular cross
sections, and is tapered as shown in Figure 6
...
1(a)
...
3
...
Sub( ) ϭ 0(
4
stituting ( ) into the differential equation in (24),
we see that the deflection in this case is determined
from the BVP

ϭ0

9

2

Љϩ9

2

2

Љ ϩ 3 Ј ϩ (2 Ϫ 1) ϭ 0

Јϩ2 ϭ0

In Problems 25–30, ϭ 0 is a regular singular point of
the given differential equation
...
Use the method
of Frobenius to obtain at least one series solution about
ϭ 0
...
Form the general solution on (0, ϱ)
...
Use the results of Problem 33
to find the critical loads
for the tapered column
...

Use a CAS to plot the graph of the first buckling
mode 1 ( ) corresponding to the Euler load 1
when ϭ 11 and ϭ 1
...
Show that the indicial
roots of the singularity differ by an integer
...
How many solutions did you find? Next
use the recurrence relation with the smaller root 2
...
Show that the substitution ϭ 1 yields the DE
2
2

ϩ

2

ϩ ␭ ϭ 0,
Tapered column in Problem 34

which now has a regular singular point at ϭ 0
...

Express each series solution of the original equation
in terms of elementary functions
...


Each of the differential equations
In Example 4 of
Section 5
...


(24)

The assumption here is that the column is hinged at both
ends
...


3

Љϩ

ϭ0

and

2

Љ ϩ (3 Ϫ 1) Ј ϩ

ϭ0

has an irregular singular point at ϭ 0
...
Discuss
and explain your findings
We have seen that ϭ 0 is a regular singular point of
2
any Cauchy-Euler equation
Љϩ
Ј ϩ ϭ 0
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

●

Sections 6
...
3
In the

at the end of Section 2
...
Perhaps a better title for this field of applied mathematics might be
because many of the functions studied bear proper names: Bessel functions,
Legendre functions, Airy functions, Chebyshev polynomials, Hermite polynomials, Jacobi polynomials, Laguerre polynomials, Gauss’ hypergeometric function, Mathieu functions, and so on
...
In
effect, a special function was determined or defined by the differential equation and many properties
of the function could be discerned from the series form of the solution
...
2 and 6
...
They are called,
respectively,
␯, named after the German mathematician and astronomer
(1784–1846), and
, named after the
French mathematician
(1752–1833)
...


Because ϭ 0 is a regular singular point of
Bessel’s equation, we know that there exists at least one solution of the form
ϩ
ϭ ͚ϱϭ0

...


(3)

ϩ2

From (3) we see that the indicial equation is 2 Ϫ ␯ 2 ϭ 0, so the indicial roots are
1 ϭ ␯ and 2 ϭ Ϫ␯
...
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...
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...
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...


●

Therefore by the usual argument we can write (1 ϩ 2␯)
( ϩ 2)( ϩ 2 ϩ 2 ␯ )
or

ϩ2

ϭ

ϩ2

1

ϭ0

ϩ

Ϫ
,
( ϩ 2)( ϩ 2 ϩ 2␯)

ϭ 0 and

ϭ 0, 1, 2,
...

we find, after letting ϩ 2 ϭ 2 , ϭ 1, 2, 3,
...

22 ( ϩ ␯)

(5)

0

22 ؒ 1 ؒ (1 ϩ ␯)
2

22 ؒ 2(2 ϩ ␯)
4

22 ؒ 3(3 ϩ ␯)

ϭ

0

24 ؒ 1 ؒ 2(1 ϩ ␯)(2 ϩ ␯)

ϭϪ

0

26 ؒ 1 ؒ 2 ؒ 3(1 ϩ ␯)(2 ϩ ␯)(3 ϩ ␯)

(Ϫ1) 0
,
!(1 ϩ ␯)(2 ϩ ␯) и и и ( ϩ ␯)

It is standard practice to choose

0

ϭ 1, 2, 3,
...
See Appendix I
...
For example,
⌫(1 ϩ ␯ ϩ 1) ϭ (1 ϩ ␯)⌫(1 ϩ ␯)
⌫(1 ϩ ␯ ϩ 2) ϭ (2 ϩ ␯)⌫(2 ϩ ␯) ϭ (2 ϩ ␯)(1 ϩ ␯)⌫(1 ϩ ␯)
...


and ϭ ␯, a series solution of (1) is
denoted by ␯ ( ):
ϱ

␯(

)ϭ͚

ϭ0

Using the coefficients 2 just obtained
ϭ ͚ϱϭ0 2 2 ϩ␯
...


(7)

If ␯ 0, the series converges at least on the interval [0, ϱ)
...


(8)

The functions ␯ ( ) and Ϫ␯ ( ) are called
of order ␯
and Ϫ␯, respectively
...
*
*

When we replace by | |, the series given in (7) and (8) converge for 0 Ͻ | | Ͻ ϱ
...
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...
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...
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...
8
0
...
4
0
...
2
_ 0
...
When ␯ ϭ 0,
it is apparent that (7) and (8) are the same
...
3 that ␯ ( ) and Ϫ␯ ( ) are
linearly independent solutions of (1) on (0, ϱ), and so the general solution on the
interval is ϭ 1 ␯ ( ) ϩ 2 Ϫ␯ ( )
...
3 that
when 1 Ϫ 2 ϭ 2␯ is a positive integer, a second series solution of (1)
exist
...
When ␯ ϭ ϭ positive integer, Ϫ ( )
defined by (8) and ( ) are not linearly independent solutions
...
In addition, 1 Ϫ 2 ϭ 2␯
can be a positive integer when ␯ is half an odd positive integer
...
In other words, the general
solution of (1) on (0, ϱ) is
1 ␯(

ϭ

Bessel functions of
the first kind for ϭ 0, 1, 2, 3, 4

The graphs of ϭ

0(

)ϩ

2 Ϫ␯ (

) and ϭ

1(

),

integer
...
4
...


By identifying ␯ 2 ϭ 1 and ␯ ϭ 1, we can see from (9) that the general solution of the
4
2
equation 2 Љ ϩ Ј ϩ 2 Ϫ 1 ϭ 0 on (0, ϱ) is ϭ 1 1/ 2 ( ) ϩ 2 Ϫ1/ 2 ( )
...
5
_ 0
...
5
_2
_2
...
Thus another form of
the general solution of (1) is ϭ 1 ␯ ( ) ϩ 2 ␯ ( ), provided that ␯ integer
...
However, it can be shown
by L’Hôpital’s Rule that lim␯ : ␯ ( ) exists
...
Hence
value of ␯ the general solution of (1) on (0, ϱ) can be written as

8

ϭ
␯ ( ) is called the
the graphs of 0 ( ) and

1 ␯(

)ϩ

2 ␯(

)
...
Figure 6
...
2 shows

1(

)
...

Sometimes it is possible to
transform a differential equation into equation (1) by means of a change of variable
...
For example, if we let ϭ ␣ , ␣ Ͼ 0, in
2

Љϩ

Ј ϩ (a2

2

Ϫ ␯ 2 ) ϭ 0,

(12)

then by the Chain Rule,
2

ϭ

ϭ␣

and

2

ϭ

΂ ΃

ϭ ␣2

2
2


...
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...
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...
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...


The last equation is Bessel’s equation of order ␯ with solution ϭ 1 ␯ ( ) ϩ 2 ␯ ( )
...


(13)

Equation (12), called the
␯, and its general
solution (13) are very important in the study of certain boundary-value problems
involving partial differential equations that are expressed in cylindrical coordinates
...


(14)

This DE can be solved in the manner just illustrated for (12)
...


ϩ

Because solutions of the last DE are ␯ ( ) and ␯ ( ),
(14) are ␯ ( ) and ␯ ( )
...


(15)

See Problem 21 in Exercises 6
...

Analogous to (10), the
␯ integer is defined to b
␯(

y

of order

)ϭ

␲
2

Ϫ␯ (

) Ϫ ␯( )
,
sin ␯␲

(16)

and for integer ␯ ϭ ,

3
2
...
5
1
0
...


Because ␯ and ␯ are linearly independent on the interval (0, ϱ) for any value of
, the general solution of (14) on that interval is

I2
x

1

3
2
...
5
1
0
...


(17)

The graphs of ϭ 0( ), ϭ 1( ), and ϭ 2( ) are given in Figure 6
...
3 and
the graphs of ϭ 0( ), ϭ 1( ), and ϭ 2( ) are given in Figure 6
...
4
...
Figures 6
...
3 and 6
...
4 also illustrate
the fact that the modified Bessel functions ( ) and ( ), ϭ 0, 1, 2,
...
Also notice that the modified Bessel functions of
the second kind ( ) like the Bessel functions of the second kind ( ) become
unbounded as : 0 ϩ
...


The general solution of the last equation on the interval (0, ϱ) is
ϭ

1 ␯ (a

)ϩ

2

␯ (a

)
...
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...
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...
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...


(18)

Although we shall not supply the details, the general solution of (18),

΄

ϭ

(

1

)ϩ

2

(

΅

) ,

(19)

can be found by means of a change in both the independent and the dependent
, ( )ϭ

variables: ϭ
replaced by

Ϫ

΂΃

/

( )
...


Find the general solution of

Љ ϩ 3 Ј ϩ 9 ϭ 0 on (0, ϱ)
...


The first and third equations imply that ϭ Ϫ1 and ϭ 1
...
From (19)
we find that the general solution of the given DE on the interval (0, ϱ) is
ϭ Ϫ1 [ 1 2 (6 1/2) ϩ 2 2 (6 1/2)]
...
1 we saw that one mathematical model for the free undamped
motion of a mass on an aging spring is given by
Љ ϩ Ϫ␣ ϭ 0, ␣ Ͼ 0
...
It is left as a problem
2
Ϫ␣ / 2
to show that the change of variables ϭ
transforms the differential
␣B
equation of the aging spring into
2

2

2

ϩ

ϩ

2

ϭ 0
...
The general solution of the new
equation is ϭ 1 0( ) ϩ 2 0( )
...
4
...


The other model that was discussed in Section 5
...
By dividing through by , we see that
the equation

Љϩ

ϭ 0 is Airy’s equation Љ ϩ ␣ 2

ϭ 0
...
2
...
See Problems 34, 35, and 40 in Exercises 6
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

We list below a few of the more useful properties of Bessel
functions of order , ϭ 0, 1, 2,
...


:0

Note that Property ( ) indicates that ( ) is an even function if
is an even
integer and an odd function if is an odd integer
...
4
...
This
last fact is not obvious from (10)
...
3
...
3 is 1( ) ϭ 0( ), whereas (22) of that
section is
ϱ

2(

)ϭ

0(

)ln Ϫ

͚
ϭ1

΂

΃΂2΃
...
57721566
...
That is,

΂

΃΂2΃

(Ϫ1)
1
1
1ϩ ϩиииϩ
2
( !)
2

2

,

Because of the presence of the
( ) is discontinuous at ϭ 0
...
4
...
Some additional function values of these four functions
1
are given in Table 6
...
2
...
4048
5
...
6537
11
...
9309

1(

)

0
...
8317
7
...
1735
13
...
8936
3
...
0861
10
...
3611

Numerical Values of

1
1(

)

2
...
4297
8
...
7492
14
...
0000
0
...
2239
Ϫ0
...
3971
Ϫ0
...
1506
0
...
1717
Ϫ0
...
2459
Ϫ0
...
0477
0
...
1711
Ϫ0
...
0000
0
...
5767
0
...
0660
Ϫ0
...
2767
Ϫ0
...
2346
0
...
0435
Ϫ0
...
2234
Ϫ0
...
1334
0
...
0883
0
...
3769
Ϫ0
...
3085
Ϫ0
...
0259
0
...
2499
0
...
1688
Ϫ0
...
0782
0
...
2055

1(

)

—
Ϫ0
...
1070
0
...
3979
0
...
1750
Ϫ0
...
1581
0
...
2490
0
...
0571
Ϫ0
...
1666
0
...
In the next
example we derive a

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...
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...
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...


●

Derive the formula

Ј
␯(

) ϭ ␯ ␯( ) Ϫ

␯ϩ1 (

)
...


The result in Example 5 can be written in an alternative form
...


This last expression is recognized as a linear first-order differential equation in ␯ ( )
...


It can be shown in a similar manner that
[

␯

␯(

)] ϭ

␯

␯ Ϫ1 (

(21)

)
...
4
...
Observe that
when ␯ ϭ 0, it follows from (20) that
Ј
0(

) ϭ Ϫ 1( )

and

Ј
0(

) ϭ Ϫ 1 ( )
...
4
...
, Bessel functions of the first and second kinds can be
2
2
2
expressed in terms of the elementary functions sin , cos , and powers of
...
From (7)
2
ϱ

1/2( ) ϭ

͚
ϭ0

(Ϫ1)
!⌫ 1 ϩ 1 ϩ
2

(

) ΂2΃

2 ϩ1/2


...
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...
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...

2
2
2
2
6
7
6 2!
3!

ϱ

Hence

1/2 (

) ϭ͚

ϭ0

y

(Ϫ1)
(2 ϩ 1)!
2
! 2 ϩ1
1␲
2
!

΂΃

1/ 2 (

J 1/2
x

0

4

6

8

10

12

Ϫ1/ 2 (

14

Bessel functions of
order 1 (blue) and order Ϫ1 (red)
2
2

)ϭ

We leave it as an exercise to show that

0
...

B␲

2
B␲

(Ϫ1)
ϩ 1)!

ϱ

͚ (2
ϭ0

2 ϩ1


...
1 you should recognize that the infinite series in the last line is
the Maclaurin series for sin , and so we have shown that

J 1/ 2
-

0
...

ϩ1
!

(

⌫1ϩ1ϩ
2

In general,

1

()

)ϭ

(23)

2
cos
...
4
...
4
...
Because cos( ϩ 1) p ϭ 0
2
2
and sin( ϩ 1)p ϭ cos p ϭ (Ϫ1) , we see from (10) that
ϩ1>2( ) ϭ
2
(Ϫ1) ϩ1 Ϫ( ϩ1>2)( )
...
In view of (23) and (24) these results are the same as
1>2(

and

Ϫ1>2(

p
B2

)ϭ

2
cos
Bp

2
sin
...


For example, for

1>2(

)ϭ

1>2(

)ϭϪ

(27)
and ( )
ϭ 0 the

cos
p
2
cos ϭ Ϫ

...
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...
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...
4
...

Spherical Bessel functions arise in the solution of a special partial differential
equation expressed in spherical coordinates
...
4 and
Problem 13 in Exercises 13
...

Since ϭ 0 is an ordinary point of
, shift summation inLegendre’s equation (2), we substitute the series ϭ ͚ϱϭ0
dices, and combine series to get
(1 Ϫ

2

) ЉϪ2

Ј ϩ ( ϩ 1) ϭ [ ( ϩ 1)

0

ϩ 2 2 ] ϩ [( Ϫ 1)( ϩ 2)

1ϩ

6 3]

ϱ

ϩ ͚ [( ϩ 2)( ϩ 1)

ϩ2

ϩ ( Ϫ )( ϩ ϩ 1) ]

ϭ0

ϭ2

which implies that

( ϩ 1) 0 ϩ 2
( Ϫ 1)( ϩ 2) 1 ϩ 6
( ϩ 2)( ϩ 1) ϩ2 ϩ ( Ϫ )( ϩ ϩ 1)

( ϩ 1)
0
2!
( Ϫ 1)( ϩ 2)
3 ϭ Ϫ
1
3!
( Ϫ )( ϩ ϩ 1)
,
ϩ2 ϭ Ϫ
( ϩ 2)( ϩ 1)

or

2

ϭ0
3 ϭ 0
ϭ0
2

ϭϪ

(28)

ϭ 2, 3, 4,
...
, the recurrence relation (28) yields
( Ϫ 2)( ϩ 3)
( Ϫ 2) ( ϩ 1)( ϩ 3)
2 ϭ
0
4ؒ3
4!
( Ϫ 3)( ϩ 4)
( Ϫ 3)( Ϫ 1)( ϩ 2)( ϩ 4)
5 ϭ Ϫ
3 ϭ
1
5ؒ4
5!
( Ϫ 4)( ϩ 5)
( Ϫ 4)( Ϫ 2) ( ϩ 1)( ϩ 3)( ϩ 5)
6 ϭ Ϫ
4 ϭ Ϫ
0
6ؒ5
6!
( Ϫ 5)( ϩ 6)
( Ϫ 5)( Ϫ 3)( Ϫ 1)( ϩ 2)( ϩ 4)( ϩ 6)
7 ϭ Ϫ
5 ϭ Ϫ
1
7ؒ6
7!
and so on
...


Notice that if is an even integer, the first series terminates, whereas
infinite series
...


4

; that is,
of

Legendre’s equation
...
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...
For ϭ 0 we choose
0 ϭ 1, and for ϭ 2, 4, 6,
...


ϭ (Ϫ1)(

Ϫ1) /2

1ؒ3иии

...


These specific th-degree polynomial solutions are
called
and are denoted by ( )
...
5

P1

Remember, 0 ( ), 1 ( ),
differential equations

P2

ϭ
ϭ
ϭ
ϭ

x
-0
...
5

for

0
...


),
...
4
...

You are encouraged to verify the following properties using the
Legendre polynomials in (30)
...
4
...


even
( ) is an even or odd

Recurrence relations that relate Legendre polynomials
of different degrees are also important in some aspects of their applications
...
In (30) we listed the first six Legendre polynomials
...
This relation expresses 6 ( )
in terms of the known 4 ( ) and 5 ( )
...
4
...
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...
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...

for these polynomials is
( )ϭ

1

(

2 !

2

Ϫ 1) ,

ϭ 0, 1, 2,
...
4
...
Any solution of
Legendre’s equation is called a
If is
a nonnegative
integer, then both Legendre functions 1 ( ) and 2 ( ) given in (29) are infinit
series convergent on the open interval (Ϫ1, 1) and divergent (unbounded) at
ϭ Ϯ1
...
You should be aware of the fact that Legendre’s
equation possesses solutions that are bounded on the
interval [Ϫ1, 1]
only in the case when ϭ 0, 1, 2,
...
See Problem 47 in
Exercises 6
...


2

In Problems 1–6 use (1) to find the general solution of the
given differential equation on (0, ϱ)
...

Љϩ2 Јϩ4 ϭ0

ϭ0

Љϩ3 Јϩ

Ϫ 4) ϭ 0
2

Ϫ 4΃ ϭ 0
9

In Problems 11 and 12 use the indicated change of variable
to find the general solution of the given differential equation
on (0, ϱ)
...

2

2

2

ЉϪ5 Јϩ
2

4

Љϩ(
2

2

ϭ0

Ϫ 2) ϭ 0

Љ ϩ (16

2

ϩ 1) ϭ 0

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...
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...


●

3

Љϩ3 Јϩ
2

9

(

ϭ0
6

Љϩ9 Јϩ(

Ϫ 36) ϭ 0

Use the series in (7) to verify that
real function
...
Use this assumption to
express the general solution of the given differential
equation in terms the modified Bessel functions
and
...


Ј ( ) ϭ Ϫ␯ ␯ ( ) ϩ
␯

11

21

Verify by direct substitution that ϭ 1 1(2 1 )
is a particular solution of the DE in the case ␭ ϭ 1
...
]

Љϩ ϭ0
Љϩ4

(

Use the Table 6
...
1 to find the first three positive eigenvalues and corresponding eigenfunctions of the
boundary-value problem

In Problems 23 – 26 first use (18) to express the general solution of the given differential equation in terms of Bessel functions
...


2

)

Show that ϭ 1 / 2 2 ␣ 3 / 2 is a solution of Airy’s
3
differential equation Љ ϩ ␣ 2 ϭ 0, Ͼ 0, whenever
is a solution of Bessel’s equation of order 1, that
3
is, 2 Љ ϩ Ј ϩ 2 Ϫ 1 ϭ 0, Ͼ 0
...
]
3
Use the result of Problem 34 to express the general
solution of Airy’s differential equation for Ͼ 0 in
terms of Bessel functions
...


␯Ϫ1(

)
...
]

(21␭ ) ϩ

1

1

(21␭ )
...

Use the formula obtained in Example 5 along with
part (a) of Problem 27 to derive the recurrence relation
2␯ ␯ ( ) ϭ

␯ϩ1 (

)ϩ

␯Ϫ1(

)
...


͵

0

Use a CAS to graph
Ϫ5/2 ( )
...

Use the recurrence relation in Problem 28 along with
(23) and (24) to express 3/2( ), Ϫ3/2 ( ), 5/2 ( ) and
Ϫ5/2 ( ) in terms of sin , cos , and powers of
2
Ϫ␣ / 2
Use the change of variables ϭ
to show
␣B
that the differential equation of the aging spring
Љ ϩ Ϫ␣ ϭ 0, ␣ Ͼ 0, becomes
2

2
2

ϩ

ϩ

2

ϭ 0
...
1

ϭ 0,

(0) ϭ 1,

Ј(0) ϭ Ϫ1
...
4
...

Use the general solution obtained in Problem 35 to
solve the IVP
4 Љϩ

ϭ 0,

(0
...
1) ϭ Ϫ1
...

A uniform
thin column of length , positioned vertically with one

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...
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...


●

Use the information in Problem 37 to find a solution of

end embedded in the ground, will deflect, or bend away,
from the vertical under the influence of its own weight
when its length or height exceeds a certain critical value
...

Use Table 6
...
1 to find the Euler load 1 for the
column
...
For simplicity
assume that 1 ϭ 1 and ϭ 1
...
3, suppose that
the rod holding the mass at one end is replaced by a
flexible wire or string and that the wire is strung over a
pulley at the point of support in Figure 5
...
3
...
In other words, the
length ( ) of the pendulum varies with time
...
3, it can be shown* that the differential equation
for the displacement angle ␪ is now
2

2

2

␪
2

ϩ ␦ ( Ϫ )␪ ϭ 0, ␪ (0) ϭ 0, ␪ Ј( ) ϭ 0,

where is Young’s modulus, is the cross-sectional
moment of inertia, ␦ is the constant linear density, and
is the distance along the column measured from its base
...
4
...
The column will bend only for those
values of for which the boundary-value problem has a
nontrivial solution
...

Use the general solution found in part (a) to find a
solution of the BVP and an equation which define
the critical length , that is, the smallest value of
for which the column will start to bend
...
05 in
...
28 lb/in
...
6 ϫ 107 lb/in
...

4

ϭ 0,

ϩ

(0) ϭ 0,

␪ Љ ϩ 2 Ј␪ Ј ϩ sin ␪ ϭ 0
...


Make the change of variables
show that (34) becomes
2

␪
2

( )

ϩ

2 ␪

ϭ( 0ϩ )

(34)
and

␪ ϭ 0
...

Use the general solution obtained in part (c) to solve
the initial-value problem consisting of equation (34)
and the initial conditions ␪(0) ϭ ␪ 0 , ␪Ј(0) ϭ 0
...

B
Also, recall that (20) holds for both 1 ( ) and 1( )
...
2 we saw that when a constant vertical
compressive force, or load,
was applied to a thin
column of uniform cross section and hinged at both
ends, the deflection ( ) is a solution of the BVP:

1(

) 2( ) Ϫ

2

2(

)ϭ2

) 1( ) ϭ Ϫ

2
␲

will be helpful
...


If the bending stiffness factor
is proportional
to , then ( ) ϭ , where is a constant of
proportionality
...


(
*See

)

Mary Boas, John Wiley
& Sons, Inc
...
Also see the article by Borelli, Coleman, and Hobson
in
vol
...
2, March 1985
...
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...
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...
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...
Experiment with the graph using different time intervals such as [0, 10], [0, 30],
and so on
...

Show that the general solution of the equation is
( ) ϭ 0 1( ) ϩ 1 2 ( ), where
)ϭ1ϩ

2(

) ϭ ϩ ͚ (Ϫ1)

ϱ

ϭ1

Use the explicit solutions 1 ( ) and 2 ( ) of
Legendre’s equation given in (29) and the appropriate choice of 0 and 1 to find the Legendre polynomials 6 ( ) and 7 ( )
...

Use the recurrence relation (32) and 0 ( ) ϭ 1,
to generate the next six Legendre polynomials
...

Find the first three positive values of ␭ for which the
problem

2 a(a Ϫ 2)
...
(a Ϫ 2 ϩ 1)
(2 ϩ 1)!

When a ϭ is a nonnegative integer, Hermite’s
differential equation always possesses a polynomial
solution of degree
...
Then use 2( ) to find polynomial solutions
for ϭ 1, ϭ 3, and ϭ 5
...
Use the polynomial solutions in part (a) to show that the first six
Hermite polynomials are
0(

2(

2(

),
...
, 7 ( )
...
Form a conjecture about the location of the zeros of any Legendre
polynomial ( ), and then investigate to see whether
it is true
...


The differential equation
(1 Ϫ

2

Ј ϩ a2 ϭ 0,

) ЉϪ

where a is a parameter, is known as
after the Russian mathematician
(1821–1894)
...
Find a
fifth degree polynomial solution of this differential
equation
...


Use a CAS to graph
interval [Ϫ1, 1]
...


(1 Ϫ 2) Љ Ϫ 2 Ј ϩ ␭ ϭ 0,
(0) ϭ 0,
( ), Ј( ) bounded on [Ϫ1,1]

For purposes of this problem ignore the list of Legendre
polynomials given on page 266 and the graphs given
in Figure 6
...
3
...
, 7( )
...
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...
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...

The general solution of
ϭ 1 1 ( ) ϩ 2 Ϫ1 ( )
...


Because ϭ 0 is an irregular singular point of
3
Љ Ϫ Ј ϩ ϭ 0, the DE possesses no solution that
is analytic at ϭ 0
...


(Ϫϱ, ϱ)

ϭ 0 is an ordinary point of a certain linear differential
equation
...
Discuss whether
the series converges at Ϫ7, 0, 7, 10, and 11
...

cos
In Problems 7 and 8 construct a linear second-order differential equation that has the given properties
...
[
: Recall the Maclaurin
series for cos and
...
Use the
assumption ϭ ͚ϱϭ0
to find the general solution
ϭ ϩ that consists of three power series centered
at ϭ 0
...

However, a solution can be expressed in terms of Bessel
functions
...

Use (18) in Section 6
...

Use (20) and (21) in Section 6
...


Јϩ


...


ϭ 0
...


A regular singular point at
singular point at ϭ 0

(0) ϭ 3, Ј(0) ϭ Ϫ2

Even though ϭ 0 is an ordinary point of the differential equation, explain why it is not a good idea to try to
find a solution of the IV

(Ϫ1, ϱ)
[Ϫ1, 1]

[Ϫ1, 1]
2 2

Ј ϩ 2 ϭ 0,

Љϩ

and

)ϭ

Ј
␯(

)ϭϪ

␯(

␯

)Ϫ

␯(

␯ϩ1(

)ϩ

)

␯Ϫ1 (

)

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

as an aid to show that a one-parameter family of
solutions of
ϭ 2 ϩ 2 is given by
ϭ

( )
( )

( 1 2)
...
4 and Problem 32 of Exercises 6
...
4 to show that
1/ 2 (

)ϭ

that a second Legendre function satisfying the DE
for Ϫ1 Ͻ Ͻ 1 is

and

Ϫ1/ 2 (

␲
B2

)ϭ

΃

2
cosh
...
4 and part (b) to show that
1/ 2 (

)ϭ

ϭ

΂

΃
...
4
that when ϭ 1, Legendre’s differential equation
(1 Ϫ 2 ) Љ Ϫ 2 Ј ϩ 2 ϭ 0 possesses the polynomial solution ϭ 1( ) ϭ Use (5) of Section 4
...

1Ϫ

Use a graphing utility to graph the logarithmic
Legendre functions given in parts (a) and (b)
...


From (30) and (31) of Section 6
...
Use (5) of Section 4
...


Use the result obtained in part (a) to show that
(1) ϭ 1 and (Ϫ1) ϭ (Ϫ1)
...


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...
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...
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...


The Laplace Transform

7
...
2 Inverse Transforms and Transforms of Derivatives
7
...
1 Inverse Transforms
7
...
2 Transforms of Derivatives
7
...
3
...
3
...
4 Operational Properties II
7
...
1 Derivatives of a Transform
7
...
2 Transforms of Integrals
7
...
3 Transform of a Periodic Function
7
...
6 Systems of Linear Differential Equations
Chapter 7 in Review

In the linear mathematical models for a physical system such as a spring/mass
system or a series electrical circuit, the right-hand member, or input, of the
differential equations
2

2
2

ϩb

ϩ

ϭ ()

or

2

ϩ

ϩ

1

ϭ ()

is a driving function and represents either an external force ( ) or an impressed
voltage ( )
...
1 we considered problems in which the functions and
were continuous
...

For example, the impressed voltage on a circuit could be piecewise continuous and
periodic, such as the “sawtooth” function shown on the left
...


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...
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...
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...


●

●
●

Improper integrals with infinite limits of integratio
Integration by parts and partial fraction decomposition

In elementary calculus you learned that differentiation and integration are
; this means, roughly speaking, that these operations transform a function into another
function
...


Moreover, these two transforms possess the
that the transform of a linear
combination of functions is a linear combination of the transforms
...
In this section we will examine a special type of
integral transform called the
In addition to possessing the linearity property the
Laplace transform has many other interesting properties that make it very useful in solving linear
initial-value problems
...

For example, by holding constant, we see that ͵2 2 2
ϭ 3 2
...
If ( ) is defined for
Ն 0, then the improper integral ͵ϱ ( , ) ( ) is defined as a limit
0

͵

ϱ

( , ) ()

ϭ lim

:ϱ

0

᭤

͵

( , ) ()


...


The function ( , ) in (1) is called the
of the transform
...
Then the integral

͵

ϱ

ᏸ { ( )} ϭ

Ϫ

()

(2)

0

is said to be the

of , provided that the integral converges
...

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

When the defining integral (2) converges, the result is a function of In general
discussion we shall use a lowercase letter to denote the function being transformed
and the corresponding capital letter to denote its Laplace transform—for example,

ᏸ { ( )} ϭ ( ),

ᏸ { ( )} ϭ ( ),

ᏸ { ( )} ϭ ( )
...


Evaluate ᏸ {1}
...
In other words, when
0, the exponent Ϫ
Ϫ
: 0 as : ϱ
...


is negative, and

The use of the limit sign becomes somewhat tedious, so we shall adopt the
notation ͉ϱ as a shorthand for writing lim : ϱ ( ) ͉ 0
...


: 0 as : ϱ for

Ϫ

0
...

From Definition 7
...
1 we have ᏸ { } ϭ ͵ϱ Ϫ

...


ϱ
Ϫ

0

ᏸ{ 5 }

}

In each case we use Definition 7
...
1
...

ϩ3

0

The last result is valid for Ͼ Ϫ3 because in order to have lim
must require that ϩ 3 Ͼ 0 or Ͼ Ϫ3
...
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...
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...
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...

Ϫ5

In contrast to part (a), this result is valid for
demands Ϫ 5 Ͼ 0 or
5
...

From Definition 7
...
1 and two applications of integration by parts we
obtain

͵

ϱ

ᏸ{sin 2 } ϭ

0

2
ϭ –
lim

Ϫ

:ϱ

Ϫ

͵

Ϫ Ϫ sin 2
ϭ ––––––––––––

sin 2

͉

ϱ

2
ϩ–
0

͵

ϱ
Ϫ

0

cos 2

ϱ
Ϫ

0

cos 2 ϭ 0,

cos 2

,

0

0

[

͉

2 Ϫ Ϫ cos 2
ϭ – ––––––––––––

ϱ

2
Ϫ –
0

͵

Laplace transform of sin 2

]

ϱ
Ϫ

0

sin 2

4
2
ϭ –– Ϫ –– ᏸ{sin 2 }
...

Solving for that quantity yields the result
2
ᏸ {sin 2 } ϭ 2
,
0
...


(3)

Because of the property given in (3), ᏸ is said to be a

In this example we use the results of the preceding examples to illustrate the linearity of the Laplace transform
...


5,

} Ϫ 10ᏸ {sin2 } ϭ

4
Ϫ
Ϫ5

2

20

...
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...
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...

ϩ3
Ϫ3

We state the generalization of some of the preceding examples by means of the
next theorem
...


ᏸ {1} ϭ
ᏸ{ } ϭ

!

,

ϭ 1, 2, 3,
...
1
...


Then for ϭ 1, 2, and 3, we have, respectively,

ᏸ{ } ϭ
ᏸ { 2} ϭ
ᏸ { 3} ϭ

1
2
3

1 1
1
ؒ ϭ 2

ؒ ᏸ {1} ϭ
ؒᏸ{}ϭ

2

ؒ ᏸ { 2} ϭ

ؒ

1
2

ϭ

2ؒ1
3

3 2ؒ1 3ؒ2ؒ1
ؒ 3 ϭ
4

If we carry on in this manner, you should be convinced that

ᏸ{ } ϭ
()

1

2

3

Piecewise continuous
function


...


The integral that define
ᏸ
the Laplace transform does not have to converge
...
Sufficient conditions guaranteeing the existence of ᏸ { ( )} are that
be piecewise continuous on [0, ϱ) and that be of exponential order for
Recall
that a function is
on [0, ϱ) if, in any interval 0 Յ Յ Յ ,
there are at most a finite number of points , ϭ 1, 2,
...
See
Figure 7
...
1
...


A function is said to be of
,
0, and
0 such that ͉ ( )͉ Յ

if there exist constants
for all

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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

()

(

0)

()

If is an
function, then the condition ͉ ( )͉ Յ
,
, simply
states that the graph of on the interval ( , ϱ) does not grow faster than the graph
of the exponential function
, where is a positive constant
...
1
...

The functions ( ) ϭ , ( ) ϭ Ϫ , and ( ) ϭ 2 cos are all of exponential order
because for ϭ 1, ϭ 1, ϭ 0 we have, respectively, for
0
͉ ͉Յ ,

is of exponential
order

͉

Ϫ

͉Յ ,

͉ 2 cos ͉ Յ 2
...
1
...

()

()

()

2
2 cos

Three functions of exponential order

A positive integral power of is always of exponential order, since, for
͉ ͉Յ

()

͉ ͉Յ

or

0,

for

is equivalent to showing that lim : ϱ > is finite for ϭ 1, 2, 3,
...
A function such as ( ) ϭ is not of
2
exponential order since, as shown in Figure 7
...
4, grows faster than any positive
linear power of for > > 0
...

2

is not of exponential

order

If is piecewise continuous on [0, ϱ) and of exponential order, then ᏸ { ( )}
exists for
By the additive interval property of definite integrals we can writ

ᏸ { ( )} ϭ

͵

()

ϩ

0

The integral
on which Ϫ
constants ,
͉ 2͉ Յ

1

͵

ϱ

Ϫ

Ϫ

()

ϭ

1

ϩ 2
...
Now since is of exponential order, there exist
0,
0 so that ͉ ( ) ͉ Յ
for
We can then write

͵

ϱ

͉

Ϫ

( )͉

͵

ϱ

Յ

͵

ϱ

Ϫ

ϭ

Ϫ( Ϫ )

Ϫ( Ϫ )

ϭ

Ϫ

Ϫ( Ϫ )
for
Since ͵ϱ
converges, the integral ͵ϱ ͉ Ϫ ( ) ͉ converges
by the comparison test for improper integrals
...
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...
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...
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...


Ϫ

()

exists

The function , shown in Figure 7
...
5, is piecewise continuous and of
exponential order for
0
...


We conclude this section with an additional bit of theory related to the types of
functions of that we will, generally, be working with
...


:ϱ
If
is piecewise continuous on [0, ϱ) and of exponential order and
( ) ϭ ᏸ{ ( )}, then lim ( ) ϭ 0
...
If denotes the
maximum of the set { 1, 2} and denotes the maximum of {0, g}, then
͉ ( )͉ Յ
for

͵

ϱ

0

Ϫ

͉ ( )͉

͵

ϱ

Յ

0

͵

ϱ

Ϫ

ϭ

Ϫ( Ϫ )

ϭ

0

Ϫ

As : ϱ, we have ͉ ( ) ͉ : 0, and so ( ) ϭ ᏸ{ ( )} : 0
...
We note, however,
that these two conditions are sufficient but not necessary for the existence of a
Laplace transform
...
The function ( ) ϭ 2 cos
is not of exponential order, but it can be shown that its Laplace transform
exists
...
1
...
1
...
But you should not conclude from this that 1( ) and 2 ( ) are
Laplace
transforms
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 1 – 18 use Definition 7
...
1 to find ᏸ{ ( )}
...

( ) ϭ sin 2 cos 2

( ) ϭ cos 2

( ) ϭ sin(4 ϩ 5)

( ) ϭ 10 cos

΂ Ϫ 6΃

We have encountered the
⌫(a) in our
study of Bessel functions in Section 6
...

One definition of this function is given by the improper
integral

(2, 2)

1

͵

ϱ

⌫(␣) ϭ

1

Graph for Problem 7
()

aϪ1 Ϫ

,

0
...

Use Problem 41 and a change of variables to obtain the
generalization
⌫(␣ ϩ 1)
ᏸ { ␣} ϭ
, ␣ Ͼ Ϫ1,
␣ϩ1

(2, 2)

1

In Problems 43–46 use Problems 41 and 42 and the fact
that ⌫ 1 ϭ 1p to find the Laplace transform of the given
2
function
...
1
...


Graph for Problem 8

()

()
1

1/2

()ϭ

()ϭ

()ϭ

1

Ϫ1/2
3/2

()ϭ2

1/2

ϩ 8 5/2

Graph for Problem 9
()

Make up a function ( ) that is of exponential order but
where ( ) ϭ Ј( ) is not of exponential order
...

Graph for Problem 10

()ϭ

ϩ7

()ϭ

Ϫ2 Ϫ5

()ϭ

4

()ϭ

2 Ϫ2

()ϭ

Ϫ

()ϭ

cos

sin

( ) ϭ cos

( ) ϭ sin

In Problems 19 – 36 use Theorem 7
...
1 to find ᏸ{ ( )}
...
When does
ᏸ{ 1( ) ϩ 2( )} ϭ

1(

)ϩ

1
2(

and that

)?

Figure 7
...
4 suggests, but does not prove, that the func2
tion ( ) ϭ
is not of exponential order
...
1
...
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...
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...
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...
]

and 2 ϭ Ϫ1
...

Assume that and are positive constants
...

Show that the function ( ) ϭ 1> does not possess a
Laplace transform
...


1

This result is known as the

function

Explain why the function
,
0Յ Ͻ2
( ) ϭ 4,
2Ͻ Ͻ5
1>( Ϫ 5),
5

ᏸ {1> 2} ϭ

2

cos } exists
...


2

2

ϩ 2
...
]

●
●

Partial fraction decomposition
See the

In this section we take a few small steps into an investigation of how
the Laplace transform can be used to solve certain types of equations for an unknown function
...
After some important preliminary background material on the Laplace
transform of derivatives Ј( ), ЈЈ( ),
...


If ( ) represents the Laplace transform of a function
( ), that is, ᏸ { ( )} ϭ ( ), we then say ( ) is the
of
( ) and write ( ) ϭ ᏸ Ϫ1{ ( )}
...
1 we have, respectively,

ᏸ {1} ϭ
ᏸ{ } ϭ
ᏸ{

Ϫ3

1
1

Ϫ1

2

}ϭ

Ά1·
1
ϭᏸ Ά ·
ϭᏸ Ά

1 ϭ ᏸ Ϫ1

1
ϩ3

Ϫ3

2

Ϫ1

·

1
ϩ3

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

We shall see shortly that in the application of the Laplace transform to equations we are not able to determine an unknown function ( ) directly; rather, we are
able to solve for the Laplace transform ( ) of ( ); but from that knowledge
we ascertain by computing ( ) ϭ ᏸ Ϫ1{ ( )}
...

ϩ4
We shall show how to solve this problem in Example 2
...
1
...


Ά1·

1 ϭ ᏸ Ϫ1

Ά ! ·,

ϭ ᏸ Ϫ1

sin

sinh

Ά

ϭ ᏸ Ϫ1

2

Ά

ϭ ᏸ Ϫ1

Ά

ϭ 1, 2, 3,
...
It may be necessary to “fix up” the function of by multiplying and dividing
by an appropriate constant
...

ϩ7

To match the form given in part (b) of Theorem 7
...
1, we identify
ϩ 1 ϭ 5 or ϭ 4 and then multiply and divide by 4!:
To match the form given in part (d) of Theorem 7
...
1, we identify
ϭ 17
...


17
1
ϭ
sin 17
...
1,
(1) extends to any finite linear combination of Laplace transforms

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...
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...
Editorial review has
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...


●

ΆϪ2 ϩϩ46·
...


}

(2)

parts (e) and (d)
of Theorem 7
...
1 with ϭ 2

Partial fractions play an important role in finding inverse
Laplace transforms
...
Indeed, some CASs have packages that implement Laplace transform and
inverse Laplace transform commands
...
Although we shall examine each of these cases as this chapter develops, it
still might be a good idea for you to consult either a calculus text or a current precalculus text for a more comprehensive review of this theory
...

Ϫ 1)( Ϫ 2)( ϩ 4)

There exist unique real constants , , and

so that

2

ϩ6 ϩ9
ϭ
ϩ
ϩ
( Ϫ 1)( Ϫ 2)( ϩ 4)
Ϫ1
Ϫ2
ϩ4
ϭ

( Ϫ 2)( ϩ 4) ϩ ( Ϫ 1)( ϩ 4) ϩ ( Ϫ 1)( Ϫ 2)

...
(3)

By comparing coefficients of powers of on both sides of the equality, we know that
(3) is equivalent to a system of three equations in the three unknowns , , and
However, there is a shortcut for determining these unknowns
...
Hence the partial fraction decomposition is

2

ϩ6 ϩ9
16>5
25>6
1> 30
,
ϭϪ
ϩ
ϩ
( Ϫ 1)( Ϫ 2)( ϩ 4)
Ϫ1
Ϫ2
ϩ4

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...
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...
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...


(4)

●

and thus, from the linearity of ᏸ Ϫ1 and part (c) of Theorem 7
...
1,

Ά(

ᏸ Ϫ1

·

2

Ά

·

ϩ6 ϩ9
16
ϭ Ϫ ᏸ Ϫ1
Ϫ 1)( Ϫ 2)( ϩ 4)
5
ϭϪ

16
5

ϩ

Ά

1
25
ϩ ᏸ Ϫ1
Ϫ1
6

25
6

2

ϩ

1
30

·

Ά

1
1
ϩ ᏸ Ϫ1
Ϫ2
30


...

To that end we need to evaluate quantities such as ᏸ { > } and ᏸ { 2 > 2}
...


Here we have assumed that

͵

(6)

( ) : 0 as : ϱ
...


(7)

In like manner it can be shown that

ᏸ { ٞ( )} ϭ

3

2

( )Ϫ

(0) Ϫ

(8)

Ј(0) Ϫ Љ(0)
...
The next theorem gives the
Laplace transform of the th derivative of The proof is omitted
...
, ( Ϫ1) are continuous on [0, ϱ) and are of exponential order and if
( )
( ) is piecewise continuous on [0, ϱ), then

ᏸ{

( )

( )} ϭ

( )Ϫ

Ϫ1

(0) Ϫ

Ϫ2

Ј(0) Ϫ и и и Ϫ

( Ϫ1)

(0),

where ( ) ϭ ᏸ { ( )}
...
2
...
This property makes the Laplace transform ideally suited
for solving linear initial-value problems in which the differential equation has
Such a differential equation is simply a linear combination of terms
, Ј, Љ,
...
,

0

( Ϫ1)

ϭ ( ),

(0) ϭ

Ϫ1,

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

where the , ϭ 0, 1,
...
, Ϫ1 are constants
...


(9)

From Theorem 7
...
2, (9) becomes
[

( )Ϫ

ϩ

Ϫ1[

Ϫ1

Ϫ1

(0) Ϫ и и и Ϫ

( )Ϫ

Ϫ2

( Ϫ1)

(0)]

(0) Ϫ и и и Ϫ

( Ϫ2)

(0)] ϩ и и и ϩ

0

( )ϭ

( ),

(10)

where ᏸ { ( )} ϭ ( ) and ᏸ { ( )} ϭ ( )
...
, and the prescribed initial conditions 0 , 1,
...
* Typically, we put the two terms in (11) over the least
common denominator and then decompose the expression into two or more
partial fractions
...

The procedure is summarized in the diagram in Figure 7
...
1
...


We first take the transform of each member of the differential
equation:

ᏸ

Ά · ϩ 3ᏸ { } ϭ 13ᏸ {sin 2 }
...
3 with the
usual symbol replaced by

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...
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...
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...


●

From (6), ᏸ { > } ϭ ( ) Ϫ (0) ϭ ( ) Ϫ 6, and from part (d) of Theorem 7
...
1,
ᏸ {sin 2 } ϭ 2>( 2 ϩ 4), so (12) is the same as
( )Ϫ6ϩ3 ( )ϭ

2

26
ϩ4

or

( ϩ 3) ( ) ϭ 6 ϩ

2

26

...

ϩ 3 ( ϩ 3)( 2 ϩ 4) ( ϩ 3)( 2 ϩ 4)

( )ϭ

(13)

Since the quadratic polynomial 2 ϩ 4 does not factor using real numbers, its
assumed numerator in the partial fraction decomposition is a linear polynomial in :
6 2 ϩ 50
ϭ
ϩ
( ϩ 3)( 2 ϩ 4)
ϩ3

2

ϩ

...
Setting ϭ Ϫ3 then
immediately yields ϭ 8
...
Thus
( )ϭ

6 2 ϩ 50
8
Ϫ2 ϩ 6
ϭ
ϩ 2

...
This was done by termwise division in Example 2
...

ϩ4

It follows from parts (c), (d), and (e) of Theorem 7
...
1 that the solution of the initialvalue problem is ( ) ϭ 8 Ϫ3 Ϫ 2 cos 2 ϩ 3 sin 2

Solve Љ Ϫ 3 Ј ϩ 2 ϭ

Ϫ4

,

(0) ϭ 1,

Ј(0) ϭ 5
...
We take the sum
of the transforms of each term, use (6) and (7), use the given initial conditions, use (c) of
Theorem 7
...
1, and then solve for ( ):

Ά · Ϫ 3ᏸ Ά · ϩ 2 ᏸ { } ϭ ᏸ {
2

ᏸ
2

( )Ϫ

2

(0) Ϫ Ј(0) Ϫ 3[ ( ) Ϫ (0)] ϩ 2 ( ) ϭ
(

( )ϭ

2

ϩ2
ϩ
Ϫ3 ϩ2 (

2

2

Ϫ4

}

1
ϩ4
1
ϩ4

Ϫ 3 ϩ 2) ( ) ϭ ϩ 2 ϩ

2
1
ϩ6 ϩ9

...
In view of the results in (4) and (5) we have the solution of the
initial-value problem
( ) ϭ ᏸ Ϫ1{ ( )} ϭ Ϫ

16
5

ϩ

25
6

2

ϩ

1
30

Ϫ4


...
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...
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...
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...
3 and 4
...
6
...
Yes, there is a lot of algebra inherent in the use of the Laplace transform,
observe that we do not have to use variation of parameters or worry about the
cases and algebra in the method of undetermined coefficients
...

The Laplace transform has many operational properties
...


( ) The inverse Laplace transform of a function ( ) may not be unique; in other
words, it is possible that ᏸ { 1( )} ϭ ᏸ { 2( )} and yet 1
2
...
If 1 and 2 are piecewise continuous
on [0, ϱ) and of exponential order, then 1 and 2 are
the same
...
2
...

( ) This remark is for those of you who will be required to do partial fraction
decompositions by hand
...
Let us illustrate by reexamining Example 3
...
Since the coefficients of
the right-hand side of the equality are zero, we get
2

ϩ6 ϩ9
( Ϫ 2)( ϩ 4)

͉

or

ϭ
ϭ1

(15)
and

on

16
ϭϪ
...
Now to obtain and , we simply evaluate
the left-hand side of (15) while covering up, in turn, Ϫ 2 and ϩ 4:
2ϩ6 ϩ9
––––––––––––––––––––––
( Ϫ 1)( Ϫ 2)( ϩ 4)

and

2ϩ6 ϩ9
––––––––––––––––––––––
( Ϫ 1)( Ϫ 2)( ϩ 4)

͉

͉

ϭ2

ϭϪ4

25
–
ϭ –– ϭ
6
1
–
ϭ –– ϭ
...
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...
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...
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...
This special technique for
determining coefficients is naturally known as the
( ) In this remark we continue our introduction to the terminology of
dynamical systems
...
The polynomial
( )ϭ
ϩ Ϫ1 Ϫ1 ϩ и и и ϩ 0 in (11) is the total coefficient of ( ) in
(10) and is simply the left-hand side of the DE with the derivatives
͞
replaced by powers , ϭ 0, 1,
...


(16)

In this manner we have separated, in an additive sense, the effects on the
response that are due to the initial conditions (that is, ( ) ( )) from those
due to the input function (that is, ( ) ( ))
...
Hence the
response ( ) of the system is a superposition of two responses:
( ) ϭ ᏸ Ϫ1{ ( ) ( )} ϩ ᏸ Ϫ1{ ( ) ( )} ϭ

0(

)ϩ

1(

)
...

This solution is called the
of the system
...
Now if the
initial state of the system is the zero state (all the initial conditions are zero), then
( ) ϭ 0, and so the only solution of the initial-value problem is 1( )
...
Both 0 ( ) and 1( ) are
particular solutions: 0 ( ) is a solution of the IVP consisting of the associated homogeneous equation with the given initial conditions, and 1( ) is a solution of the
IVP consisting of the nonhomogeneous equation with zero initial conditions
...


Verify that the sum of 0 ( ) and 1( ) is the solution ( ) in Example 5 and that
Ј
Ј
0 (0) ϭ 1, 0 (0) ϭ 5 , whereas 1(0) ϭ 0, 1(0) ϭ 0
...
2
...
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...
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...
9
Ά( Ϫ 0
...
2)·
Ά΂ Ϫ 13Ϫ 3ϩ 13 ΃·
΃΂
Ά( Ϫ 2)( Ϫ 3)( Ϫ 6)·
Ά ( Ϫ 1)( ϩ 1 Ϫ 2)·
ϩ 1)(
1
Ά ϩ5·
Ϫ
Ά( ϩ2 )( 4ϩ 1)·
1
Ά( ϩ 1)( ϩ 4)·

Ά
Ά

ᏸ Ϫ1

2

ᏸ Ϫ1

2

ᏸ Ϫ1

2

ᏸ Ϫ1

2

ϩ1
Ϫ4
1
ϩ

·

2 ٞϩ3 ЉϪ3 ЈϪ2 ϭ Ϫ,
Љ(0) ϭ 1
ٞ ϩ 2 Љ Ϫ Ј Ϫ 2 ϭ sin 3 ,
Љ(0) ϭ 1

Ϫ 20·

Ά(
Ά(

ᏸ Ϫ1

ᏸ Ϫ1

ᏸ Ϫ1

ᏸ Ϫ1

Ά(
Ά
Ά

ᏸ Ϫ1

ϩ 2)(

ᏸ Ϫ1

2

2

ᏸ Ϫ1

4

ᏸ Ϫ1

2

2

ᏸ Ϫ1

4

2

·

ϩ 4)

·

1
Ϫ9

·

6 ϩ3
ϩ52ϩ4

In Problems 31– 40 use the Laplace transform to solve the
given initial-value problem
...


In Problems 41 and 42 use the Laplace transform and these
inverses to solve the given initial-value problem
...
1 are

ᏸ Ϫ1

ᏸ Ϫ1

(0) ϭ 0,

Јϩ ϭ

Ϫ3

cos 2 ,

Љ Ϫ 2 Ј ϩ 5 ϭ 0,

(0) ϭ 0
(0) ϭ 1,

Ј(0) ϭ 3

With a slight change in notation the transform in (6)
is the same as

ᏸ { Ј( )} ϭ ᏸ { ( )} Ϫ (0)
...
1
...

Proceed as in part (a), but this time discuss how to
use (7) with ( ) ϭ sin in conjunction with (d)
and (e) of Theorem 7
...
1 to evaluate ᏸ { sin }
...
Do not think profound thoughts
...
Find the
zero-input and the zero-state response for the IVP in
Problem 36
...
1 to justify
(0) ϭ lim

:ϱ

where
cos

( ),

( ) ϭ ᏸ{ ( )}
...
1
...
For example, the integration by parts involved in evaluating, say,
ᏸ { 2 sin 3 } is formidable, to say the least
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Evaluating transforms such as ᏸ { 5 3} and ᏸ { Ϫ2 cos 4 } is
straightforward provided that we know (and we do) ᏸ { 3} and ᏸ {cos 4 }
...
This result is known as the
or

If ᏸ { ( )} ϭ ( ) and is any real number, then

ᏸ{

( )} ϭ ( Ϫ )
...
1
...


0

If we consider a real variable, then the graph of ( Ϫ ) is the graph of ( )
shifted on the -axis by the amount ͉ ͉
...
See Figure 7
...
1
...


ᏸ{

Evaluate

5 3

}

ᏸ{

Ϫ2

( ) of ( ) we replace the

cos 4 }
...
1
...
3
...


ᏸ{
ᏸ{

5 3

} ϭ ᏸ { 3}͉

Ϫ2

: Ϫ5

ϭ

3!
4

cos 4 } ϭ ᏸ {cos 4 }͉

͉

ϭ
: Ϫ5

: Ϫ(Ϫ2)

6
( Ϫ 5)4

ϭ

2

ϩ 16

͉

ϭ
: ϩ2

ϩ2
( ϩ 2)2 ϩ 16

To compute the inverse of ( Ϫ ), we
must recognize ( ), find ( ) by taking the inverse Laplace transform of ( ), and
then multiply ( ) by the exponential function
...

The first part of the next example illustrates partial fraction decomposition in the
case when the denominator of ( ) contains

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Ά(2 Ϫϩ3)5 ·

Evaluate

ᏸ Ϫ1

Ά

ᏸ Ϫ1

2

2

>2 ϩ 5>3

...
Recall that if ( Ϫ ) appears in the denominator of a
rational expression, then the assumed decomposition contains partial fractions
with constant numerators and denominators Ϫ , ( Ϫ )2,
...

ϭ
ϩ
( Ϫ 3)2
Ϫ 3 ( Ϫ 3)2
By putting the two terms on the right-hand side over a common denominator, we
obtain the numerator 2 ϩ 5 ϭ ( Ϫ 3) ϩ , and this identity yields ϭ 2 and
ϭ 11
...
In this situation we
2

·

(3)

shifted three units to the right
...

Ϫ3
( Ϫ 3)2

1
1
ϭ ᏸ Ϫ1 2
Ϫ 3)2

ᏸ Ϫ1

(2)

2

3


...


(4)

ϩ 4 ϩ 6 has no real zeros and
:

> 2 ϩ 5> 3
>2 ϩ 5>3

...
What we are
trying to do is analogous to working part (b) of Example 1 backwards
...

However, we must fix up the numerator by manipulating the constants:
1
5
1
5
2
1
2
2 ϩ 3 ϭ 2 ( ϩ 2) ϩ 3 Ϫ 2 ϭ 2 ( ϩ 2) ϩ 3
...
2
...

Ϫ1

: ϩ2

Ϫ2

12
ϩ2

2

·

͉

: ϩ2

·

(6)
(7)

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Solve Љ Ϫ 6 Ј ϩ 9 ϭ

2 3

,

(0) ϭ 2,

Ј(0) ϭ 17
...
After using linearity, Theorem 7
...
1, and the
initial conditions, we simplify and then solve for ( ) ϭ ᏸ { ( )}:

ᏸ { Љ} Ϫ 6ᏸ { Ј} ϩ 9ᏸ { } ϭ ᏸ { 2
2

( )Ϫ

3

}

2
( Ϫ 3)3

(0) Ϫ Ј(0) Ϫ 6[ ( ) Ϫ (0)] ϩ 9 ( ) ϭ
2

Ϫ 6 ϩ 9) ( ) ϭ 2 ϩ 5 ϩ

2
( Ϫ 3)3

( Ϫ 3)2 ( ) ϭ 2 ϩ 5 ϩ

(

2
( Ϫ 3)3

( )ϭ

2 ϩ5
2

...

ϩ
ϩ
2
Ϫ 3 ( Ϫ 3)
( Ϫ 3)5

( )ϭ
Thus

Ά

( ) ϭ 2ᏸ Ϫ1

·

Ά

·

Ά

·

1
1
2
4!

...
3
...


(0) ϭ 0,

Ј(0) ϭ 0
...


ϩ 4 ϩ 6) ( ) ϭ
( )ϭ

1

ϩ

Ϫ

}

1
ϩ1

2 ϩ1
( ϩ 1)
2 ϩ1
( ϩ 1)( 2 ϩ 4 ϩ 6)

Since the quadratic term in the denominator does not factor into real linear factors, the
partial fraction decomposition for ( ) is found to be
( )ϭ

1>6

ϩ

1> 3
> 2 ϩ 5> 3

...
So in view of the
results in (6) and (7) we have the solution

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...
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...
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...


1
1
ϩ2
2
12
Ϫ ᏸ Ϫ1
Ϫ
ᏸ Ϫ1
ϩ1
2
( ϩ 2)2 ϩ 2
312
( ϩ 2)2 ϩ 2
●

Ά·

cos 12 Ϫ

Ά

1
1
1
( ) ϭ ᏸ Ϫ1
ϩ ᏸ Ϫ1
6
3
ϭ

1 1
ϩ
6 3

Ϫ

Ϫ

1
2

Ϫ2

·

12
3

Ά

Ϫ2

sin 12
...
” For example, an external force acting on a mechanical
system or a voltage impressed on a circuit can be turned off after a period of time
...
This function is called
the
or the
, named after the English polymath
(1850–1925)
...


Notice that we define ᐁ ( Ϫ ) only on the nonnegative -axis, since this is all
that we are concerned with in the study of the Laplace transform
...
3
...
In the case
when ϭ 0, we take ᐁ ( ) ϭ 1 for Ն 0
...
For example, consider
the function ( ) ϭ 2 Ϫ 3
...
See Figure 7
...
3
...
For example, if we consider 0 Յ Ͻ 2, 2 Յ Ͻ 3, and Ն 3
and the corresponding values of ᐁ ( Ϫ 2) and ᐁ ( Ϫ 3), it should be apparent
that the piecewise-defined function shown in Figure 7
...
4 is the same as
( ) ϭ 2 Ϫ 3 ᐁ ( Ϫ 2) ϩ ᐁ ( Ϫ 3)
...


(10)

Similarly, a function of the type

()ϭ

1

Function is
( ) ϭ 2 Ϫ 3ᐁ ( Ϫ 2) ϩ ᐁ ( Ϫ 3)

Ά0,
1,

Ά

0,
0Յ Ͻ
( ),
Յ Ͻ
0,
Ն

(11)

can be written
( ) ϭ ( )[ ᐁ ( Ϫ ) Ϫ ᐁ ( Ϫ )]
...
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...
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...
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...
Graph
...
3
...
Now from (9) and (10) with
ϭ 5, ( ) ϭ 20 , and ( ) ϭ 0 we get ( ) ϭ 20 Ϫ 20 ᐁ ( Ϫ 5)
...
The piecewise-define

()

Ն
()

If ( ) ϭ ᏸ { ( )} and
Ϫ

Ά0,( Ϫ

0Յ Ͻ
(13)
),
Ն
plays a significant role in the discussion that follows
...
3
...

We saw in Theorem 7
...
1 that an exponential multiple of ( ) results in a translation of the transform ( ) on the -axis
...
3
...
This result, presented next in its direct transform
version, is called the
or
( Ϫ ) ᐁ( Ϫ ) ϭ

0, then

ᏸ { ( Ϫ ) ᐁ ( Ϫ )} ϭ

Ϫ

( )
...


one for
Ն

ϭ

Ϫ ( ϩ )

in the last integral, then

͵

ϱ

( )

ϭ

Ϫ

0

Ϫ

( )

ϭ

Ϫ

ᏸ { ( )}
...
This can be
from either Definition 7
...
1 or Theorem 7
...
2
...
3
...


(14)

Find the Laplace transform the function in Figure 7
...
4
...
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...
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...
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...


ϩ

If ( ) ϭ ᏸ Ϫ1{ ( )}, the inverse form of
Theorem 7
...
2,

0, is

ᏸ Ϫ1{

Ά

Evaluate

ᏸϪ1{ ( )} ϭ

4

1
Ϫ4

ᏸ Ϫ1

·

Ϫ2

Ά

ᏸ Ϫ1

2

Ϫ

ϩ9

With the three identifications
, we have from (15)

Ά

ᏸ Ϫ1
With

( )} ϭ ( Ϫ ) ᐁ ( Ϫ )
...


/2

ϭ 2,

( ) ϭ 1͞( Ϫ 4), and

ᐁ ( Ϫ 2)
...


/2

Ϫ

ϩ9

The last expression can be simplified somewhat by using the addition formula for the
cosine
...


We are frequently confronted with
the problem of finding the Laplace transform of a product of a function and a unit
step function ᐁ ( Ϫ ) where the function lacks the precise shifted form ( Ϫ )
in Theorem 7
...
2
...
For example, if
we wanted to use Theorem 7
...
2 to find the Laplace transform of 2 ᐁ ( Ϫ 2), we
would have to force ( ) ϭ 2 into the form ( Ϫ 2)
...
Therefore

ᏸ { 2 ᐁ ( Ϫ 2)} ϭ ᏸ {( Ϫ 2)2 ᐁ ( Ϫ 2) ϩ 4( Ϫ 2) ᐁ ( Ϫ 2) ϩ 4ᐁ ( Ϫ 2)},
where each term on the right-hand side can now be evaluated by Theorem 7
...
2
...
3
...
Using Definition 7
...
1, the definitio
of ᐁ ( Ϫ ), and the substitution ϭ Ϫ , we obtain

͵

ϱ

ᏸ { ( ) ᐁ ( Ϫ )} ϭ
That is,

͵

ϱ

Ϫ

()

ϭ

Ϫ ( ϩ )


...


(16)

Evaluate ᏸ{cos ᐁ ( Ϫ )}
...
Hence by (16),

ᏸ {cos ᐁ ( Ϫ )} ϭ Ϫ

Ϫ

ᏸ {cos } ϭ Ϫ

2

Ϫ

ϩ1


...
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...
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...
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...


The function can be written as ( ) ϭ 3 cos ᐁ ( Ϫ ), so by linearity, the results of Example 7, and the usual partial fractions, we have

ᏸ { Ј} ϩ ᏸ { } ϭ 3ᏸ {cos ᐁ ( Ϫ )}
( ) Ϫ (0) ϩ ( ) ϭ Ϫ3

Ϫ

2

ϩ1
3
( ϩ 1) ( ) ϭ 5 Ϫ 2
ϩ1

( )ϭ

΄

5
3
1
Ϫ
Ϫ
ϩ1 2
ϩ1

Ϫ

ϩ

2

1
ϩ1

Ϫ

Ϫ

ϩ

2

Ϫ

ϩ1

Now proceeding as we did in Example 7, it follows from (15) with
inverses of the terms inside the brackets are

Ά

ᏸ Ϫ1

1
ϩ1

Ϫ

·ϭ

Ϫ( Ϫ )

Ά

and

5
4
3
2
1

Ά

ᐁ ( Ϫ ),

ᏸ Ϫ1

2

Ϫ

ϩ1

ᏸ Ϫ1

2

1
ϩ1

· ϭ cos( Ϫ

Ϫ

· ϭ sin( Ϫ


...


Thus the inverse of (17) is

y

()ϭ5

Ϫ

ϭ5

Ϫ

t

_1
_2

2

3

Graph of function (18)

Ά

5

ϭ

5

3 Ϫ( Ϫ )
3
3
ᐁ ( Ϫ ) Ϫ sin( Ϫ ) ᐁ ( Ϫ ) Ϫ cos( Ϫ ) ᐁ ( Ϫ )
2
2
2
3
ϩ [ Ϫ( Ϫ ) ϩ sin ϩ cos ] ᐁ ( Ϫ )
; trigonometric identities
2

Ϫ
Ϫ

ϩ

,

0Յ Ͻ
3
ϩ
2

Ϫ( Ϫ

3
3
)
ϩ sin ϩ cos ,
2
2

(18)

Ն
...
3
...


in Example 9

In Section 5
...
The Laplace transform is particularly useful in solving (19) when ( )
is piecewise-defined
...
Note, too, that the next
example is a boundary-value problem rather than an initial-value problem
...
3
...
Find the
deflection of the beam when the load is given b

Embedded beam with
variable load in Example 10

( )ϭ

Ά

0

0,

΂1 Ϫ 2 ΃,

>2 Ͻ
0Ͻ

Ͻ >2
Ͻ
...
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...
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...
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...
Now by (10) we can express
( ) in terms of the unit step function:

΂1 Ϫ 2 ΃ Ϫ ΂1 Ϫ 2 ΃ ᐁ΂

( )ϭ

0

2

ϭ

0

΄2 Ϫ

΂

ϩ

Ϫ

2

΃ ᐁ΂

Ϫ

2

Transforming (19) with respect to the variable gives
΂

4

( )Ϫ

3

(0) Ϫ

2

If we let

1

ϭ Љ(0) and

2

( )ϭ

Љ(0) Ϫ ٞ (0)΃ ϭ

Ј(0) Ϫ
4

or

( )Ϫ

Љ(0) Ϫ ٞ(0) ϭ

ϭ ٞ(0), then
1
3

2
4

ϩ

2

ϩ

0

ϭ

1

2!

0

2

΃

΃
...


,

>2
1
1
Ά2!· ϩ 3! ᏸ Ά3!· ϩ 2 ΄ 4! ᏸ Ά4!· Ϫ 5! ᏸ Ά5!· ϩ 5! ᏸ Ά5!
ϩ
ϩ
΄52 Ϫ ϩ ΂ Ϫ 2΃ ᐁ΂ Ϫ 2΃
...


2

·

Ά

ᏸ Ϫ1

2

2

Ά(

·

ᐁ
2΃ ΂
5

Ϫ

Ά

ᏸ Ϫ1

·

ᏸ Ϫ1

ϩ4 ϩ5

Ά

Ϫ

2

΃
...


2

ϭ0

·

Ϫ1
ϩ 1)3

ᏸ Ϫ1

·
·

Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 21– 30 use the Laplace transform to solve the
given initial-value problem
...
Consider two cases:
1͞
and ϭ 1͞

(0) ϭ 0, Ј(0) ϭ 1

Љ Ϫ 4 Ј ϩ 4 ϭ 3,

(0) ϭ 1, Ј(0) ϭ 0

Љ Ϫ 6 Ј ϩ 13 ϭ 0,

(0) ϭ 0, Ј(0) ϭ Ϫ3

2 Љ ϩ 20 Ј ϩ 51 ϭ 0,
ЉϪ Јϭ

cos ,

(0) ϭ 2, Ј(0) ϭ 0

(0) ϭ 0, Ј(0) ϭ 0

ЉϪ2 Јϩ5 ϭ1ϩ ,

In Problems 37 – 48 find either ( ) or ( ), as indicated
...


(20)

See Section 5
...
Use the Laplace transform to find ( )
when ϭ 1 h, ϭ 20 ⍀, C ϭ 0
...
What is the current ( )?
Consider a battery of constant voltage 0 that charges
the capacitor shown in Figure 7
...
9
...
The weight
is released from rest 18 inches above the equilibrium
position, and the resulting motion takes place in a
medium offering a damping force numerically equal to
7
8 times the instantaneous velocity
...


2

ᏸ{

Ά

΂ Ϫ 2 ΃·
Ϫ2 2

ᏸ Ϫ1

3

Ϫ

ᏸ Ϫ1

ᐁ ( Ϫ 2)}

Ά(1 ϩϩ 2 ) ·
Ά ϩ 4·
Ά ( Ϫ 1)·

Ϫ2

(0) ϭ 0, Ј(p) ϭ 0

2

2Ϫ

ᏸ {( Ϫ 1) ᐁ ( Ϫ 1)}

ᏸ {cos 2 ᐁ ( Ϫ )}

In Problems 31 and 32 use the Laplace transform and
the procedure outlined in Example 10 to solve the given
boundary-value problem
...
The graph of ( ) is given in
Figure 7
...
10
...

Graph for Problem 49

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

()

Ά0, ,
sin

()ϭ
Graph for Problem 50

()ϭ

Ά0,,

()ϭ

sin
Ά0, ,

0 Յ Ͻ 3 >2
Ն 3 >2

0Յ Ͻ2
Ն2

0Յ Ͻ2
Ն2
()

()

1

rectangular pulse

Graph for Problem 51

Graph for Problem 61
()

()

3
2
1
1

Graph for Problem 52

2

3

4

staircase function

Graph for Problem 62

()

In Problems 63–70 use the Laplace transform to solve the
given initial-value problem
...
Find the Laplace transform of the given function
...
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...
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...
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...
If the
weight is released from rest at the equilibrium position,
find the equation of motion ( ) if an impressed force
( ) ϭ 20 acts on the system for 0 Յ Ͻ 5 and is then
removed (see Example 5)
...

Use a graphing utility to graph ( ) on the interval [0, 10]
...

In Problems 73 and 74 use the Laplace transform to find the
charge ( ) on the capacitor in an
series circuit subject to
the given conditions
...
5 ⍀,
Figure 7
...
19

ϭ 0
...
01 f, and ( ) is as
given in Figure 7
...
22
...
Use a computer graphing
program to graph ( ) for 0 Յ Յ 6
...

()
0

( ) given in

1

3

( ) in Problem 76

()
5

A cantilever beam is embedded at its left end and free
at its right end
...


Solve Problem 77 when the load is given by
(0) ϭ 0 ,
ϭ 10 ⍀,
Figure 7
...
20

ϭ 0
...
5

( ) in Problem 74

Use the Laplace transform to find the current
( ) in a single-loop
-series circuit when
(0) ϭ 0, ϭ 1 h, ϭ 10 ⍀, and ( ) is as given
in Figure 7
...
21
...
Use the graph to estimate max and
min , the maximum and minimum values of the
current
...


Find the deflection ( ) of a cantilever beam embedded
at its left end and free at its right end when the load is as
given in Example 10
...
Find the deflection ( ) when the load is
as given in Problem 77
...
1 on the cooling of a cake that is taken out of an
oven
...

Use the Laplace transform to solve the initial-value
problem in part (a)
...
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...
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...
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...
Show that ᏸ {
deduce

Discuss how you would fix up each of the following
functions so that Theorem 7
...
2 could be used directly
to find the given Laplace transform
...


ᏸ {(2 ϩ 1) ᐁ ( Ϫ 1)}
ᏸ {cos ᐁ ( Ϫ )}

2

●

Ϫ 2
( 2 ϩ 2)2
2
}ϭ 2

...


Assume that Theorem 7
...
1 holds when the symbol is replaced by , where is a real number

●

} can be used to

Definition 7
...

Theorems 7
...
1 and 7
...
2

In this section we develop several more operational properties of the
Laplace transform
...
The last two transform properties allow us to solve some equations that we have
not encountered up to this point: Volterra integral equations, integrodifferential equations, and ordinary differential equations in which the input function is a periodic piecewise-defined function
...
To
motivate this result, let us assume that ( ) ϭ ᏸ { ( )} exists and that it is possible
to interchange the order of differentiation and integration
...


We can use the last result to find the Laplace transform of

ᏸ { 2 ( )} ϭ ᏸ { ؒ

()

Ϫ

0

ᏸ { ( )} ϭ Ϫ

΂Ϫ

2

( ):

΃

ᏸ { ( )} ϭ

The preceding two cases suggest the general result for ᏸ {

2
2

ᏸ { ( )}
...


If ( ) ϭ ᏸ { ( )} and ϭ 1, 2, 3,
...


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...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Evaluate ᏸ { sin

}
...
4
...


If we want to evaluate ᏸ { 2 sin } and ᏸ { 3 sin } , all we need do, in turn, is
take the negative of the derivative with respect to of the result in Example 1 and
then take the negative of the derivative with respect to of ᏸ { 2 sin }
...
3
...
4
...
For example,
Theorem 7
...
1: ᏸ {

3

Theorem 7
...
1: ᏸ {

3

} ϭ ᏸ{ }
}ϭϪ

Solve Љ ϩ 16 ϭ cos 4 ,

: Ϫ3

ᏸ{

(0) ϭ 0,

3

1

ϭ

2

͉

: Ϫ3

ϭ

1

...

ϭ ( Ϫ 3)Ϫ2 ϭ
Ϫ3
( Ϫ 3)2

}ϭϪ

Ј(0) ϭ 1
...
The mass starts with an initial velocity of
1 ft /s in the downward direction from the equilibrium position
...


Now we just saw in Example 1 that

Ά(

ᏸ Ϫ1

2

2
ϩ

·ϭ

2 2

)

sin

,

(1)

and so with the identification ϭ 4 in (1) and in part (d) of Theorem 7
...
1, we obtain
()ϭ
ϭ

Ά

1 Ϫ1
ᏸ
4

2

·

Ά

4
1
ϩ ᏸ Ϫ1
ϩ 16
8
(

2

·

8
ϩ 16)2

1
1
sin 4 ϩ
sin 4
...


(3)

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...
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...
Editorial review has
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...


●

It is left as an exercise to show that

͵

͵

(␶) ( Ϫ ␶) ␶ ϭ

0

( Ϫ ␶ ) (␶ ) ␶ ;

0

that is, ‫ ء‬ϭ ‫
...

It is
true that the integral of a product of functions is the product of the
integrals
...
The result that follows is known as the

If ( ) and
order, then

( ) are piecewise continuous on [0, ϱ) and of exponential

ᏸ { ‫ } ء‬ϭ ᏸ { ( )} ᏸ { ( )} ϭ ( ) ( )
...


0
ϱ

and

( ) ϭ ᏸ { ( )} ϭ

0

Proceeding formally, we have

΂͵
͵͵
͵ ͵
ϱ

( ) ( )ϭ

0

ϱ

ϭ

0

: to

ϱ

Ϫ ( ␶ ϩ ␤)

ϱ

(␤) ␤

΃

(␶) (␤ ) ␶ ␤
Ϫ (␶ ϩ␤)

(␶) ␶

0

(␤) ␤
...


␶

In the t-plane we are integrating over the shaded region in Figure 7
...
1
...
} ء‬

·

sin( Ϫ ␶) ␶
...

ϩ 1 ( Ϫ 1)( 2 ϩ 1)

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

The convolution theorem is sometimes
useful in finding the inverse Laplace transform of the product of two Laplace
transforms
...
4
...
ء‬

Many of the results in the table of Laplace transforms in Appendix III can be derived
using (4)
...

ϩ 2 )2

(5)

·
...


͵

·

1
1
ϭ 2 sin ␶ sin ( Ϫ ␶) ␶
...

2 3
Multiplying both sides by 2 3 gives the inverse form of (5)
...

(7)

Ά͵

0

The inverse form of (7),

͵

·

Ά ( )·,

(␶) ␶ ϭ ᏸ Ϫ1

0

(8)

can be used in lieu of partial fractions when is a factor of the denominator and
( ) ϭ ᏸ Ϫ1{ ( )} is easy to integrate
...

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

The convolution theorem and the result in (7)
are useful in solving other types of equations in which an unknown function appears
under an integral sign
...


(9)

0

The functions ( ) and ( ) are known
...


Ϫ␶

0

In the integral we identify ( Ϫ t) ϭ Ϫt so that ( ) ϭ
...
4
...

ϩ1
Ϫ1

Ϫ

After solving the last equation for
decomposition, we find
( )ϭ

6
3

( ) and carrying out the partial fraction
6

Ϫ

4

ϩ

1

2

...


In a single-loop or series circuit, Kirchhoff’s second law states
that the sum of the voltage drops across an inductor, resistor, and capacitor is equal
to the impressed voltage ( )
...
It follows that the current in a
circuit, such as that shown in Figure 7
...
2, is governed by the
-series circuit

ϩ

()ϩ

1

͵

(␶) ␶ ϭ ( )
...
1 f, (0) ϭ 0, and the impressed voltage is

ϭ 0
...

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...
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...
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...


●

With the given data equation (10) becomes

͵

0
...


0

Now by (7), ᏸ{͵0 (␶) ␶} ϭ ( )͞ , where ( ) ϭ ᏸ { ( )}
...
1 ( ) ϩ 2 ( ) ϩ 10

ϭ 120

΄1 Ϫ 1
2

2

Multiplying this equation by 10 , using
for ( ) gives
( ) ϭ 1200

΄(

2

Ϫ

1

Ϫ


...
3

ϩ 20 ϩ 100 ϭ ( ϩ 10)2, and then solving

1
1
Ϫ
ϩ 10)2
( ϩ 10)2

1
( ϩ 10)2

Ϫ

΄1>100 Ϫ 1>100 Ϫ ( 1>10 Ϫ 1>100
ϩ 10
ϩ 10)

Ϫ

Ϫ

Ϫ


...


Ϫ

From the inverse form of the second translation theorem, (15) of Section 7
...
5

1

1
...
5

Graph of current ( )
in Example 6

᭤

Ϫ 1080( Ϫ 1)

Ϫ10

Ϫ

Ϫ10( Ϫ1)

Ϫ10( Ϫ1)

ᐁ ( Ϫ 1)]

ᐁ ( Ϫ 1)
...


Using this last expression and a CAS, we graph ( ) on each of the two intervals and
then combine the graphs
...
4
...

By applying the Laplace transform
to the initial-value problem
Љϩ
where and

Јϩ

ϭ ( ),

(0) ϭ 0, Ј(0) ϭ 0,

are constants, we find that the transform of ( ) is
( )ϭ

()
2

ϩ

ϩ

,

where ( ) ϭ ᏸ { ( )}
...
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...
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...
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...
On the other hand, we
ϩ ϩ
know from (10) of Section 4
...

By comparing (11) and (12) we see that the Green’s function for the differential
1
equation is related to ᏸ Ϫ1 2
ϭ ( ) by
ϩ ϩ
( , ␶) ϭ ( Ϫ ␶)
...

2
ϩ4

Thus from (13) we see that the Green’s function for the DE Љ ϩ 4 ϭ ( ) is
( , ␶) ϭ ( Ϫ ␶) ϭ 1 sin 2( Ϫ ␶)
...
8
...
The next theorem shows that the Laplace transform of a periodic
function can be obtained by integration over one period
...

1Ϫ Ϫ 0

͵

Write the Laplace transform of as two integrals:

ᏸ { ( )} ϭ

͵

()

ϩ

0

When we let ϭ

͵

ϱ

ϩ , the last integral becomes

͵

ϱ

Ϫ

Therefore

()

ϭ

Ϫ ( ϩ )

͵

ϱ

Ϫ

Ϫ

͵

()


...


0

Ϫ

()

ϩ

Ϫ

ᏸ { ( )}
...


()

Find the Laplace transform of the periodic function shown in Figure 7
...
4
...
For

0Յ Ͻ1
1Յ Ͻ2

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...
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...


●

and outside the interval by ( ϩ 2) ϭ ( )
...
4
...


(14)

The differential equation for the current ( ) in a single-loop

-series circuit is

ϭ ( )
...
4
...

If we use the result in (14) of the preceding example, the Laplace transform of the DE is
()ϩ

()ϭ

1
(1 ϩ

Ϫ

or

)

1>
1
ؒ
( ϩ > ) 1ϩ

( )ϭ

Ϫ


...
With the identification ϭ Ϫ ,
1
1ϩ

ϭ1Ϫ

2

ϩ

Ϫ

3

From
we can then rewrite (16) as
( )ϭ
ϭ

΂

Ϫ

΂

Ϫ

1 1
1 1

1
ϩ >

Ϫ

΃ (1 Ϫ
Ϫ2

ϩ

Ϫ

ϩ

Ϫ2

Ϫ

΃

Ϫ3

ϩиии Ϫ

Ϫ

1
1ϩ

becomes

ϩиии

Ϫ3

1

΂

Ϫ

ϭ1Ϫ

Ϫ

1
>
>
ϭ
Ϫ
( ϩ > )
ϩ >

ϩ

Ϫ2

Ϫ

Ϫ3

ϩиии
...


By applying the form of the second translation theorem to each term of both series,
we obtain
()ϭ

1

(1 Ϫ ᐁ ( Ϫ 1) ϩ ᐁ ( Ϫ 2) Ϫ ᐁ ( Ϫ 3) ϩ и и и)

1
Ϫ (

Ϫ /

Ϫ

Ϫ ( Ϫ1)/

ᐁ ( Ϫ 1) ϩ

Ϫ ( Ϫ2)/

ᐁ ( Ϫ 2) Ϫ

Ϫ ( Ϫ3)/

ᐁ ( Ϫ 3) ϩ и и и)

or, equivalently,
()ϭ

1

(1 Ϫ

Ϫ /

)ϩ

1

ϱ

͚ (Ϫ1)
ϭ1

(1Ϫ

Ϫ ( Ϫ )/

) ᐁ ( Ϫ )
...
In this case
()ϭ1Ϫ

Ϫ

Ϫ (1 Ϫ

Ϫ1

) ᐁ ( Ϫ 1) ϩ (1 Ϫ

Ϫ( Ϫ2)

) ᐁ ( Ϫ 2) Ϫ (1 Ϫ

ϭ 1,

Ϫ( Ϫ3)

ϭ 1,

) ᐁ ( Ϫ 3);

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...
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...
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...


●

2
1
...
5

i

in other words,

Ά

1 Ϫ Ϫ,
Ϫ Ϫ ϩ Ϫ( Ϫ1),
()ϭ
1 Ϫ Ϫ ϩ Ϫ( Ϫ1) Ϫ Ϫ( Ϫ2),
Ϫ Ϫ ϩ Ϫ( Ϫ1) Ϫ Ϫ( Ϫ2) ϩ

t
2

1

3

4

Graph of current ( ) in
Example 8

ᏸ{ 3 }
ᏸ{ sinh 3 }
ᏸ { 2 cos }
ᏸ { Ϫ3 cos 3 }

In Problems 9–14 use the Laplace transform to solve
the given initial-value problem
...


In Problems 19 – 30 use Theorem 7
...
2 to evaluate the given
Laplace transform
...


ᏸ {1 ‫}3 ء‬
ᏸ{
ᏸ

Ϫ

Ά͵

‫ء‬
␶

0

ᏸ

Ά͵

Ϫ␶

ᏸ{ 2 ‫ء‬
cos }

ᏸ{

·

ᏸ

␶

(0) ϭ 0

Ј Ϫ ϭ sin ,
Љ ϩ 9 ϭ cos 3 ,
Љ ϩ ϭ sin ,

(0) ϭ 0
(0) ϭ 2,
(0) ϭ 1,

Љ ϩ 16 ϭ ( ),

(0) ϭ 0,

()ϭ
Љ ϩ ϭ ( ),

Ά

cos 4 ,
0,

(0) ϭ 1,

Ά

1,
()ϭ
sin ,

ᏸ

0

Ј(0) ϭ 5
Ј(0) ϭ Ϫ1
Ј(0) ϭ 1, where
0Յ Ͻ
Ն

0 Յ Ͻ >2
Ն >2

Ј(0) ϭ 0, where

In Problems 15 and 16 use a graphing utility to graph the
indicated solution
...
In Problems 17 and 18 use Theorem 7
...
1 to reduce
the given differential equation to a linear first-order DE
in the transformed function ( ) ϭ ᏸ { ( )}
...

Љ Ϫ Ј ϭ 2 2,
2 Љϩ

(0) ϭ 0

Ј Ϫ 2 ϭ 10,

(0) ϭ Ј(0) ϭ 0

␶

·

ᏸ

‫ ء‬sin }

Ά͵ cos ␶ ␶·

Ά͵ ␶ sin ␶ ␶·
0

·

ᏸ

Ά ͵ sin ␶ cos ( Ϫ ␶) ␶·
0

Ά ͵ sin␶ ␶·

ᏸ

0

Ά ͵␶

Ϫ␶

␶

0

·

In Problems 31 – 34 use (8) to evaluate the given inverse
transform
...


Յ
Յ
Յ
Յ

The graph of ( ) for 0 Յ Ͻ 4, given in Figure 7
...
5, was obtained with the help
of a CAS
...
4
...


ᏸ { Ϫ10 }
ᏸ{ cos 2 }
ᏸ { 2 sinh }
ᏸ { 2 sin 6 }

0
1
2
Ϫ( Ϫ3)
, 3

·

8 3

...
Use a CAS as an aid in evaluating
the convolution integral
...
Could you have
obtained the result in a different manner?
Use the Laplace transform and the results of Problem 35
to solve the initial-value problem
Љϩ

ϭ sin ϩ sin ,

(0) ϭ 0,

Ј(0) ϭ 0
...


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...
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...
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...


●

In Problems 37 – 46 use the Laplace transform to solve the
given integral equation or integrodifferential equation
...
Use a graphing utility to graph
the solution for 0 Յ Յ 3
...
1 h, ϭ 3 ⍀, ϭ 0
...
005 h, ϭ 1 ⍀, ϭ 0
...
4
...


3

4

half-wave rectification of sin

Graph for Problem 54

()
1
2

3

4

1

In Problems 55 and 56 solve equation (15) subject to
(0) ϭ 0 with ( ) as given
...

( ) is the meander function in Problem 49 with
amplitude 1 and ϭ 1
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 57 and 58 solve the model for a driven spring/
mass system with damping

Show that the square wave function ( ) given in
Figure 7
...
4 can be written
ϱ

2
2

ϩ␤

ϭ ( ),

ϩ

(0) ϭ 0,

()ϭ

Ј(0) ϭ 0,

͚ (Ϫ1)
ϭ0

ᐁ ( Ϫ )
...
Use a graphing
utility to graph ( ) for the indicated values of
...


ϭ 1, b ϭ 1, ϭ 5, is the meander function in
2
Problem 49 with amplitude 10, and ϭ p, 0 Յ Ͻ 2p
...


ϭ 1, b ϭ 2, ϭ 1, is the square wave in Problem 50
with amplitude 5, and ϭ p, 0 Յ Ͻ 4p
...


Discuss how Theorem 7
...
1 can be used to fin

Use this result to find the Laplace transform of the given
function
...


Ά

·

Ϫ3

...
4 we saw that Љ ϩ Ј ϩ ϭ 0 is Bessel’s
equation of order n ϭ 0
...
4
...
Use
this result and the procedure outlined in the instructions
to Problems 17 and 18 to show that

ᏸ { 0 ( )} ϭ

1

...
2
...

The point of this problem to guide you through the
formal steps leading to the Laplace transform of
( ) ϭ ln ,
0
...


Љ ϩ (1 Ϫ ) Ј ϩ

ϭ0

is known to possess polynomial solutions when is
a nonnegative integer
...
Find ϭ ( ), for ϭ 0, 1, 2, 3, 4 if it is
known that (0) ϭ 1
...
Conclude that
()ϭ

Ϫ
,
ϭ 0, 1, 2,
...
See (33) in Section 6
...

Ϫ2

The Laplace transform ᏸ { } exists, but without find
2
ing it solve the initial-value problem Љ ϩ ϭ Ϫ ,
(0) ϭ 0, Ј(0) ϭ 0
...


If ᏸ {ln } ϭ ( ), use Theorem 7
...
1 with
show that part (a) becomes
ϩ

ϭ 1 to

1
ϭϪ
...

Finally, the integral definition of
(sometimes called the
)
͵0ϱ Ϫ ln , where ␥ ϭ 0
...

is ␥ ϭ Ϫ
Use (1) ϭ Ϫ␥ in the solution in part (b) to show that

␥ ln
,
ᏸ {ln } ϭ Ϫ Ϫ

0
...
In
the Laplace transform of
a function ( ) is obtained using
In line two of the syntax we replace
by the symbol (

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...
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...
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...


●

Appropriately modify the procedure of Problem 68 to
find a solution o

Mathematica,
)
Consider the initial-value problem
Љ ϩ 6 Ј ϩ 9 ϭ sin ,

(0) ϭ 2,

Ј(0) ϭ Ϫ1
...
Precisely reproduce
and then, in turn, execute each line in the following
sequence of commands
...

‫؉ ؅ ؉ ؆ ؍‬
‫؍‬
؊Ͼ
؅ ؊Ͼ؊

‫؍؍‬

؊Ͼ
‫؍‬

ٞ ϩ 3 Ј Ϫ 4 ϭ 0,
(0) ϭ 0, Ј(0) ϭ 0,
The charge ( ) on a capacitor in an
given by

Љ(0) ϭ 1
...


Appropriately modify the procedure of Problem 68 to
find ( )
...


‫؍‬

In the last paragraph on page 279, we indicated that as an immediate consequence of Theorem 7
...
3, ( ) ϭ 1 cannot be the Laplace transform of a function that is piecewise
continuous on [0, ϱ) and of exponential order
...
We
shall see that there does indeed exist a function—or, more precisely, a
—whose
Laplace transform is ( ) ϭ 1
...
For example, a vibrating airplane wing could be struck by
lightning, a mass on a spring could be given a sharp blow by a ball peen hammer, and
a ball (baseball, golf ball, tennis ball) could be sent soaring when struck violently by
some kind of club (baseball bat, golf club, tennis racket)
...
5
...
The graph
of the piecewise-defined functio

A golf club applies a
force of large magnitude on the ball for a
very short period of time

Ά

0,
1
␦ ( Ϫ 0) ϭ
,
2
0,

0Յ Ͻ
Ϫ

Ϫ

Յ Ͻ

0

ϩ

Ն

0

0

0

ϩ ,

(1)

0, 0 0, shown in Figure 7
...
2(a), could serve as a model for such a force
...
The behavior of d ( Ϫ 0 ) as : 0
is illustrated in Figure 7
...
2(b)
...

In practice it is convenient to work with another
type of unit impulse, a “function” that approximates d ( Ϫ 0 ) and is defined by
the limit

␦ ( Ϫ 0 ) ϭ lim ␦ ( Ϫ 0 )
...
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...
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...
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...


The unit impulse d( Ϫ 0 ) is called the
It is possible to obtain the Laplace transform of the Dirac delta function by the formal assumption that ᏸ {␦ ( Ϫ 0 )} ϭ lim : 0 ᏸ {␦ ( Ϫ 0 )}
...
3:

␦ ( Ϫ 0) ϭ

0)

Ϫ

(3)


...

2

By linearity and (14) of Section 7
...


Ϫ

(4)

: 0, we apply L’Hôpital’s Rule:

ᏸ {␦ ( Ϫ 0 )} ϭ lim ᏸ {␦ ( Ϫ 0 )} ϭ
Now when

Ϫ

ϭ

0

lim

:0

΂

Ϫ
2

Ϫ

΃ϭ

Ϫ


...

The last result emphasizes the fact that d( ) is not the usual type of function that we have
been considering, since we expect from Theorem 7
...
3 that ᏸ{ ( )} : 0 as : ϱ
...


The two initial-value problems could serve as models for describing the motion of a
mass on a spring moving in a medium in which damping is negligible
...
In (a) the mass is released from rest 1 unit below the
equilibrium position
...

From (3) the Laplace transform of the differential equation is
2

( )Ϫ ϩ ( )ϭ4

Ϫ2

or

( )ϭ

2

ϩ1

ϩ

4
2

Ϫ2


...

Since sin( Ϫ 2p) ϭ sin , the foregoing solution can be written as
()ϭ

Άcos ,ϩ 4 sin ,
cos

0Յ Ͻ2
Ն2
...
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...
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...
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...
5
...
The influence of the unit impulse is to
increase the amplitude of vibration to 117 for
2p
...


(6)

The graph of (6) in Figure 7
...
4 shows, as we would expect from the initial conditions
that the mass exhibits no motion until it is struck at ϭ 2p
...
Because the
0
0
Dirac delta function did not “behave” like an ordinary function, even though its
users produced correct results, it was met initially with great scorn by mathematicians
...
Although we shall not pursue this
topic any further, suffice it to say that the Dirac delta function is best characterized by its effect on other functions
...
This result is known as the
since d( Ϫ 0 ) has the effect of sifting the value ( 0 ) out of the
set of values of on [0, ϱ)
...

( ) In ( ) in the
at the end of Section 7
...

coefficients is ( ) ϭ 1͞ ( ), where ( ) ϭ
The transfer function is the Laplace transform of function ( ), called the
of a linear system
...
For simplicity let us consider a second-order
linear system in which the input is a unit impulse at ϭ 0:
2

Љϩ

1

Јϩ

0

ϭ ␦ ( ),

(0) ϭ 0,

Ј(0) ϭ 0
...


From this we can see, in general, that the weight function ϭ ( ) of an th-order
linear system is the zero-state response of the system to a unit impulse
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

In Problems 1 – 12 use the Laplace transform to solve the
given initial-value problem
...
The beam is embedded at its left end and

●

Ϫ 1 ΃,
2

0

Љ ϩ ϭ d( Ϫ 2p) ϩ d( Ϫ 4p),
Љ ϩ 2 Ј ϭ d( Ϫ 1),

0␦΂

Solve the differential equation in Problem 13 subject to
(0) ϭ 0, Ј(0) ϭ 0, ( ) ϭ 0, Ј( ) ϭ 0
...
See Figure 7
...
5
...


(0) ϭ 0, Ј(0) ϭ 1

Љ ϩ 16 ϭ d( Ϫ 2p),

(

is free at its right end
...
Do you agree or disagree? Defend
your answer
...
We solve the system of
algebraic equations for each of the transformed functions and then find the inverse Laplace transforms in the usual manner
...
In turn the two
springs are attached as shown in Figure 7
...
1 on page 316
...
When
the system is in motion, spring is subject to both an elongation and a compression;
hence its net elongation is 2 Ϫ 1
...
If no external force is impressed on the system and if no damping force is present, then the net
force on 1 is Ϫ 1 1 ϩ 2 ( 2 Ϫ 1)
...


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...
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...
Editorial review has
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...


●

Similarly, the net force exerted on mass
; that is, Ϫ 2 ( 2 Ϫ 1)
...


1

1
1

2

In other words, the motion of the coupled system is represented by the system of
simultaneous second-order differential equations

1

2( 2
2

2
2

2

1 1

2

0

1)

1 Љ
1

2

2
2

2

2( 2

1)

2

ϭϪ

1 1

Љ ϭ Ϫ 2(
2

2( 2

ϩ
Ϫ

2

1)

Ϫ

(1)

1)
...


Coupled spring/mass
system

Љ ϩ 10
1

1

Ϫ4

2

ϭ0

Ϫ4

Solve

1

ϩ Љϩ4
2

2

ϭ0

2 (0)

ϭ 0, Ј (0) ϭ Ϫ1
...
The preceding system is the

ϩ 10)

1(

)Ϫ

Ϫ4

1(

)ϩ(

x1
0
...
2

1(

)ϭ

(

and therefore
5

7
...
5 1 5
1(

x2

)ϭϪ

0
...
4
5

7
...
5 1 5

Displacements of the
two masses in Example 1

ϩ 4)

2(

) ϭ Ϫ1
...
2
2
...

10
5

Ά

1
ᏸ Ϫ1
512

ϭϪ

0
...
2
2
...
4

Ϫ Ј (0) ϩ 4
2

12
3
Ϫ
ᏸ Ϫ1
ϩ2
5112

ϩ6
2> 5
ϭϪ 2
Ϫ
ϩ 2)( 2 ϩ 12)
ϩ2

13
12
sin 12 Ϫ
sin 213
...
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...
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...
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...

2( ) ϭ Ϫ
5
10
1(

The graphs of
each mass
...
6
...
3 we saw the currents 1( ) and 2 ( ) in the
network shown in Figure 7
...
3, containing an inductor, a resistor, and a capacitor,
were governed by the system of first-order di ferential equations

2

1

ϩ

2

ϭ ()
(5)

2

ϩ

2

Ϫ

1

ϭ 0
...


Solve the system in (5) under the conditions ( ) ϭ 60 V,
ϭ 10Ϫ4 f, and the currents 1 and 2 are initially zero
...

Applying the Laplace transform to each equation of the system and simplifying
gives
1(

)ϩ

50 2( ) ϭ

60

Ϫ200 1( ) ϩ ( ϩ 200) 2( ) ϭ 0,
where 1( ) ϭ ᏸ { 1( )} and 2( ) ϭ ᏸ { 2( )}
...

ϩ 100 ( ϩ 100)2

Taking the inverse Laplace transform, we find the currents to b
1(

)ϭ

6 6
Ϫ
5 5

Ϫ100

Ϫ 60

2(

)ϭ

6 6
Ϫ
5 5

Ϫ100

Ϫ 120

Ϫ100

Ϫ100


...
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...
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...
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...
Furthermore, since the current through the capacitor is
5
Ϫ100
, we observe that 3( ) : 0 as : ϱ
...
6
...
We assume that the system
oscillates in a vertical plane under the influence of gravity, that the mass of each rod
is negligible, and that no damping forces act on the system
...
6
...
The positive direction is to the
right; the negative direction is to the left
...
3, the system of differential equations describing the
motion is nonlinear:

1
1
2
2
2

Double pendulum

(

1

ϩ

2
Љ
2 ) 1 ␪1

ϩ

Љ
2 1 2␪2

2
Љ
2 2 ␪2

ϩ

cos (␪1 Ϫ ␪2 ) ϩ
Љ
2 1 2␪1 cos

Ј)2
2 1 2(␪2

sin (␪1 Ϫ ␪2 ) ϩ (

Ј)2
2 1 2(␪1

(␪1 Ϫ ␪2 ) Ϫ

1

2) 1

ϩ

sin (␪1 Ϫ ␪2 ) ϩ

sin ␪1 ϭ 0

2 2

sin ␪2 ϭ 0
...


ϩ

(7)

It is left as an exercise to fill in the details of using the Laplace transform to solve
system (7) when 1 ϭ 3, 2 ϭ 1, 1 ϭ 2 ϭ 16, u1(0) ϭ 1, u 2 (0) ϭ Ϫ1, ␪ 1(0) ϭ 0,
Ј
and ␪Ј(0) ϭ 0
...

2
2
13

␪1( ) ϭ

(8)

With the aid of a CAS the positions of the two masses at ϭ 0 and at subsequent
times are shown in Figure 7
...
5
...
6
...
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...
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...

ϭϪ ϩ

ϭ2 ϩ

ϭ2

ϭ8 Ϫ

(0) ϭ 0,
ϭ

(0) ϭ 1

(0) ϭ 1,

Ϫ2

(0) ϭ 1

Ϫ

ϩ

Ϫ

2

(0) ϭ 2

(0) ϭ 0,

ϩ

Ϫ2

ϩ

ϭ

0

ϩ

2

ϭ1

Ϫ

ϩ

ϭ0
Coupled springs in Problem 14

Show that the system of differential equations for
the currents 2 ( ) and 3 ( ) in the electrical network
shown in Figure 7
...
7 is

(0) ϭ 1

2

2

ϭ0

Ϫ

2

2

ϩ

ϭ0

ϩ

ϩ

(0) ϭ 0,
(0) ϭ 0,
2

ϭ0

Ϫ

2
2

ϩ

2

Ϫ

2

2

ϭ

2

Ϫ4

3

Ϫ4 ϩ

ϭ4

2

ϭ 6 sin

3

2

ϩ

2

ϩ

3

ϭ ()

3

ϭ0

3

2

2

ϩ

(0) ϭ 1, Ј(0) ϭ 0,
(0) ϭ Ϫ1, Ј(0) ϭ 5

Ј(0) ϭ Ϫ2,
Ј(0) ϭ 1

2

1

2
2

2

3

ϩ2 ϭ0

(0) ϭ 0,
ϩ

0

2

(0) ϭ 0

ϩ

2

1

(0) ϭ 0

Ϫ3 Ϫ3 ϭ2

(0) ϭ 0,

ϭ1

1

1

(0) ϭ Ϫ1,

2

Derive the system of differential equations describing the
straight-line vertical motion of the coupled springs shown
in Figure 7
...
6
...

2

ϭ1

ϩ3 ϩ

ϭ5 Ϫ

Solve system (1) when 1 ϭ 3, 2 ϭ 2, 1 ϭ 1,
and 1(0) ϭ 0, 1(0) ϭ 1, 2 (0) ϭ 1, Ј (0) ϭ 0
...


Solve the system in part (a) if ϭ 5 ⍀, 1 ϭ 0
...

2 ϭ 0
...


ϭ0

3

(0) ϭ 8,
(0) ϭ 0,

ϩ2 Ϫ2

Ј(0) ϭ 0,
Ј(0) ϭ 0

(0) ϭ 0,
Ј(0) ϭ 0,

1

3

2

(0) ϭ 0,
Љ(0) ϭ 0

1

2

2
2

ϩ3

ϩ3 ϭ0
Network in Problem 15

2

ϩ3 ϭ

2

(0) ϭ 0,

Ј(0) ϭ 2,

Ϫ

(0) ϭ 0

In Problem 12 in Exercises 3
...
6
...


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...
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...
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...


●

Solve the system if
ϭ 0
...

Determine the current 1( )
...
3 when
1 ϭ 6 ⍀, 2 ϭ 5 ⍀, 1 ϭ 1 h, 2 ϭ 1 h, ( ) ϭ 50 sin V,
2 (0) ϭ 0, and 3 (0) ϭ 0
...


ϭ 50 ⍀,

Solve (5) when
ϭ 60 V,
ϭ 2 h,
ϭ 10Ϫ4 f, 1(0) ϭ 0, and 2 (0) ϭ 0
...
6
...


Find the charge on the capacitor when
ϭ 1 f,
1 ϭ 1 ⍀, 2 ϭ 1 ⍀,
()ϭ
3 (0)

Ά0,
50

Ϫ

,

ϭ 1 h,

0Ͻ Ͻ1
Ն 1,

ϭ 0, and (0) ϭ 0
...

()ϭ

Ά2, Ϫ ,

Ά

0,
( ) ϭ 1,
0,

0Յ Ͻ1
Ն1

0Յ Ͻ2
2Յ Ͻ4
Ն4

Use the Laplace transform and the information
given in Example 3 to obtain the solution (8) of the
system given in (7)
...
Which mass has extreme displacements of
greater magnitude? Use the graphs to estimate the
first time that each mass passes through its equilibrium position
...

Graph u1( ) and u 2 ( ) in the u1u2-plane as parametric
equations
...
6
...
Note that we have used
1 radian Ϸ 57
...
Use a calculator or a table
application in a CAS to construct a table of values
of the angles u1 and u2 for ϭ 1, 2,
...
Then
plot the positions of the two masses at these times
...
Plot
the positions of the two masses at these times
...
6
...
Use
the animation capability of your CAS to make a
“movie” of the motion of the double pendulum
from ϭ 0 to ϭ 10 using a time increment of
0
...
[
: Express the coordinates ( 1( ), 1( ))
and ( 2 ( ), 2( )) of the masses 1 and 2, respectively, in terms of u1( ) and u 2 ( )
...

If is not piecewise continuous on [0, ϱ), then ᏸ { ( )}
will not exist
...

_______
( ) ϭ 2 ͞( 2 ϩ 4) is not the Laplace transform of a
function that is piecewise continuous and of exponential
order
...
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...
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...
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...
_______

ᏸ{

} ϭ _______

Ϫ7

ᏸ{

ᏸ {sin 2 } ϭ _______

ᏸ{

Ϫ7
Ϫ3

} ϭ _______
sin 2 } ϭ _______

0

ᏸ { sin 2 } ϭ _______

Graph for Problem 26

ᏸ {sin 2 ᐁ ( Ϫ )} ϭ _______

Ά20· ϭ _______
Ά3 1 1· ϭ _______
Ϫ
1
Ά( Ϫ 5) · ϭ _______
1
Ά Ϫ 5· ϭ _______
Ά Ϫ 10 ϩ 29· ϭ _______
Ά · ϭ _______
Ά ϩ · ϭ _______
ϩ
1
Ά ϩ · ϭ _______

ᏸ Ϫ1

6

ᏸ Ϫ1

ᏸ Ϫ1

0

Graph for Problem 27

3

ᏸ Ϫ1

2

ᏸ Ϫ1

2

0

1

Ϫ5

ᏸ

Ϫ1

ᏸ Ϫ1

2

ᏸ Ϫ1

2 2

ᏸ{

Ϫ5

2

Ϫ

2

2

} exists for

If ᏸ { ( )} ϭ

ᏸ{͵0

␶

͵0

In Problems 29 – 32 express in terms of unit step functions
...

()
1

_______
...


If ᏸ { ( )} ϭ ( ) and
0, then
ᏸ { ( Ϫ ) ᐁ ( Ϫ )} ϭ _______
...
R
...


3

4

Graph for Problem 29
()

(␶) ␶} ϭ _______ whereas
(␶) ␶} ϭ _______
...
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...
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...

ЉϪ2 Јϩ ϭ ,

(0) ϭ 0, Ј(0) ϭ 5

Љ Ϫ 8 Ј ϩ 20 ϭ

,

(0) ϭ 0, Ј(0) ϭ 0

Љ ϩ 6 Ј ϩ 5 ϭ Ϫ ᐁ ( Ϫ 2),

(0) ϭ 1, Ј(0) ϭ 0

Ј Ϫ 5 ϭ ( ), where

Ά0,,
2

()ϭ

0Յ Ͻ1
,
Ն1

(0) ϭ 1

(0) ϭ 1, where

Ј ϩ 2 ϭ ( ),
Figure 7
...
10
...


(␶) cos( Ϫ ␶) ␶,

(0) ϭ 1

0

(␶) ( Ϫ ␶) ␶ ϭ 6

2

3

2
Љϩ Љϭ
2 Ј ϩ ЉϭϪ 2
(0) ϭ 0,
(0) ϭ 0,
Ј(0) ϭ 0,
Ј(0) ϭ 0

Јϩ ϭ
4 ϩ Јϭ0
(0) ϭ 1, (0) ϭ 2

The current ( ) in an
-series circuit can be determined from the integral equation

͵

(␶) ␶ ϭ ( ),

0

where ( ) is the impressed voltage
...
5 f, and ( ) ϭ 2( 2 ϩ )
...
01 f,
2
respectively
...


4
4

ϩ

ϭ ( ),

where ϭ ( ͞4 )1/4
...
Find
the deflection ( ) of a beam that is supported on an
elastic foundation when
the beam is simply supported at both ends and a constant load 0 is uniformly distributed along its length,
the beam is embedded at both ends and ( ) is a
concentrated load 0 applied at ϭ p͞2
...
]
( )

In Problems 41 and 42 use the Laplace transform to solve
each system
...
See Figure 7
...
11
...
Find the
deflection ( ) if the load per unit length is given by

Ά10,
0,

0Յ Ͻ5
Ն5

is applied to the circuit
...


0
elastic foundation

Beam on elastic foundation in Problem 46

Suppose two identical pendulums are coupled by
means of a spring with constant
...
R
...

Under the same assumptions made in the discussion
preceding Example 3 in Section 7
...


Use the Laplace transform to solve the system when
u1(0) ϭ u 0 , u1 ϭ 0, u 2 (0) ϭ c 0 , u 2 ϭ 0, where
Ј(0)
Ј(0)
u0 and c 0 constants
...

Use the solution in part (a) to discuss the motion
of the coupled pendulums in the special case when

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

the initial conditions are u1(0) ϭ u0 , u1
Ј(0) ϭ 0,
u 2 (0) ϭ u0 , u2 ϭ 0
...

Ј(0)
Ј(0)

2
1

Show that each successive oscillation is 2 >v2
shorter than the preceding one
...

A projectile, such as the canon ball shown in Figure 7
...
13, has weight ϭ
and initial velocity
0 that is tangent to its path of motion
...
9 that
motion of the projectile is describe by the system of
linear differential equations
2
2

2

Coupled pendulums in Problem 47

In Problem 27 in Chapter 5 in Review we examined a spring/mass system in
which a mass
slides over a dry horizontal surface
whose coefficient of kinetic friction is a constant ␮
...
You
were asked to show that the piecewise-defined differential equation for the displacement ( ) of the mass is
given by
2
2

ϩ

Ά

,
ϭ
Ϫ ,

ϭϪ

2

Ј Ͻ 0 (motion to left)
Ј Ͼ 0 (motion to right)
...
Show that the times 0, >2,
, 3 >2,
...


Explain why, in general, the initial displacement
&

...

Use the Laplace transform and the concept of the
meander function to solve for the displacement
( ) for Ն 0
...
5 that on the interval [0, 2p) your solution
agrees with parts (a) and (b) of Problem 28 in
Chapter 5 in Review
...


Use the Laplace transform to solve this system subject to the initial conditions (0) ϭ 0, Ј(0) ϭ 0 cos u,
(0) ϭ 0, Ј(0) ϭ 0 sin u, where 0 ϭ ͉ 0 ͉ is constant and u is the constant angle of elevation shown
in Figure 7
...
13
...

Use ( ) in part (a) to eliminate the parameter in
( )
...
Then the differential equations
describing the motion of the mass are

Љ ϩ v2 ϭ ,

ϭ0

2
0

sin 2u
...
Show
that the same range—less than the maximum—can
be attained by firing the gun at either of two complementary angles u and p>2 Ϫ u
...

Suppose ϭ 32 ft/s2, u ϭ 38Њ, and 0 ϭ 300 ft/s
...
Find the time when the projectile hits the
ground
...
Repeat with u ϭ 52Њ
and 0 ϭ 300 ft/s
...


0

Horizontal Range

Projectile in Problem 49

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...
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...


●

Now
suppose that air resistance is a retarding force tangent to the path but acts opposite to the motion
...
9 that motion of the projectile is describe by the system of linear differential equations
2
2

ϭ Ϫb

2

ϭϪ

2

Ϫb

,

where b 0
...


Suppose ϭ 1 slug, ϭ 32 ft/s2, b ϭ 0
...
Use a CAS to find the time when
the projectile hits the ground and then compute its
corresponding horizontal range
...
Does the property in part (c) of Problem 49
hold?
Use the parametric equations ( ) and ( ) in part (a)
along with the numerical data in part (b) to plot the
ballistic curve of the projectile
...

Superimpose both curves on the same coordinate
system
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


Systems of Linear First-Order
Differential Equations

8
...
2 Homogeneous Linear Systems
8
...
1 Distinct Real Eigenvalues
8
...
2 Repeated Eigenvalues
8
...
3 Complex Eigenvalues
8
...
3
...
3
...
4 Matrix Exponential
Chapter 8 in Review

We encountered systems of ordinary differential equations in Sections 3
...
9, and
7
...
In this chapter we are going to concentrate
only on

...
We will see that this general
theory and solution procedure is similar to that of linear higher-order differential
equations considered in Chapter 4
...
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...
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...
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...
It is imperative that
you review either Appendix II or a linear algebra text if you unfamiliar with these concepts
...
9 we illustrated how to solve systems of
differential equations in unknowns of the form
ϩ
...
ϩ
21( ) 1 ϩ 22( ) 2 ϩ

...


...
ϩ
1
11(

)

1

ϩ

12(

)

2

1

( )

ϭ

1(

)

2

( )

ϭ

2(

)

( )

ϭ

linear


...


...
In this chapter
where the
we confine our study to systems of first-order DEs that are special cases of systems that have the
normal form
1
––– ϭ

1(

,

1,

2
––– ϭ

...


...
,

)

2,


...
,

)
...


...


first-order equations is called a

When each of the functions 1,
in the dependent variables 1, 2 ,
...


...

––– ϭ

21(

)

1

ϩ

22(

1(

)

1

ϩ

)

2

ϩ
...
ϩ

...


...
ϩ
2( ) 2 ϩ
)

2

2,


...


(3)

We refer to a system of the form given in (3) simply as a
We
as well as the functions are continuous on a
assume that the coefficients
common interval When ( ) ϭ 0, ϭ 1, 2,
...
,

...

()

()ϭ

11(

( ), and ( ) denote the respective

)
...

22( )

)
21( )

...


...


()
()

1
2


...


...
,

...

()

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

then the system of linear first-order di ferential equations (3) can be written as

() (
1

––

2


...


...


...


...


12(


...


...

1

()

)( ) ( )
1(

)
()
2

...


...
ϩ

...


()

ϩ
...


Јϭ

΂΃

΂3
5

΃

4
Ϫ7


...

6

ϭ2 ϩ9 Ϫ ϩ 6

A

on an interval is any column matrix

()
1(

ϭ

)
2( )

...


...


A solution vector of (4) is, of course, equivalent to
scalar equations
ϭ f 1( ), 2 ϭ f 2 ( ),
...
In the important case ϭ 2 the equations
1 ϭ f1( ), 2 ϭ f 2 ( ) represent a curve in the 1 2-plane
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Verify that on the interval (Ϫϱ, ϱ)
1

ϭ

1
΂Ϫ1΃

΂Ϫ ΃
Ϫ2

Ϫ2

ϭ

and

Ϫ2

are solutions of

Јϭ

΂1 3΃
5 3

΂ ΃
1 3
ϭ΂
ϭ
5 3΃΂Ϫ ΃ ΂5
1 3 3
3
ϭ΂
΃΂5 ΃ ϭ ΂15
5 3
Ϫ2
2

From Ј ϭ
1

Ϫ2

ϭ

΂3΃
5

6

(6)

Ј ϭ
2

18
30

6
6

Ϫ2

Ϫ2

Ϫ2

Ϫ2

Ϫ2

Ϫ2

Ϫ2

6

and

2

6


...

ϩ 15
30

and

Ϫ2

2

6

6

6

6

6

6

6

1

2

Much of the theory of systems of linear first-order differential equations is
similar to that of linear th-order differential equations
...


...


and

( 0)

where the g , ϭ 1, 2,
...
Then the problem
:

Јϭ () ϩ ()
( 0) ϭ 0

:
is an

()

␥1
␥2
ϭ
...


...


Let the entries of the matrices ( ) and ( ) be functions continuous on a common
interval that contains the point 0
...

In the next several definitions and theorems we are
concerned only with homogeneous systems
...


Let 1, 2 ,
...


, ϭ 1, 2,
...
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...
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...
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...
1
...


You should practice by verifying that the two vectors

1

΂

΃

cos
ϭ Ϫ1 cos ϩ 1 sin
2
2
Ϫcos Ϫ sin

΂΃
0

and

2 ϭ

0

are solutions of the system
Јϭ

΂

1
1
Ϫ2

΃

0
1
0

1
0
Ϫ1

(8)


...

We are primarily interested in linearly independent solutions of the homogeneous system (5)
...
,
be a set of solution vectors of the homogeneous system (5)
on an interval We say that the set is
on the interval if
there exist constants 1, 2 ,
...
If the set of vectors is not linearly dependent on the
interval, it is said to be
The case when ϭ 2 should be clear; two solution vectors 1 and 2 are linearly
dependent if one is a constant multiple of the other, and conversely
...

As in our earlier consideration of the theory of a single ordinary differential equation, we can introduce the concept of the
determinant as a test for linear independence
...


() ()
11
21

Let

1

ϭ


...


...


...


()
1

,

2


...


...


(

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


)

●

be solution vectors of the homogeneous system (5) on an interval Then the set
of solution vectors is linearly independent on if and only if the

(

1,

2,


...


...



...


...


...


Ϳ

0

(9)

for every in the interval
...
,
are solution vectors of (5), then for every
in either ( 1, 2 ,
...
, ) ϭ 0
...

Notice that, unlike our definition of the Wronskian in Section 4
...


In Example 2 we saw that

1

ϭ

1
΂Ϫ1΃

Ϫ2

and

2

ϭ

΂3΃
5

6

are solutions of

system (6)
...
In addition, we have
(

1,

2)

ϭ

͉

Ϫ2
Ϫ2

Ϫ

3
5

6
6

͉

ϭ8

4

0

for all real values of

Any set 1, 2 ,
...


There exists a fundamental set of solutions for the homogeneous system (5) on an
interval
The next two theorems are the linear system equivalents of Theorems 4
...
5
and 4
...
6
...
,
be a fundamental set of solutions of the homogeneous
system (5) on an interval Then the
of the system on the
interval is
ϭ

1

1

ϩ

2

2

ϩиииϩ

,

where the , ϭ 1, 2,
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

From Example 2 we know that

1

ϭ

1
΂Ϫ1΃

Ϫ2

and

2

ϭ

΂3΃
5

6

are linearly

independent solutions of (6) on (Ϫϱ, ϱ)
...
The general solution of the system on the interval
is then

ϭ

1

1

ϩ

2

2

ϭ

1

1
΂Ϫ1΃

Ϫ2

ϩ

2

΂3΃
5

6

(10)


...
1)
...
Thus the general solution of the system on the interval is the
linear combination ϭ 1 1 ϩ 2 2 ϩ 3 3; that is,

ϭ

1

΂

΃ ΂΃ ΂

cos
Ϫ1 cos ϩ 1 sin
2
2
Ϫcos Ϫ sin

ϩ

0
1
0

2

ϩ

3

sin
Ϫ1 sin Ϫ 1 cos
2
2
Ϫsin ϩ cos

΃


...


Let
be a given solution of the nonhomogeneous system (4) on an interval
and let
ϭ

1

1

ϩ

2

2

ϩиииϩ

denote the general solution on the same interval of the associated homogeneous system (5)
...


of the associated homogeneous system (5) is
of the nonhomogeneous system (4)
...
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...
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...
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...
(Verify this
...
Hence by Theorem 8
...
6

ϩ

΂3΃
5

2

6

3
΂Ϫ5 Ϫ 4΃
ϩ6

ϩ

is the general solution of (11) on (Ϫϱ, ϱ)
...

ϭ3 Ϫ5

ϭ4 Ϫ7

ϭ4 ϩ8

ϭ5

΂

7
Јϭ 4
0

ϭ

Ϫ

ϭ6 Ϫ

ϭ

ϩ2

ϭ 10 ϩ 4 ϩ 3

ϭϪ ϩ

ϭ

Ϫ

ϭ2 ϩ
ϭ

ϩ

ϩ ϩ Ϫ1
Ϫ Ϫ32
2

ϩ ϩ

ϭ Ϫ3 ϩ 4 ϩ
ϭ5 ϩ9 ϩ4

Ϫ ϩ2

Ϫ

Ϫ

sin 2

Ϫ9
1
3

1
3
Ϫ2

Ϫ1
Ϫ4
5

΂΃ ΂
ϭ

ϭ Ϫ3 ϩ 4 Ϫ 9

΃ ΂΃ ΂΃

5
1
Ϫ2

΂ ΃ ϭ ΂3
1

ϩ6 Ϫ

Јϭ

4
΂Ϫ1 2΃
3

1
΂Ϫ1΃

Ϫ2

1
ϩ 2
2

Ϫ

3
Ϫ Ϫ1
1

΃΂ ΃ ϩ ΂4΃ sin ϩ ΂2 Ϫ 4΃
8
ϩ1

In Problems 11–16 verify that the vector
the given system
...


5

Ϫ7
1

ϭ Ϫ2 ϩ 4 ;
ϭ

0
ϩ 2
1

Јϭ

΂

Јϭ

2
΂Ϫ1 1΃
0

΃

1
Ϫ1
4
1 Ϫ1

;

;

5
΂3 cos cos sin ΃
Ϫ
ϭ

ϭ

΂Ϫ1΃
2

΂1΃
3

ϩ

Ϫ3 /2

4
΂Ϫ4΃

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...
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...
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...


●

Јϭ

΂
΂

1
6
Ϫ1

΃

2
Ϫ1
Ϫ2

1
Јϭ
1
Ϫ2

0
1
0

1
0
Ϫ1

΃

1
0
Ϫ1

;

ϭ

΂ ΃

Јϭ

΂

sin
1
ϭ Ϫ2 sin Ϫ 1 cos
2
Ϫsin ϩ cos

;

΂΃

1

1

ϭ

2

,

2

ϭ

1

1
ϭ Ϫ2 ϩ
4

Ϫ2

1
΂Ϫ1΃

΂ ΃

,

1
ϭ
Ϫ1

΂2΃
6
2

ϩ

3

1

1
2 ϭ Ϫ2
Ϫ1

Ϫ4

ϭ

●
●

Ϫ1
4 sin 3 ;
3

ϩ

1
΂Ϫ1΃

ϭ

΂ ΃
sin 3
0
cos 3

΃

6
0
1

0
1
0

1

΂΃
6
Ϫ1
Ϫ5

Ϫ

ϩ

2

΂΃
Ϫ3
1
1

Ϫ2

ϩ

3

΂΃
2
1
1

3


...

ϭ

3
0
0

΂1΃
7

ϭ

0
Јϭ 1
1

2
4
4

1
ϭ
6 ,
Ϫ13

΂Ϫ5΃;
2

Prove that the general solution of

1
ϭ Ϫ2 ,
4

3
ϭ Ϫ6 ϩ
12

Ϫ

ϩ

Ϫ6

΂ ΃ ΂΃ ΂ ΃
΂ ΃ ΂΃
΂ ΃ ΂΃
΂΃
1
2 ,
2

΃

1
Ϫ1

1
Ј ϭ Ϫ4
Ϫ6

In Problems 17–20 the given vectors are solutions of a
system Ј ϭ

...

1
ϭ
1

΂2 1΃
3 4

΂

΃

΂2
1

Јϭ

1
6
Ϫ13

2
΂Ϫ1΃ ϩ ΂5΃
1

1

ϩ

΂1΃
0

2

ϩ

ϩ

2

2

ϩ

4
΂Ϫ6΃ ϩ ΂Ϫ1΃
5

1
΂Ϫ1 ϩ 12΃

Ϫ 12

΂Ϫ2΃ ϩ ΂1΃
...
3 of Appendix II
Also the
We saw in Example 5 of Section 8
...


have the form

ϭ

΂΃
1
2

,

ϭ 1, 2,
(

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...
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...
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...


)

●

where 1, 2, l1, and l2 are constants, we are prompted to ask whether we can always find a solution
of the form

()
1

2

ϭ


...


...

If (1) is to be a solution vector of the homogel
neous linear system (2), then Ј ϭ l l , so the system becomes l l ϭ

...
Since
ϭ , the last equation is the same as
(

)

Ϫ

ϭ
...
ϩ

...


...


...


...

ϭ 0
...
But for (3) to have solutions other than the obvious solution
ϭ 0, we must have
1 ϭ 2 ϭ и и и ϭ
det(

) ϭ 0
...
A solution
of (3) corresponding to
an eigenvalue l is called an
of
...

In the discussion that follows we examine three cases: real and distinct eigenvalues (that is, no eigenvalues are equal), repeated eigenvalues, and, finall , complex
eigenvalues
...
, l , then a
set of linearly independent eigenvectors 1, 2 ,
...
,

ϭ

is a fundamental set of solutions of (2) on the interval (Ϫϱ, ϱ)
...
, l be distinct real eigenvalues of the coefficient matrix of the
homogeneous system (2) and let 1, 2,
...
Then the
of (2) on the interval (Ϫϱ, ϱ) is given by
ϭ

1

1

1

ϩ

2

2

2

ϩиииϩ


...
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...


We first find the eigenvalues and eigenvectors of the matrix of
coefficients
From the characteristic equation
det(

Ϫ

)ϭ

͉2 Ϫ
2

͉ϭ

3
1Ϫ

2

Ϫ 3 Ϫ 4 ϭ ( ϩ 1)( Ϫ 4) ϭ 0

we see that the eigenvalues are l1 ϭ Ϫ1 and l 2 ϭ 4
...
When

2

ϩ3

2

ϭ0

2

x

1
1

ϩ2

2

ϭ 0
...

2

Since the matrix of coefficients is a 2 ϫ 2 matrix and since we have found two linearly independent solutions of (4),

2

t

_2

1

ϭ

_4
_6
_3 _2

_1

1

ϭϪ

Ϫ

2

1
΂Ϫ1΃

and

Ϫ

2

ϭ

΂3΃
2

4

,

we conclude that the general solution of the system is

3

ϩ

ϭ

1

1

ϩ

2

2

ϭ

1

1
΂Ϫ1΃

Ϫ

ϩ

2

΂3΃
2

4


...


2

2

y

6

4
2

ϭ

1

2
...
5 1 0 1 2
...
9, that is, listing the individual functions and the relationship between the constants
...


As was pointed out in Section 8
...
2
...
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...
A collection of representative trajectories in the phase
plane, as shown in Figure 8
...
2, is said to be a
of the given linear
system
...
2
...
For example, the Cartesian
equations ϭ 2 , Ͼ 0, and ϭ Ϫ , Ͼ 0, of the half-lines in the first and fourth
3
quadrants were obtained by eliminating the parameter in the solutions ϭ 3 4 ,
ϭ 2 4 , and ϭ Ϫ , ϭ Ϫ Ϫ , respectively
...
The
3
1
eigenvector 2 ϭ
lies along ϭ 2 in the first quadrant, and 1 ϭ
lies
3
2
Ϫ1
along ϭ Ϫ in the fourth quadrant
...

The origin is not only a constant solution ϭ 0, ϭ 0 of every 2 ϫ 2 homogeneous linear system Ј ϭ
, but also an important point in the qualitative study of
such systems
...
2
...
Observe that the arrowheads, with
the exception of only those on the half-lines in the second and fourth quadrants,
indicate that a particle moves away from the origin as time increases
...

We note in passing that Figure 8
...
2 represents a phase portrait that is typical of
2 ϫ 2 homogeneous linear systems Ј ϭ
with real eigenvalues of opposite
signs
...
2
...
Consequently, we call the origin a
in the case l1 Ͼ 0, l 2 Ͼ 0 and an
in the case l1 Ͻ 0, l 2 Ͻ 0
...
2
...
2
...
Investigation of the remaining case
when l ϭ 0 is an eigenvalue of a 2 ϫ 2 homogeneous linear system is left as an
exercise
...
2
...


ϭ

Using the cofactors of the third row, we fin

det(

Ϫ

)ϭ

Ϫ4 Ϫ
p 1
0

1
5Ϫ
1

1
Ϫ1
Ϫ3 Ϫ

p

ϭ Ϫ( ϩ 3)( ϩ 4)( Ϫ 5) ϭ 0,

and so the eigenvalues are l1 ϭ Ϫ3, l2 ϭ Ϫ4, and l3 ϭ 5
...
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...
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...
The choice
sponding solution vector

΂΃

(

Ϳ)

(

2

ϭ Ϫ 3
...


1 0 Ϫ10 0
0 1
1 0
0 0
0 0

Finally, when l3 ϭ 5, the augmented matrices
( ϩ5 Ϳ )ϭ

Ϳ)

1 0 Ϫ1 0
0 1
0 0
...


Ϳ)

1 0 Ϫ1 0
0 1 Ϫ8 0
0 0
0 0


...


Software packages such as MATLAB,
and DERIVE can be real time savers in finding eigenvalues and eigenvectors
of a matrix
...
, l of an ϫ matrix need be
distinct; that is, some of the eigenvalues may be repeated
...
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...
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...
For this value we find the single eigenvecto

1

ϭ

΂3΃,
1

so

1

ϭ

΂3΃
1

(11)

Ϫ3

is one solution of (10)
...

In general, if is a positive integer and (l Ϫ l1) is a factor of the characteristic
equation while (l Ϫ l1) ϩ1 is not a factor, then l1 is said to be an
The next three examples illustrate the following cases:
()

For some ϫ matrices it may be possible to find linearly independent eigenvectors 1, 2 ,
...
In this case the general solution of the system
1 of multiplicity
contains the linear combination
1

( )

1

ϩ

1

2

2

1

ϩиииϩ

1


...


...

ϭ

11
21

1

1

ϩ

1

22

Ϫ1

1 ––––––––
( Ϫ 1)!

Ϫ2

1

ϩ

2 ––––––––
( Ϫ 2)!

1

ϩ
...


We begin by considering eigenvalues of
multiplicity two
...


΂

1
Solve Ј ϭ Ϫ2
2

΃

2
Ϫ2
1

Ϫ2
1
Ϫ2


...
We see that l1 ϭ l 2 ϭ Ϫ1 and l 3 ϭ 5
...

0
0 0 0

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...
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...


●

The first row of the last matrix means 1 Ϫ 2 ϩ 3 ϭ 0 or 1 ϭ 2 Ϫ 3
...
Thus two
eigenvectors corresponding to l1 ϭ Ϫ1 are

΂΃

1
ϭ 1
0

1

΂΃

0
ϭ 1
...
Lastly, for l3 ϭ 5 the reduction

(

Ϳ)

Ϫ4 Ϫ2
2 0
( ϩ 5 Ϳ ) ϭ Ϫ2 Ϫ4 Ϫ2 0
2 Ϫ2 Ϫ4 0
implies that 1 ϭ 3 and
third eigenvector is

2

ϭ Ϫ 3
...

1

We conclude that the general solution of the system is

ϭ

1

΂΃
1
1
0

Ϫ

ϩ

2

΂΃
0
1
1

Ϫ

ϩ

3

΂΃
1
Ϫ1
1

5


...
An ϫ matrix is said to be
if its transpose
(where the rows and columns are interchanged) is the same as — that is, if
ϭ
...
,
, and the general solution of such a system is as given in
Theorem 8
...
1
...

Now suppose that l1 is an eigenvalue of multiplicity two
and that there is only one eigenvector associated with this value
...


...


1

and

2

ϭ
...


...
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...


1

Since this last equation is to hold for all values of , we must have
(
(

and

Ϫ
Ϫ

1

)

1

(13)

ϭ

) ϭ


...

1
By solving (13), we find one solution 1 ϭ

...


Find the general solution of the system given in (10)
...
Identifying ϭ
1
2

΂΃

(

΂ ΃

ϩ3 ) ϭ

6
2

or

1
1

Ϫ 18
Ϫ 6

2
2

ϭ3
ϭ 1
...
For example, by choosing 1 ϭ 1, we find 2 ϭ 1
...
Hence

ϭ

΂0΃
...
The general solution of (10) is

Ϫ3

ϩ

΂0΃ ΅
...
A phase portrait of (10) is given in
Figure 8
...
3
...
Because the single
3
eigenvalue is negative and Ϫ3 : 0 as : ϱ on
trajectory, we have
( ( ), ( )) : (0, 0) as : ϱ
...
2
...
Moreover, a moving particle or trajectory
ϭ 3 1 Ϫ3 ϩ 2(3 Ϫ3 ϩ 1 Ϫ3 ), ϭ 1 Ϫ3 ϩ 2 Ϫ3 , 2 0, approaches (0, 0)
2
tangentially to one of the half-lines as : ϱ
...
See Problem 21
in Exercises 8
...
Analogous to Figure 8
...
2, Figure 8
...
3 is typical of
2ϫ2
homogeneous linear systems Ј ϭ
that have two repeated negative eigenvalues
...
2
...
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...
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...
,

...



...


...


...


and

By substituting (15) into the system Ј ϭ
and must satisfy

, we find that the column vectors

, ,

(

1

)

ϭ

(16)

(

Ϫ

1

) ϭ

(17)

(

and

Ϫ
Ϫ

1

)

(18)

ϭ
...


΂

2
Solve Ј ϭ 0
0

1
2
0

΃

6
5
2

1


...
By solving ( Ϫ 2 ) ϭ , we find the single
eigenvector

΂΃

1
ϭ 0
...

5

and

1
5

Using (12) and (15), we see that the general solution of the system is
ϭ

1

΂ ΃ ΄΂ ΃ ΂ ΃ ΅ ΄΂ ΃ ΂ ΃ ΂ ΃ ΅
1
0
0

2

ϩ

2

1
0
0

2

0
ϩ 1
0

2

ϩ

3

1 2
0
2
0

2

0
ϩ 1
0

2

0
ϩ Ϫ6
5

2


...
Hence the two cases listed on page 338 are not all the possibilities under
which a repeated eigenvalue can occur
...
See Problems 31 and 50 in Exercises 8
...


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...
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...


●

If l1 ϭ a ϩ b and l 2 ϭ a Ϫ b , b Ͼ 0, 2 ϭ Ϫ1 are complex eigenvalues of the
coefficient matrix , we can then certainly expect their corresponding eigenvectors
to also have complex entries
...


From the quadratic formula we find l1 ϭ 5 ϩ 2 , l 2 ϭ 5 Ϫ 2
...


Since 2 ϭ (1 Ϫ 2 ) 1,† the choice
corresponding solution vector:
1

ϭ

1

ϭ 1 gives the following eigenvector and

1
΂1 Ϫ 2 ΃,

1

ϭ

1
΂1 Ϫ 2 ΃

(5ϩ2 )


...


In like manner, for l 2 ϭ 5 Ϫ 2 we fin
2

ϭ

1
΂1 ϩ 2 ΃,

We can verify by means of the Wronskian that these solution vectors are linearly
independent, and so the general solution of (19) is
ϭ

1

1
΂1 Ϫ 2 ΃

(5ϩ2 )

ϩ

2

1
΂1 ϩ 2 ΃

(5Ϫ2 )


...
The conjugate of l1 is, of course, l 2
...
We have illustrated the following general
result
...
Then
1

1

and

1

1

are solutions of (2)
...

†
Note that the second equation is simply (1 ϩ 2 ) times the first

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...
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...


●

It is desirable and relatively easy to rewrite a solution such as (20) in terms of
real functions
...


Then, after we multiply complex numbers, collect terms, and replace
and ( 1 Ϫ 2 ) by 2 , (20) becomes
ϭ

1

1

ϩ

ϩ

2

by

2,

2

1

(21)

where

1

ϭ

0
΄΂1΃ cos 2 Ϫ ΂Ϫ2΃ sin 2 ΅
1

5

and

2

ϭ

0
΄΂Ϫ2΃ cos 2 ϩ ΂1΃ sin 2 ΅
1

5


...
Consequently, we are justified in ignoring the relationship between 1, 2 and 1, 2, and we can regard 1 and
2 as completely arbitrary and real
...
Moreover, with the real form given in (21) we
are able to obtain a phase portrait of the system in (19)
...


By plotting the trajectories ( ( ), ( )) for various values of 1 and 2, we obtain the
phase portrait of (19) shown in Figure 8
...
4
...
This is why the arrowheads in Figure 8
...
4 point away from the
origin; a particle on any trajectory spirals away from the origin as : ϱ
...

The process by which we obtained the real solutions in (21) can be generalized
...
Then the solution vectors in
Theorem 8
...
2 can be written as
1

1

ϭ

1

1

1

ϭ

1

␣

␤

ϭ

␣ Ϫ␤

1

ϭ

␣

(cos ␤ ϩ sin ␤ )
␣

1

(cos ␤ Ϫ sin ␤ )
...
1
...
Therefore, the entries in the column vectors 1(
2
1
1) are real numbers
...

complex
1) and

1),

we are led to the following theorem
...
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...
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...
Then
1

ϭ[

1 cos

␤ Ϫ

2 sin

␤]

␣

2

ϭ[

2 cos

␤ ϩ

1 sin

␤]

␣

(23)

are linearly independent solutions of (2) on (Ϫϱ, ϱ)
...
For example, (21) follows from (23) with
1

1

ϭ Re(

1)

ϭ

ϭ

(24)

parts of the eigenvector

1
0
΂1 Ϫ 2 ΃ ϭ ΂1΃ ϩ ΂Ϫ2΃,
1

΂1΃
1

and

ϭ Im(

2

1)

0
΂Ϫ2΃
...


(25)

First we obtain the eigenvalues from
det(

)ϭ

Ϫ

Ϫ
͉2 Ϫ1

The eigenvalues are l1 ϭ 2 and

2

(2 Ϫ 2 )
Ϫ
gives

1

ϭ

ϭ

1

͉ϭ

2

ϩ 4 ϭ 0
...
For l1 the system

1

ϩ

8

2

ϭ0

1

ϩ (Ϫ2 Ϫ 2 )

2

ϭ0

ϭ Ϫ(2 ϩ 2 ) 2
...

0

Now from (24) we form
1

ϭ Re(

1)

ϭ

2
΂Ϫ1΃

and

2

ϭ Im(

1)

ϭ

΂2΃
...

2
2

(26)

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...
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...


●

y

x
(2, _1)

A phase portrait of (25)
in Example 6

Some graphs of the curves or trajectories defined by solution (26) of the system
are illustrated in the phase portrait in Figure 8
...
5
...
Thus the solution

΂ ΃

to the problem is

ϭ

2 Ϫ
΂2 cos Ϫcos 2 sin 2 ΃
...
2
...
Note that this curve passes through (2, Ϫ1)
...
But often the mathematical
model of a dynamical physical system is a homogeneous second-order system
whose normal form is Љ ϭ

...
6,
1 Љ
1

ϭϪ

2 Љ
2

ϭ Ϫ 2(

can be written as
where

΂0

2

2

΃,

ϭ

΂Ϫ

1

Ϫ

1)

Ϫ

(27)

1),

Ϫ

,

Љϭ

0

1

ϭ

2( 2

ϩ

1 1

2

2

Ϫ

2

2

΃,

and

Љϭ

΂

1

2

2

1

Ϫ

1

1

2

2

Ϫ

2

΂ (( ))΃
...


Ϫ1


...
If we let Ј ϭ 3 and Ј ϭ 4, then Ј ϭ 1 and
1
2
3
Ј ϭ Љ and so (27) is equivalent to a system of
linear first-order DEs:
4
2
Јϭ
1

3

Јϭ
2

4

΂

ЈϭϪ
3
Јϭ
4

2
2

1

1

Ϫ

΃

2

ϩ

1

1
2
2

2

1

ϩ

2
1

2

or

΂

Јϭ Ϫ

0
0
1

2
2

0
1

2

0

0

1

1

1
0

2

Ϫ

0
0
1

0

0

Ϫ

2
2

΃


...
See Problem 48(a) in Exercises 8
...


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...
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...


●

• Second, because (27) describes free undamped motion, it can be argued
that real-valued solutions of the second-order system (28) will have
the form
cos ␻

ϭ

and

sin ␻ ,

ϭ

(30)

where
is a column matrix of constants
...
(Verify
...
It can
be shown that the eigenvalues ϭ Ϫ␻ 2, ϭ 1, 2 of are negative, and
so ␻ ϭ 1Ϫ is a real number and represents a (circular) frequency
of vibration (see (4) of Section 7
...
By superposition of solutions the
general solution of (28) is then
ϭ

1

1 cos

ϭ ( 1 cos ␻1 ϩ

2

1 sin

2 sin

␻1 ϩ

␻1 )

␻1 ϩ
1

2 cos

3

␻2 ϩ

ϩ ( 3 cos ␻ 2 ϩ

4

4 sin

2 sin

␻2

␻2 )

2,

(31)

where 1 and 2 are, in turn, real eigenvectors of corresponding to l1
and l2
...
If Ϫ␻ 1 , Ϫ␻ 2 ,
...
,
are corresponding real
eigenvectors of the ϫ coefficient matrix , then the homogeneous
second-order system Љ ϭ
has the general solution
ϭ
where
and
Exercises 8
...


In Problems 1–12 find the general solution of the given
system
...
See Problem 48(b) in

Јϭ

΂
΂
΂
΂

΃

1
2
3

Ϫ1
1
0

1
Јϭ 0
1

ϩ3

ϭϪ

5
ϩ2
2

͚(
ϭ1

0
1
Ϫ1

΃

0
1
0

1
0
1

΃
΃

Ϫ1 Ϫ1
3
Ϫ3
4
2
1
8

1
4

0
3
Ϫ1
2

Ϫ1
4
Јϭ
0

4
Ϫ1
0

2
Ϫ2
6

Јϭ

In Problems 13 and 14 solve the given initial-value problem
...
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...

In Problems 15 and 16 use a CAS or linear algebra software
as an aid in finding the general solution of the given system
...
9
Ј ϭ 0
...
1

Јϭ

΂

΃

2
...
5
1
...
8

3
...
2
3
...
1
2
1
0

2
0
Ϫ3
Ϫ3
...
8
Ϫ1
0
4
1
...
If possible, include
arrowheads as in Figure 8
...
2
...

Obtain the Cartesian equations of each of the four
half-lines in part (a)
...


has an eigenvalue l1 of multiplicity 5
...


Find phase portraits for the systems in Problems 2 and 4
...


Find phase portraits for the systems in Problems 20
and 21
...


In Problems 19 – 28 find the general solution of the given
system
...


ϭ3 Ϫ

ϭ Ϫ6 ϩ 5

ϭ6 Ϫ

ϭ

ϭ9 Ϫ3

ϭ Ϫ5 ϩ 4

ϭ5 ϩ2

ϭ Ϫ2 Ϫ

ϭ5 ϩ

ϭ4 ϩ5

ϭ Ϫ2 ϩ 3

ϭ Ϫ2 ϩ 6

Јϭ

΂Ϫ1 3΃
Ϫ3 5

ϭ3 Ϫ

Јϭ

΂12
4

΃

Ϫ9
0

ϭ3 ϩ2 ϩ4

Ϫ

Јϭ

ϭ

ϩ

Ϫ

ϭ2 ϩ2

ϭ

Ϫ

ϩ

ϭ4 ϩ2 ϩ3

΂

Ϫ4
0
2

΂

0
2
1

5
Јϭ 1
0
1
Јϭ 2
0

΃

1
Јϭ 0
0

΃

4
Јϭ 0
0

0
2
5
0
Ϫ1
0

΂

0
3
Ϫ1

΂

1
4
0

΃

0
1
1

΃

0
1
4

΂4
5

΃

Ϫ5
Ϫ4

Јϭ

ϩ

΂1
1

΃

Ϫ8
Ϫ3

ϭ

ϭ2 ϩ

ϭϪ

ϭ3 ϩ6

ϭ

ϭ Ϫ4 Ϫ 3

΂

΃

1 Ϫ1 2
Ј ϭ Ϫ1
1 0
Ϫ1
0 1

Јϭ

΂

4
0
Ϫ4

ϩ2

0
6
0

1
0
4

΃

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...
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...
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...


●

΂

΃

΂

2
5 1
Ј ϭ Ϫ5 Ϫ6 4
0
0 2

΃

2
4
4
Ј ϭ Ϫ1 Ϫ2
0
Ϫ1
0 Ϫ2

and ϭ cos v Find the eigenvalues and eigenvectors of a 2 ϫ 2 matrix
...
6
...


΂

΃

1 Ϫ12
Јϭ 1
2
1
1

΂

Ϫ14
Ϫ3
Ϫ2

΃

6 Ϫ1
Јϭ
5
4

,

,

΂΃

4
(0) ϭ
6
Ϫ7

΂ ΃

Ϫ2
(0) ϭ
8

Find phase portraits for the systems in Problems 36, 37,
and 38
...
6 using the first method
outlined in the
(page 345)—that is, express
(2) of Section 7
...
Use a CAS or linear algebra software
as an aid in finding eigenvalues and eigenvectors of
a 4 ϫ 4 matrix
...
6
...
6 using the second method outlined in the
—that is, express (2) of Section 7
...
Assume solutions of the form ϭ sin v

●
●

Solve each of the following linear systems
...
What is the geometric significance of the line ϭ Ϫ in each portrait?
Consider the 5 ϫ 5 matrix given in Problem 31
...
Use
the general solution as a basis for a discussion of how the
system can be solved using the matrix methods of this
section
...

Obtain a Cartesian equation of the curve define
parametrically by the solution of the linear system in
Example 6
...
2
...
[
: Compute 2, 2, and ]
Examine your phase portraits in Problem 47
...
4 (Undetermined Coefficients
Section 4
...
1 we saw that the general solution of a nonhomogeneous linear
system Ј ϭ
ϩ ( ) on an interval is ϭ
ϩ , where
ϭ 1 1ϩ 2 2ϩиииϩ
is the
or general solution of the associated homogeneous linear system
Јϭ
and
is any
of the nonhomogeneous system
...
2 we saw
how to obtain
when the coefficient matrix was an ϫ matrix of constants
...

The methods of
and
used in Chapter 4 to
find particular solutions of nonhomogeneous linear ODEs can both be adapted to the solution of
nonhomogeneous linear systems Ј ϭ
ϩ ( )
...
However, there are instances when the method of undetermined
coefficients provides a quick means of finding a particular solutio

As in Section 4
...
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...
Not surprisingly, the matrix version of undetermined coefficient
is applicable to Ј ϭ
ϩ ( ) only when the entries of are constants and the
entries of ( ) are constants, polynomials, exponential functions, sines and cosines,
or finite sums and products of these functions

Solve the system Ј ϭ

΂Ϫ1 2΃
Ϫ1 1

ϩ

΂Ϫ8΃ on (Ϫϱ, ϱ)
...


The characteristic equation of the coefficient matrix ,
det(

)ϭ

Ϫ

͉

yields the complex eigenvalues l1 ϭ and
Section 8
...
By the procedures of

ϩ
Ϫ
΂cos cos sin ΃ ϩ ΂cosϪsin sin ΃
...
Substituting this latter assumption into the original system and equat1
1

ing entries leads to
0ϭϪ

1

ϩ2

1

Ϫ8

0ϭϪ

1

ϩ

1

ϩ 3
...
The general solution of the original system of DEs on the interval
ϭ
11
(Ϫϱ, ϱ) is then ϭ
ϩ
or

΂ ΃

ϭ

1

Solve the system Ј ϭ

ϩ
Ϫ
΂cos cos sin ΃ ϩ ΂cosϪsin sin ΃ ϩ ΂14΃
...


The eigenvalues and corresponding eigenvectors of the associated
6 1
1
,
homogeneous system Ј ϭ
are found to be l1 ϭ 2, l 2 ϭ 7, 1 ϭ
4 3
Ϫ4
1
and 2 ϭ

...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Now because ( ) can be written ( ) ϭ

6
΂Ϫ10΃ ϩ ΂0΃, we shall try to find a
4

particular solution of the system that possesses the

΂ ΃ ϩ ΂ ΃
...


2
2

ϩ4

From the last identity we obtain four algebraic equations in four unknowns
6
4

2
2

ϩ
ϩ3

ϩ 6ϭ0
2 Ϫ 10 ϭ 0

2

6
4

and

ϩ
1 ϩ 3
1

1
1

Ϫ
Ϫ

2
2

ϭ0
ϩ 4 ϭ 0
...
We then
substitute these values into the last two equations and solve for 1 and 1
...
It follows, therefore, that a particular solution vector is
7
7

ϭ

΂ ΃
Ϫ2
6

΂΃
Ϫ4
7

ϩ

10
7

The general solution of the system on (Ϫϱ, ϱ) is

ϭ

1

΂ ΃
1
Ϫ4

2

ϩ

2

΂΃
1
1


...


for the system

ϩ1

Ϫ 5 ϩ 7
...

2

1

2

1

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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

The method of undetermined coefficients for linear systems is not as straightforward as the last three examples would seem to indicate
...
4 the form
of a particular solution was predicated on prior knowledge of the complementary function
...
But there are further difficulties: The special rules governing the form of in Section 4
...
For example, if ( ) is a constant vector, as
in Example 1, and l ϭ 0 is an eigenvalue of multiplicity one, then
contains
a constant vector
...
This is not the
ϭ ΂ ΃ ϩ ΂ ΃
...


ϩ

2

1

2

2

3

1

Rather than delving into these difficulties, we turn instead to the method of
variation of parameters
...
,
on an interval , then its general solution on the inϭ 1 1ϩ 2 2ϩиииϩ
or

1


...


...


...


1
2

ϩ
...


...


ϩ
21 ϩ

2 12

1

1

2

2

ϭ

2

ϩ
...
ϩ
22 ϩ

...


...
ϩ
2

1 11

1

1

ϩ

2

)


...
In other words, the general solution (1) can be written as the product
ϭ ⌽( ) ,

(2)

where is an ϫ 1 column vector of arbitrary constants 1, 2 ,
...


...



...


1
2

)


...


...


of the system on the interval
...
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...
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...
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...

• If ⌽( ) is a fundamental matrix of the system Ј ϭ
⌽Ј( ) ϭ

, then

⌽( )
...
1
...
,
)
...
Since
⌽( ) is nonsingular, the multiplicative inverse ⌽Ϫ1( ) exists for every in the interval
...

Analogous to the procedure in Section 4
...


...


so

ϭ ⌽( ) ( )

(4)

()

is a particular solution of the nonhomogeneous system
ϩ ( )
...


(6)

Note that the order of the products in (6) is very important
...
Substituting (4) and (6)
into (5) gives
⌽( ) Ј( ) ϩ ⌽Ј( ) ( ) ϭ

⌽( ) ( ) ϩ ( )
...


or

(8)

Multiplying both sides of equation (8) by ⌽Ϫ1( ) gives
Ј( ) ϭ ⌽Ϫ1( ) ( )
Since

and so

()ϭ

͵

⌽Ϫ1( ) ( )


...


(9)

To calculate the indefinite integral of the column matrix ⌽Ϫ1( ) ( ) in (9), we integrate each entry
...


(10)

Note that it is not necessary to use a constant of integration in the evaluation of
͵⌽Ϫ1( ) ( ) for the same reasons stated in the discussion of variation of parameters in Section 4
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Solve the system
Јϭ

΂Ϫ3
2

΃

1
Ϫ4

ϩ

΂3 ΃

(11)

Ϫ

on (Ϫϱ, ϱ)
...


(12)

The characteristic equation of the coefficient matrix i
det(

)ϭ

Ϫ

͉

͉

1
Ϫ4 Ϫ

Ϫ3 Ϫ
2

ϭ ( ϩ 2)( ϩ 5) ϭ 0,

so the eigenvalues are l1 ϭ Ϫ2 and l 2 ϭ Ϫ5
...
The solution vectors of the homogeneous system (12) are then
2 ϭ
Ϫ2

΂΃

΂ ΃
1

ϭ

΂΃
1
1

΂ ΃
Ϫ2

Ϫ2

ϭ

and

Ϫ2

2

ϭ

΂ ΃
1
Ϫ2

΂Ϫ2 ΃
...
Hence
⌽( ) ϭ

΂

Ϫ2
Ϫ2

Ϫ5
Ϫ5

Ϫ2

΃

and

΂

form the second

2

1 2
3

2 2
3

⌽ ()ϭ
Ϫ1

1 5
3

Ϫ1
3

5

΃


...


Hence from (10) the general solution of (11) on the interval is
ϭ

ϭ

΂
1

Ϫ2
Ϫ2

΂΃
1
1

Ϫ5

Ϫ2
Ϫ2

Ϫ5

ϩ

΃΂ ΃
1

ϩ

2

2

΂ ΃
1
Ϫ2

΂

Ϫ5

6
5
3
5

Ϫ 27 ϩ 1
50
4

Ϫ

Ϫ 21 ϩ 1
50
2

Ϫ

ϩ

΃

΂΃ ΂ ΃ ΂΃
6
5
3
5

Ϫ

27
50
21
50

ϩ

1
4
1
2

Ϫ


...
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...
This last form is useful in solving (5) subject
to an initial condition ( 0) ϭ 0 , because the limits of integration are chosen so that
the particular solution vanishes at ϭ 0
...
Substituting this last result into
0 ϭ ⌽( 0)
(13) gives the following solution of the initial-value problem:
ϭ ⌽( )⌽Ϫ1( 0)

0

ϩ ⌽( )

͵


...


The system of differential equations for the currents
2( ) and 3( ) in the electrical network shown in
Figure 8
...
1 is

΂ ΃ ϭ ΂Ϫ
Ϫ
2

ϭ2 ϩ3 Ϫ7

3

1> 1

Ϫ 1> 1
Ϫ( 1 ϩ 2)>

1> 2

ϭ5 ϩ9 ϩ2
ϭ Ϫ ϩ 11 ϩ 6

1
1

΂ ΃

Јϭ

΂1
4

Јϭ

΂ ΃

Јϭ

΂

΂4Ϫ ϩϩ9 ΃

ϩ

΃

4 1
3
9 6

ϩ

΃

5
1

΂
΂

2

2

ϩ

΂

΃

sin
Ϫ2 cos

΃ ΂΃
΃ ΂ ΃

0 0
Јϭ 0 5
5 0

5
0
0

5
ϩ Ϫ10
40

Ϫ1
Solve Ј ϭ
3

Network in Problem 10

΂Ϫ3΃
10
1
ϩ Ϫ1
2

΂Ϫ4΃
...

2 ϭ 1 h,
Determine the current 1( )
...


΃

Ϫ2
4

In Problems 11 – 30 use variation of parameters to solve the
given system
...
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...
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...
3
...


Јϭ

΂1
1

8
Ϫ1

΃

ϩ

΂ ΃

Јϭ

3
΂Ϫ2

2
Ϫ1

Јϭ

3
΂Ϫ2

2
Ϫ1

Јϭ

΂0
1

Ϫ1
0

Јϭ

΂1
1

Ϫ1
1

Јϭ

΂1
1

Ϫ1
1

Јϭ

΂

Ϫ2
Ϫ6

Ϫ1
2

2

1

1

2

Ϫ3

ϩ
2>

Ϫ

1

΃

ϩ
ϩ

΂1΃
1

1

2>
Ϫ 2>

1

2
1

Ϫ

΃΂ ΃ ϩ ΂ >0 ΃
...
6
...

Consider the system

΂

2
Ϫ1
Јϭ
0
0

tan
1

΃ ΂΃
΃ ΂΃

0
0
3

ϩ

2

3

0

Ϫ1
Ϫ1
1

ϩ

2

In Problems 31 and 32 use (14) to solve the given initialvalue problem
...


1

Use a CAS or linear algebra software to find the
eigenvalues and eigenvectors of the coefficien
matrix
...

Use the computer to carry out the computations of:
⌽Ϫ1( ) ( ), ͵⌽Ϫ1( ) ( ) , ⌽( )͵⌽Ϫ1( ) ( ) ,
⌽( ) , and ⌽( ) ϩ ͵⌽Ϫ1( ) ( ) , where
is a
column matrix of constants 1, 2, 3, and 4
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

●

Appendix II
...
10 and II
...
Recall that the simple linear first-order differential equation
Ј ϭ , where is constant, has the general solution ϭ
, where is a constant
...


We shall now see that it is possible to define a matrix exponential

so that
(1)

ϭ

is a solution of the homogeneous system Ј ϭ

...
Note in (1) that the
matrix post multiplies
because we want
to be an ϫ matrix
...

!

The series in (2) converges for all Using this series, with 1 replaced by the identity
matrix and the constant replaced by an ϫ matrix of constants, we arrive at
a definition for the ϫ matrix
...

!

(3)

It can be shown that the series given in (3) converges to an
, 3 ϭ ( 2), and so on
...

0 3

From the various powers
2

΂20 30 ΃,
2

ϭ

2

3

΂20 30 ΃,
3

ϭ

3

4

΂20 30 ΃,
...
,

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...
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...


●

we see from (3) that
2

ϭ ϩ

ϩ

2

2!

ϩ
...
ϩ ΂20 30 ΃
0 1
0 3

ϭ

΂

2

2

2

1 ϩ 2 ϩ 22

2

2!

ϩ
...


΃

0
2

!

ϩ
...


In view of (2) and the identifications ϭ 2 and ϭ 3, the power series in the firs
and second rows of the last matrix represent, respectively, 2 and 3 and so we have

΂0

2

ϭ

΃
...
In general,
an ϫ matrix is a
if all its entries off the main diagonal are
zero, that is,

ϭ

Hence if

is any

΂

0


...


22

11

΃


...

o

diagonal matrix it follows from Example 1 that

ϫ

ϭ

΂

0

΃


...


22

11

0
0

...


The derivative of the matrix exponential is analogous to the
differentiation property of the scalar exponential

ϭ


...


Because of (4), we can now prove that (1) is a solution of Ј ϭ
vector of constants:
Јϭ

ϩ

ϭ

(

)ϭ

for every

ϫ1


...
3)
...
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...
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...
4
...
It turns out that these
two properties are sufficient for us to conclude that ⌿( ) is a fundamental matrix of the
system Ј ϭ

...
3 that the general
solution of the single linear first-order differential equation Ј ϭ ϩ ( ), where
is a constant, can be expressed as
ϭ

ϩ

ϭ

͵

ϩ


...


(5)

0

Since the matrix exponential
is a fundamental matrix, it is always nonsingular and
Ϫ
ϭ ( )Ϫ1
...
However, the practical utility of (3) is limited by the fact that
the entries in
are power series in With a natural desire to work with simple and
familiar things, we then try to recognize whether these series define a closed-form
function
...


Јϭ

0


...


,

Јϭ

(6)

}, then the Laplace transform of (6) is
(0) ϭ

()

or

(

Ϫ

) ( )ϭ
...
In other words, ᏸ{ } ϭ ( Ϫ )Ϫ1 or
ϭ ᏸϪ1{(

Ϫ

Use the Laplace transform to compute

for

First we compute the matrix
ϭ

Ϫ

(

Ϫ

Ϫ1

Ϫ
΂ Ϫ21

΂

1
ϩ2

΃

Ϫ1

...


ϭ

Ϫ

( )ϭ(

Ϫ1
Ϫ2

΃
΃

Ϫ1

ϭ

΂

ϩ2
( ϩ 1)
2
( ϩ 1)

΃

Ϫ1
( ϩ 1)

...
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...
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...

1
2
Ϫ ϩ
ϩ1

(8)

It follows from (7) that the inverse Laplace transform of (8) gives the desired result,
ϭ

΂

΃

2Ϫ Ϫ
2Ϫ2 Ϫ

Ϫ1 ϩ Ϫ

...

See Problems 27 and 28 in Exercises 8
...


In Problems 1 and 2 use (3) to compute
ϭ

΂1 0΃
0 2

ϭ

΂
΂

0
ϭ 3
5

Ϫ

Solve the system in Problem 7 subject to the initial
condition


...

6


...
Use (1) to find the general
solution of the given system
...

Јϭ

΂1 0΃
0 2

΂

1
Јϭ
1
Ϫ2

Јϭ

΃

1
1
Ϫ2

΂

1
1
Ϫ2

0
Јϭ 3
5

0
0
1

΃

0
0
0

4
΂Ϫ4

Јϭ

΂0 1΃
1 0

Јϭ

΂5
1

3
Ϫ4

΃

Јϭ

΂4
1

΃

Јϭ

0
΂Ϫ2

Ϫ9
Ϫ1

΂ ΃
0
2

΂ ΃

1
ϩ
1

΂ ΃

cosh
ϩ
sinh

ϭ

0
Јϭ
1

0
Јϭ
1

1
0

1
0

΂ ΃

0

0

...


...


...


...


0

΂ ΃

ϩ

0
2

΃

1
Ϫ2

( )
1

3
ϩ
Ϫ1

΂ ΃

1
Јϭ
0

΃

Ϫ2
1

Let
denote a matrix whose columns are eigenvectors
, 2,
...
, l of an ϫ matrix
...

1
Јϭ
0

΂4΃
...


...


(9)


...


΂΃
΂

΃

ϭ

2
΂Ϫ3 1΃
6

ϭ

΂2 1΃
1 2

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...
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...
Editorial review has
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...


●

Ϫ1

Suppose ϭ
(3) to show that
If

, where is defined as in (9)
...


is defined as in (9), then fin

In Problems 23 and 24 use the results of Problems 19 – 22 to
solve the given system
...
Use a CAS to find
...
2
...

Use (1) to find the general solution of

΂ ΃

΂ ΃

Reread the discussion leading to the result given in (7)
...

A matrix
is said to be
some positive integer such that

΂

Ϫ1
ϭ Ϫ1
Ϫ1

if there exists
ϭ
...


΂

4
Јϭ
1
Ϫ1

Consider the linear system

΂΃

6
3
Ϫ4

΃

6
2
Ϫ3


...


Consider the linear system Ј ϭ
of two differential
equations, where is a real coefficient matrix
...


΂ ΃

΂

Ϫ4
0 6
0
0 Ϫ5 0 Ϫ4
Јϭ
Ϫ1
0 1
0
0
3 0
2
Use MATLAB or a CAS to find
...
What is
the solution of the system corresponding to this
eigenvector?

΂4΃ is a solution of
5
1
4
8
Јϭ΂
Ϫ΂ ΃
2 Ϫ1΃
1
Ϫ1
1

΃

Use (1) to find the general solution o

1
1
1

is nilpotent
...
Compute
and then use (1) to
solve the system Ј ϭ

...
Use a CAS to find
...

ϭ2 ϩ

ϭ Ϫ4 ϩ 2

ϭϪ

ϭ2 Ϫ4

Јϭ

1
΂Ϫ2 2΃
1

΂

1
Јϭ 0
4

Ϫ1
1
3

Јϭ

΃

1
3
1

΂Ϫ2 5΃
Ϫ2 4

΂

0
Јϭ 1
2

2
1
2

΃

1
Ϫ2
Ϫ1

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...
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...
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...


●

Јϭ

΂2 8΃
0 4

Јϭ

΂Ϫ1 2΃
1

ϩ

΂

Јϭ

΂Ϫ1 1΃
Ϫ2 1

ϩ

1
΂cot ΃

Јϭ

3
΂Ϫ1 1΃
1

ϩ

΂Ϫ2΃
1

1
2

ϩ

2
΂16 ΃
0
tan

and l ϭ 2 is known to be an eigenvalue of multiplicity two
...
2)
...


2

Consider the linear system Ј ϭ
of three first
order differential equations, where the coefficien
matrix is
5
ϭ
3
Ϫ5

΃

1 1
Јϭ 1 1
1 1

3
5
Ϫ5

΃

3
3
Ϫ3

Verify that

ϭ

΂΃
1

is a solution of the linear system

2

Јϭ
for arbitrary constants
portrait of the system
...


By hand, draw a phase

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...
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...
Editorial review has
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...


Numerical Solutions of
Ordinary Differential Equations

9
...
2
9
...
4
9
...
In many instances we have to
be content with an approximation of the solution
...
In this chapter we continue to explore the basic
idea introduced in Section 2
...

Our concentration in this chapter is primarily on first-order initial-value problem
> ϭ ( , ), ( 0) ϭ 0
...
10 that numerical procedures
developed for first-order DEs extend in a natural way to systems of first-ord
equations
...
Chapter
concludes with a method for approximating solutions of linear second-order
boundary-value problems
...
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...
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...
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...
6

In Chapter 2 we examined one of the simplest numerical methods for
approximating solutions of first-order initial-value problems Ј ϭ ( , ), ( 0) ϭ 0
...
The recursive use of (1)
for ϭ 0, 1, 2,
...
of points on successive “tangent lines” to
the solution curve at 1, 2 , 3 ,
...
The values 1, 2 , 3 ,
...
But whatever advantage (1) has in its simplicity is lost in the crudeness of its
approximations
...
6 you were asked to use Euler’s
method to obtain the approximate value of (1
...
You should have obtained the analytic solution ϭ Ϫ1
and results similar to those given in Tables 9
...
1 and 9
...
2
...
1

Euler’s Method with ϭ 0
...
00
1
...
20
1
...
40
1
...
0000
1
...
4640
1
...
2874
2
...

error

% Rel
...
0000
1
...
5527
1
...
6117
3
...
0000
0
...
0887
0
...
3244
0
...
00
2
...
71
8
...
42
16
...
00
1
...
10
1
...
20
1
...
30
1
...
40
1
...
50

1
...
1000
1
...
3492
1
...
6849
1
...
1419
2
...
7714
3
...

error

% Rel
...
0000
1
...
2337
1
...
5527
1
...
9937
2
...
6117
3
...
4903

0
...
0079
0
...
0314
0
...
0702
0
...
1343
0
...
2403
0
...
00
0
...
47
2
...
11
4
...
93
5
...
92
7
...
08

In this case, with a step size ϭ 0
...
5) is totally unacceptable
...
05
...
For some kinds of computation the accumulation of errors might reduce the
accuracy of an approximation to the point of making the computation useless
...

One source of error that is always present in calculations is
This error results from the fact that any calculator or computer can represent numbers
using only a finite number of digits
...
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...
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...
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...
3333 and 9 is represented as 0
...
If we use this
calculator to compute 2 Ϫ 1 ͞ Ϫ 1 for ϭ 0
...
3334) Ϫ 0
...
1112 Ϫ 0
...

0
...
3333
0
...
3333
With the help of a little algebra, however, we see that
2

Ϫ1
9
ϭ
Ϫ1
3

(

(

Ϫ1
3

)(

ϩ1
3

Ϫ

) (

1
3

)ϭ

1
ϩ ,
3

)

so when ϭ 0
...
3334 ϩ 0
...
6667
...
One way to reduce the effect of round-off error is to minimize the number of
calculations
...
In general, round-off error is unpredictable and difficult to analyze, and we will neglect it in the error analysis that follows
...

In the sequence of values 1, 2,

...
See Figure 2
...
2
...

To derive a formula for the local truncation error for Euler’s method, we use
Taylor’s formula with remainder
...

2!

ϩ1

Euler’s method (1) is the last formula without the last term; hence the local
truncation error in ϩ1 is
2

Љ( )

2!

,

where

Ͻ

Ͻ

ϩ1
...

Ͻ Ͻ

ϩ1

In discussing errors that arise from the use of numerical methods, it is helpful to use
the notation ( )
...

Thus the local truncation error for Euler’s method is ( 2 )
...


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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Find a bound for the local truncation errors for Euler’s method applied to
Ј ϭ 2 , (1) ϭ 1
...
1 we can get an upper
where is between
bound on the local truncation error for 1 by replacing by 1
...
1)2 ]

((1
...
1)2
ϭ 0
...

2

From Table 9
...
1 we see that the error after the first step is 0
...

Similarly, we can get a bound for the local truncation error for any of the fiv
steps given in Table 9
...
1 by replacing by 1
...
Doing
this gives
[2 ϩ (4)(1
...
5)2Ϫ1)

(0
...
1920
2

(2)

as an upper bound for the local truncation error in each step
...
05 in Example 1, then the error bound is 0
...
This is expected because the local truncation error for Euler’s method is ( 2 )
...
The total error in ϩ1 is an accumulation of the errors in each of the previous
steps
...

We expect that, for Euler’s method, if the step size is halved the error will
be approximately halved as well
...
1
...
1
...
50 with ϭ 0
...
5625 and with ϭ 0
...
3171,
approximately half as large
...

For the remainder of this section and in the subsequent sections we study methods that give significantly greater accuracy than does Eule ’s method
...
from (3), we must, at each step, first use Euler’s method (4) to obtain
*
an initial estimate *ϩ1
...
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...
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...
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...
In Figure 9
...
1, observe that
*
ϭ ( 0 , 0 ) and 1 ϭ ( 1, 1 ) are slopes of the solid straight lines shown passing
*
through the points ( 0, 0 ) and ( 1, 1 ), respectively
...
With the first step, rather than advancing along the line through
( 0 , 0 ) with slope ( 0 , 0 ) to the point with -coordinate 1 obtained by Euler’s
*
method, we advance instead along the red dashed line through ( 0 , 0 ) with slope ave
until we reach 1
...

In general, the improved Euler’s method is an example of a
The value of *ϩ1 given by (4) predicts a value of ( ), whereas the value of
defined by formula (3) corrects this estimate
ϩ1
1

0
*)
1

2

( 1, *)
1
( 0 , 0)

( 1,

*
1)

1

Slope of red dashed
line is the average of 0 and 1

Use the improved Euler’s method to obtain the approximate value of (1
...
Compare the results for
ϭ 0
...
05
...
1, we firs

compute (4):
*ϭ
1

0

ϩ (0
...
1)

2

0 0

ϩ2
2

*
1 1

ϭ 1 ϩ (0
...
2
...
1)

The comparative values of the calculations for
Tables 9
...
3 and 9
...
4, respectively
...
1
Actual
value
1
...
10
1
...
30
1
...
50

1
...
2320
1
...
9832
2
...
4509

Abs
...
0000
0
...
0048
0
...
0209
0
...
00
0
...
31
0
...
80
1
...
1)(1
...
232
...
1 and

ϭ 0
...
05

% Rel
...
0000
1
...
5527
1
...
6117
3
...
1 ϭ 1
...
00
1
...
10
1
...
20
1
...
30
1
...
40
1
...
50

1
...
1077
1
...
3798
1
...
7531
1
...
2721
2
...
0038
3
...

error

% Rel
...
0000
1
...
2337
1
...
5527
1
...
9937
2
...
6117
3
...
4904

0
...
0002
0
...
0008
0
...
0020
0
...
0041
0
...
0079
0
...
00
0
...
04
0
...
08
0
...
14
0
...
22
0
...
31

A brief word of caution is in order here
...
In other words, we cannot use
the data in Table 9
...
1 to help construct the values in Table 9
...
3
...
The derivation of this result is
similar to the derivation of the local truncation error for Euler’s method
...
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...
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...
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...
This can be seen in Example 2; when the step size is halved from
ϭ 0
...
05, the absolute error at ϭ 1
...
0394 to 0
...

2
4

()

In Problems 1–10 use the improved Euler’s method to obtain
a four-decimal approximation of the indicated value
...
1 and then use ϭ 0
...

Ј ϭ 2 Ϫ 3 ϩ 1, (1) ϭ 5;
Ј ϭ 4 Ϫ 2 , (0) ϭ 2;
2

Јϭ1ϩ
Јϭ

2

Јϭ

Ϫ

, (0) ϭ 0;
, (0) ϭ 1;

, (0) ϭ 0;

(0
...
5)

ϩ 1 , (0) ϭ 1;

Јϭ ϩ

(0
...
5)

2

ϩ

(1
...
5)

Ј ϭ ( Ϫ ) 2, (0) ϭ 0
...
5)

Јϭ

(0
...
5;

(1
...
5)

Consider the initial-value problem Ј ϭ ( ϩ Ϫ 1) 2,
(0) ϭ 2
...
1 and ϭ 0
...
5
...

Although it might not be obvious from the differential
equation, its solution could “behave badly” near a point
at which we wish to approximate ( )
...
Let ( ) be the solution of the initial-value problem Ј ϭ 2 ϩ 3, (1) ϭ 1
...
4]
...
1, compare the results
obtained from Euler’s method with the results from
the improved Euler’s method in the approximation
of (1
...

Consider the initial-value problem Ј ϭ 2 , (0) ϭ 1
...

Approximate (0
...

Find a bound for the local truncation error in 1
...

Approximate (0
...


Verify that the global truncation error for Euler’s
method is ( ) by comparing the errors in parts
(a) and (d)
...

Its global truncation error is ( 2 )
...
The analytic solution is
ϭ1 Ϫ1ϩ5
2
4
4

Ϫ2


...

Its global truncation error is ( 2 )
...
The analytic solution is
( ) ϭ 1 ϩ 2 ϩ 38
9
3
9


...

Find a bound for the local truncation error in each
step if ϭ 0
...
5)
...
5) using ϭ 0
...
05 with
Euler’s method
...
6
...

Repeat Problem 17 using the improved Euler’s
method, which has a global truncation error ( 2 )
...
You might need to keep more than four
decimal places to see the effect of reducing the order
of the error
...
The analytic solution is ( ) ϭ ln( ϩ 1)
...
5)
...
6
...
See
Problem 5
...


Answer the question “Why not?” that follows the three
sentences after Example 2 on page 366
...
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...
6 (see page 78)

Probably one of the more popular as well as most accurate numerical procedures used in obtaining approximate solutions to a first-order initial-value problem Ј ϭ ( , ),
( 0 ) ϭ 0 is the
As the name suggests, there are Runge-Kutta
methods of different orders
...
1 in that the slope function is
replaced by a weighted average of slopes over the interval Յ Յ ϩ1
...


(1)

Here the weights
, ϭ 1, 2,
...
, , is the function evalu1
ated at a selected point ( , ) for which Յ Յ ϩ1
...
The number is called the
of the method
...
Hence Euler’s method is said to be a
ϩ1 ϭ
The average in (1) is not formed willy-nilly, but parameters are chosen so that
(1) agrees with a Taylor polynomial of degree As we saw in the preceding section,
if a function ( ) possesses ϩ 1 derivatives that are continuous on an open interval
containing and , then we can write
( ) ϭ ( ) ϩ Ј( )
where
ϩ1 ϭ

Ϫ
1!

ϩ Љ( )

( Ϫ )2
ϩиииϩ
2!

( ϩ1)

is some number between and
If we replace
ϩ , then the foregoing formula becomes
2

(

ϩ1)

ϭ (

ϩ )ϭ ( )ϩ

Ј( ) ϩ

( )

( Ϫ ) ϩ1
,
( ϩ 1)!

by
ϩ1

Љ( ) ϩ и и и ϩ

2!

( ϩ 1)!

and

by

( ϩ1)

( ),

where is now some number between
and ϩ1
...


To further illustrate (1), we conThis consists of finding con1, 2, a, and b so that the formula
ϩ1

where

ϩ

ϭ

ϩ (

1

ϭ ( ,

2

ϭ (

ϩ

1 1

ϩ

2 2 ),

(2)

)
,

ϩ␤

1),

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

agrees with a Taylor polynomial of degree two
...

2

(3)

This is an algebraic system of three equations in four unknowns and has infinitel
many solutions:
1

ϭ1Ϫ

2,

1
,
2 2

ϭ

where 2 0
...


Since
ϩ ϭ ϩ1 and ϩ 1 ϭ ϩ ( , ), the foregoing result is recognized to be the improved Euler’s method that is summarized in (3) and (4) of
Section 9
...

In view of the fact that 2 0 can be chosen arbitrarily in (4), there are many
possible second-order Runge-Kutta methods
...
2
...

A
consists of finding parameters so that the formul
ϩ1

where

ϩ (

ϭ

1 1

ϩ

2 2

ϩ

3 3

(5)

4 4 ),

ϩ

1

ϭ ( ,

)

2

ϭ (

ϩ

1

,

ϩ ␤1

1)

3

ϭ (

ϩ

2

,

ϩ ␤2

1

ϩ ␤3

2)

4

ϭ (

ϩ

3

,

ϩ ␤4

1

ϩ ␤5

2

ϩ ␤6

3 ),

agrees with a Taylor polynomial of degree four
...
The most commonly used set of values for the parameters
yields the following result:
ϩ1

ϭ

ϩ

1

ϭ ( ,

2

ϭ

3

ϭ

4

ϭ (

(
(

6

(

1

ϩ2

2

ϩ2

3

ϩ

4),

)
ϩ1 ,
2

ϩ1
2

ϩ1 ,
2

ϩ1
2

ϩ ,

ϩ

)
2)
1

(6)

3)
...
It is (6)
that we have in mind, hereafter, when we use the abbreviation the
You are advised to look carefully at the formulas in (6); note that 2 depends on
, 3 depends on 2, and 4 depends on 3
...

2

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...
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...
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...


●

Use the RK4 method with ϭ 0
...
5) for the solution of Ј ϭ 2 , (1) ϭ 1
...
From

(6) we fin
1

ϭ ( 0,

0)

ϭ2

0 0

ϭ2

1
( 0ϩ
0 ϩ 2 (0
...
1))( 0 ϩ 1 (0
...
31
2
2
1
1
3 ϭ ( 0 ϩ 2 (0
...
1)2
...
1))( 0 ϩ 1 (0
...
34255
2
2
1
2 (0
...
00
1
...
20
1
...
40
1
...
0000
1
...
5527
1
...
6116
3
...

error

% Rel
...
0000
1
...
5527
1
...
6117
3
...
0000
0
...
0000
0
...
0001
0
...
00
0
...
00
0
...
00
0
...
1),

ϭ 2(

ϭ 0
...
1)(

0
0

ϩ (0
...
34255)

ϩ 0
...
715361

and therefore
1

0
...
1
ϭ1 ϩ
(2 ϩ 2(2
...
34255) ϩ 2
...
23367435
...
2
...

Inspection of Table 9
...
1 shows why the fourth-order Runge-Kutta method is
so popular
...
Table 9
...
2 compares the results of applying Euler’s, the
improved Euler’s, and the fourth-order Runge-Kutta methods to the initial-value
problem Ј ϭ 2 , (1) ϭ 1
...
1
...
1
...
)

Ј ϭ 2 , (1) ϭ 1
Comparison of numerical methods with

ϭ 0
...
00
1
...
20
1
...
40
1
...
0000
1
...
4640
1
...
2874
2
...
0000
1
...
5479
1
...
5908
3
...
0000
1
...
5527
1
...
6116
3
...
0000
1
...
5527
1
...
6117
3
...
05

Euler
1
...
05
1
...
15
1
...
25
1
...
35
1
...
45
1
...
0000
1
...
2155
1
...
5044
1
...
8955
2
...
4311
2
...
1733

1
...
1077
1
...
3798
1
...
7531
1
...
2721
2
...
0038
3
...
0000
1
...
2337
1
...
5527
1
...
9937
2
...
6117
3
...
4903

1
...
1079
1
...
3806
1
...
7551
1
...
2762
2
...
0117
3
...
1 we saw that global
truncation errors for Euler’s method and for the improved Euler’s method are, respectively, ( ) and ( 2)
...
It is now obvious why Euler’s
method, the improved Euler’s method, and (6) are
and
Runge-Kutta methods, respectively
...
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...
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...
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...

By computing the fifth derivative of the known solution ( ) ϭ

2

Ϫ1

,

we get
(5)

( )

RK4 Method
Approx
...
1 3
...
32321089
0
...
49033382 9
...


Ϫ1

5!

Thus with ϭ 1
...
00028 on the local truncation error for
each of the five steps when ϭ 0
...
Note that in Table 9
...
1 the error in 1 is much
less than this bound
...
2
...
5 that are obtained from the RK4 method
...
5, we can find the error in these approximations
...
Note that when is halved, from ϭ 0
...
05, the error is divided by a factor of about 2 4 ϭ 16, as expected
...
Of course, this enhanced accuracy is usually obtained at a cost—
namely, increased computation time and greater possibility of round-off error
...
Numerical methods that use a variable step size are called
One of the more popular of the adaptive
routines is the
Because Fehlberg employed two
Runge-Kutta methods of differing orders, a fourth- and a fifth-order method, this al*
gorithm is frequently denoted as the
*

The Runge-Kutta method of order four used in RKF45 is

Use the RK4 method with ϭ 0
...
5), where ( ) is the solution of the initial-value
problem Ј ϭ ( ϩ Ϫ 1) 2, (0) ϭ 2
...
1
...
Use the resulting secondorder Runge-Kutta method to approximate (0
...
Compare this approximate value with the approximate value obtained in Problem 11 in Exercises 9
...

In Problems 3–12 use the RK4 method with ϭ 0
...

Ј ϭ 2 Ϫ 3 ϩ 1, (1) ϭ 5;
Ј ϭ 4 Ϫ 2 , (0) ϭ 2;
Јϭ1ϩ

2

, (0) ϭ 0;

(1
...


2

, (0) ϭ 1;

ϩ

, (0) ϭ 0;

(0
...
5)

2

, (0) ϭ 0;

(0
...
5;

(0
...
5)

Јϭ

2

Јϭ Ϫ

Ϫ , (1) ϭ 1;
2

, (0) ϭ 0
...
5)
(0
...


(0
...
5)

Let (0) ϭ 0, ϭ 0
...


Copyright 2012 Cengage Learning
...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

Use the RK4 method with ϭ 1 to approximate the
velocity (5)
...

Use separation of variables to solve the IVP and
then find the actual value (5)
...
128 Ϫ 0
...
*
Suppose that the initial area is 0
...

Use the RK4 method with ϭ 0
...
53

36
...
50

Consider the initial-value problem Ј ϭ 2 ϩ 3, (1) ϭ 1
...
1
...
4] with step sizes
ϭ 0
...
05
...
4]
...

The analytic solution is ( ) ϭ 2
...
1) using one step and the RK4
method
...

Compare the error in 1 with your error bound
...
1) using two steps and the RK4
method
...

Repeat Problem 16 using the initial-value problem
Ј ϭ Ϫ2 ϩ , (0) ϭ 1
...
A
...


Ϫ3( Ϫ1)

Find a formula involving and for the local truncation error in the th step if the RK4 method is used
...
1 is used to approximate (1
...

Approximate (1
...
1 and ϭ 0
...
See Problem 3
...

Repeat Problem 18 for the initial-value problem Ј ϭ Ϫ ,
(0) ϭ 0
...

Approximate (0
...
See Problem 7
...
40

Use a numerical solver to graph the solution of the
initial-value problem
...

Use separation of variables to solve the initial-value
problem and compute the actual values (1), (2),
(3), (4), and (5)
...
78

Consider the initial-value problem Ј ϭ 2 Ϫ 3 ϩ 1,
(1) ϭ 5
...


(London: Harrap, 1939)
...
Determine the number of evaluations of required for each step of Euler’s,
the improved Euler’s, and the RK4 methods
...


The RK4 method for solving an initial-value problem
over an interval [ , ] results in a finite set of points
that are supposed to approximate points on the graph of
the exact solution
...
These
interpolating functions may be polynomials or sets of
polynomials joined together smoothly
...
The interpolating function
can now be treated like any other
function built into the computer algebra system
...
Graph this solution and find its positive roots
...
1 to approximate a
solution of the initial-value problem in part (a)
...
Find
the positive roots of the interpolating function of the
interval [0, 2]
...
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...
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...
1 and 9
...
On the other hand,
or
use the values from several computed steps to obtain the value of
ϩ1
...


The multistep method that is discussed in this section is called the fourth-order
Like the improved Euler’s method it is a predictor-corrector method—that
is, one formula is used to predict a value * ϩ1, which in turn is used to obtain a corrected value ϩ1
...
The value of *ϩ1 is then substituted into the Adams-Moulton corrector
ϩ1

ϭ

Јϩ1 ϭ (

ϩ

24

ϩ1 ,

(9 Јϩ1 ϩ 19 Ј Ϫ 5 ЈϪ1 ϩ ЈϪ2 )

(2)

*ϩ1 )
...
The value of 0 is, of course, the given initial condition
...


Use the Adams-Bashforth-Moulton method with ϭ 0
...
8) for the solution of
Јϭ

ϩ

Ϫ 1,

(0) ϭ 1
...
2, (0
...
To get
started, we use the RK4 method with 0 ϭ 0, 0 ϭ 1, and ϭ 0
...
02140000,

2

ϭ 1
...
22210646
...
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...
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...
2,

2

ϭ 0
...
6, and

Ј ϭ ( 0,
0

0)

ϭ (0) ϩ (1) Ϫ 1 ϭ 0

Ј ϭ ( 1,
1

1)

ϭ (0
...
02140000) Ϫ 1 ϭ 0
...
4) ϩ (1
...
49181796

Ј ϭ ( 3,
3

3)

ϭ (0
...
22210646) Ϫ 1 ϭ 0
...


With the foregoing values the predictor (1) then gives
*ϭ
4

3

ϩ

0
...
42535975
...
8 ϩ 1
...
22535975
...
2
(9 Ј ϩ 19 Ј Ϫ 5 Ј ϩ Ј) ϭ 1
...

4
3
2
1
24

You should verify that the actual value of
(0
...
42554093
...
3
...
8) in Example 1 is

An important consideration in using numerical methods to approximate the solution of an initial-value problem is the stability of the method
...
A numerical
method is said to be
if it is not stable
...
Because
of the presence of round-off error, this value will almost certainly vary at least
slightly from the true value of the solution
...

One possible method for detecting instability in the numerical solution of a specific initial-value problem is to compare the approximate solutions obtained when
decreasing step sizes are used
...
Another way of checking stability is to
observe what happens to solutions when the initial condition is slightly perturbed
(for example, change (0) ϭ 1 to (0) ϭ 0
...

For a more detailed and precise discussion of stability, consult a numerical
analysis text
...

Many considerations enter into the choice of a method to solve a differential equation
numerically
...
However, a
major drawback is that the right-hand side of the differential equation must be evaluated many times at each step
...
On the other hand, if the function evaluations in the
previous step have been calculated and stored, a multistep method requires only
one new function evaluation for each step
...


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...
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...


●

As an example, solving Ј ϭ ( , ), ( 0 ) ϭ 0 numerically using steps by the
fourth-order Runge-Kutta method requires 4 function evaluations
...
In general the Adams-Bashforth multistep method requires slightly more than a quarter of the number of function evaluations required for the RK4 method
...
Each time
the corrector is used, another function evaluation is done, and so the accuracy is
increased at the expense of losing an advantage of the multistep method
...
This is often
the basis of the variable step size methods, whose discussion is beyond the scope
of this text
...
Compare the actual values of (0
...
4),
(0
...
8) with the approximations 1, 2, 3, and 4
...

In Problems 3 and 4 use the Adams-Bashforth-Moulton
method to approximate (0
...
Use ϭ 0
...

Ј ϭ 2 Ϫ 3 ϩ 1,
Јϭ4 Ϫ2 ,

(0) ϭ 1

In Problems 5 – 8 use the Adams-Bashforth-Moulton method
to approximate (1
...
First use ϭ 0
...
1
...

Јϭ1ϩ

●

,

ϩ 1 ,

(0) ϭ 0

Ј ϭ ϩ cos ,
Ј ϭ ( Ϫ ) 2,
Јϭ

(0) ϭ 2

●

2

(0) ϭ 1
(0) ϭ 0
(0) ϭ 1

Section 1
...
10 (second-order DE written as a system of first-order DEs

So far, we have focused on numerical techniques that can be used to approximate the solution of a first-order initial-value problem Ј ϭ ( , ), ( 0) ϭ 0
...
To do this, we begin by writing the second-order DE in normal form
by solving for Љ in terms of , , and Ј
...
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...
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...
If we let Ј ϭ , the differential equation in (1) becomes the system
Јϭ
Ј ϭ ( , , )
...
The system (2) can now be solved numerically by simply applying a particular numerical method to each first-order differential equation in the system
...
, ( Ϫ1) ) as a system of
substitutions ϭ 1, Ј ϭ 2, Љ ϭ 3 ,
...


th-order differential equation
first-order equations using the
Ϫ1)
ϭ
...
2), where ( ) is the
solution of the initial-value problem

approximate
y(0
...
2

1

Љϩ
x

2

ϭ 0,

Јϩ

(0) ϭ 1,

Ј(0) ϭ 2
...


ЈϭϪ
Thus from (3) we obtain

y

ϩ1

ϭ

ϩ

ϩ1

2

ϭ

ϩ [Ϫ

Using the step size ϭ 0
...
1)

ϭ

0

ϩ (0
...
1)

2

ϭ

1

ϩ (0
...


ϭ 2, we fin

ϭ 1 ϩ (0
...
2
0 0

Ϫ

0]

ϭ 2 ϩ (0
...
9

ϭ 1
...
1)(1
...
39
1 1

Ϫ

1]

ϭ 1
...
1)[Ϫ(0
...
9) Ϫ 1
...
761
...
2) Ϸ 1
...
2) Ϸ 1
...

With the aid of the graphing feature of a numerical solver, in Figure 9
...
1(a) we
compare the solution curve of (5) generated by Euler’s method ( ϭ 0
...
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...
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...
1)
...
4
...

If desired, we can use the method of Section 6
...
But unless this method reveals that the DE
possesses an elementary solution, we will still only be able to approximate (0
...
Reinspection of the infinite series solutions of Airy’s differential equation
Љ ϩ ϭ 0, given on page 242, does not reveal the oscillatory behavior of the
solutions 1 ( ) and 2 ( ) exhibited in the graphs in Figure 6
...
2
...
1
...


Write

ЉϪ Јϩ5 ϩ2 Љϭ
Ϫ2 ϩ Љ ϩ 2 ϭ 3

2

as a system of first-order di ferential equations
...
This
gives
Љ ϭ Ϫ9 ϩ 4 ϩ Ј ϩ Ϫ 6 2
...
If we let Ј ϭ and Ј ϭ , the expressions for Љ and Љ become, respectively,
Ј ϭ Љ ϭ Ϫ9 ϩ 4 ϩ

Ϫ6

ϩ

2

2

Јϭ Љϭ2 Ϫ2 ϩ3
...


Ϫ6

2

It might not always be possible to carry out the reductions illustrated in Example 2
...
,

)

2
––– ϭ

2(

,

1,

2,


...
,

)


...


...


...

(,

1,

2,

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

can be approximated by a version of Euler’s, the Runge-Kutta, or the
Adams-Bashforth-Moulton method adapted to the system
...


Consider the initial-value problem
Јϭ2 ϩ4
ЈϭϪ ϩ6
(0) ϭ Ϫ1,

(0) ϭ 6
...
6) and (0
...
Compare the results for
ϭ 0
...
1
...
2
...
2, 9
...
416) ϭ 96
...
2, 9
...
416) ϭ 106
...


(0ϩ1
2
(0ϩ1
2
(0ϩ1
2
(0ϩ1
2

)ϭ
1) ϭ
2) ϭ
2) ϭ

1

(0
...
2, 9
...
2
(0
...
2, 9
...
1, 3
...
7) ϭ 53
...
1, 3
...
7) ϭ 67
...
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...
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...
2

0
...
20
0
...
60

1

6
...
0683
55
...
8192

Ϫ1
...
2453
46
...
9430

ϭ

0

ϩ

ϭϪ1 ϩ

1

ϭ 0
...
00
0
...
20
0
...
40
0
...
60

6
...
8883
19
...
8539
55
...
3006
152
...
0000
2
...
3379
22
...
5103
88
...
7563

1

ϩ2

2

ϩ2

3

ϩ

4)

0
...
2) ϩ 2(53
...
88) ϭ 9
...
2
(
6

1

ϩ2

2

ϩ2

3

4)

ϩ

0
...
08) ϩ 106
...
0683,
6

where, as usual, the computed values of 1 and 1 are rounded to four decimal
places
...
2) and 1 Ϸ (0
...
The
subsequent values, obtained with the aid of a computer, are summarized in
Tables 9
...
1 and 9
...
2
...
6) ϭ 160
...
6) ϭ 152
...
4
...
The graph of the solution in a neighborhood
of ϭ 0 is shown in Figure 9
...
2; the graph was obtained from a numerical solver
using the RK4 method with ϭ 0
...


x, y

1

0
...


Numerical solution
curves for IVP in Example 3

Use Euler’s method to approximate (0
...


Use ϭ 0
...
Find the analytic solution of the problem,
and compare the actual value of (0
...

Use Euler’s method to approximate (1
...
Use ϭ 0
...
Find the analytic solution of the
problem, and compare the actual value of (1
...

In Problems 3 and 4 repeat the indicated problem using the
RK4 method
...
2 and then use ϭ 0
...

Problem 1

1

ϭ Ϫ20 1 ϩ 10 3 ϩ 100

3

ϭ 10 1 Ϫ 20 3 ,

where 1(0) ϭ 0 and 3(0) ϭ 0
...
1, 0
...
3, 0
...
5
...
1
...
Use the graphs to predict the
behavior of 1( ) and 3( ) as : ϱ
...
2), where ( )
is the solution of the initial-value problem
ЉϪ2 Јϩ2 ϭ

When ϭ 100 V, ϭ 10 ⍀, and ϭ 1 h, the system of differential equations for the currents 1 ( ) and
3 ( ) in the electrical network given in Figure 9
...
3 is

cos ,

(0) ϭ 1,

First use ϭ 0
...
1
...

Network in Problem 6

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...
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...
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...


●

In Problems 7 – 12 use the Runge-Kutta method to approximate (0
...
2)
...
2 and then use
ϭ 0
...
Use a numerical solver and ϭ 0
...

Јϭ2 Ϫ
Јϭ
(0) ϭ 6,

(0) ϭ 2

●
●
●
●

ЈϭϪ ϩ
Јϭ Ϫ
(0) ϭ Ϫ3,

(0) ϭ 5

Јϩ4 Ϫ Јϭ7
Јϩ ЈϪ2 ϭ3
(0) ϭ 1, (0) ϭ Ϫ2

Јϭ ϩ2
Јϭ4 ϩ3
(0) ϭ 1, (0) ϭ 1

Јϭ6 ϩ ϩ6
Ј ϭ 4 ϩ 3 Ϫ 10 ϩ 4
(0) ϭ 0
...
2
Јϩ Јϭ 4
Ϫ Ј ϩ Ј ϩ ϭ 6 2 ϩ 10
(0) ϭ 3, (0) ϭ Ϫ1

Section 4
...
3 (Problems 37–40)
Exercises 4
...
2
We just saw in Section 9
...


In this section we are going to examine two methods for approximating a solution of a
Љ ϭ ( , , Ј),

( ) ϭ a,

( ) ϭ b
...


The Taylor series expansion, centered
at a point , of a function ( ) is
( ) ϭ ( ) ϩ Ј( )

Ϫ
1!

ϩ Љ( )

( Ϫ )2
( Ϫ )3
ϩ ٞ( )
ϩиии
...


For the subsequent discussion it is convenient then to rewrite this last expression in
two alternative forms:
2

( ϩ ) ϭ ( ) ϩ Ј( ) ϩ Љ( )

2

3

ϩ ٞ( )

2

and

( Ϫ ) ϭ ( ) Ϫ Ј( ) ϩ Љ( )

2

6

ϩиии

(1)

ϩиии
...
since these values are negligible
...


(4)

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...
Editorial review has
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...


●

Subtracting (1) and (2) also gives
Ј( ) Ϸ

1
[ ( ϩ ) Ϫ ( Ϫ )]
...


(6)

The right-hand sides of (3), (4), (5), and (6) are called
expressions
( ϩ ) Ϫ ( ),

( ) Ϫ ( Ϫ ),

The

( ϩ ) Ϫ ( Ϫ ),

( ϩ )Ϫ2 ( )ϩ ( Ϫ )

and
are called

Specificall ,
( ) Ϫ ( Ϫ ) is a
and ( ϩ ) Ϫ 2 ( ) ϩ ( Ϫ ) are called
(5) and (6) are referred to as
Ј and Љ
...


(7)

Suppose ϭ 0 Ͻ 1 Ͻ 2 Ͻ и и и Ͻ Ϫ1 Ͻ ϭ represents a regular partition of
the interval [ , ], that is, ϭ ϩ , where ϭ 0, 1, 2,
...
,

Ϫ1

ϩ ( Ϫ 1)

ϭ

of the interval [ , ]
...


(8)

The last equation, known as a
is an approximation
to the differential equation
...
, Ϫ1 of the interval [ , ]
...
, Ϫ 1 in (8), we obtain Ϫ 1 equations in the Ϫ 1
unknowns 1, 2 ,
...
Bear in mind that we know 0 and , since
these are the prescribed boundary conditions 0 ϭ ( 0 ) ϭ ( ) ϭ a and
ϭ ( ) ϭ ( ) ϭ b
...
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...
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...
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...

To use (8), we identify ( ) ϭ 0,
ϭ (1 Ϫ 0)> 4 ϭ 1
...
25 ϩ

ϩ1

Ϫ1

( ) ϭ Ϫ4,

( ) ϭ 0, and
(9)

ϭ 0
...
25 1 ϩ 0 ϭ 0
Ϫ 2
...
25

3

ϩ

0

ϭ 0 and

4

ϭ 5 the foregoing system becomes

1

ϩ

2

1

Ϫ 2
...
25

3

ϭ Ϫ5
...
25

2

ϭ 0
...
7256, 2 ϭ 1
...
9479
...
The other boundary condition
gives 2
...
Thus the actual values (rounded to four decimal places) of
this solution at the interior points are as follows: (0
...
7184, (0
...
6201,
and (0
...
9354
...
It is left as an exercise to show that with ϭ 1,
8
approximations to (0
...
5), and (0
...
7202, 1
...
9386, respectively
...
5
...


In this case we identify ( ) ϭ 3,
ϭ (2 Ϫ 1)͞10 ϭ 0
...
15

ϩ1

Ϫ 1
...
85

Ϫ1

ϭ 0
...


Now the interior points are 1 ϭ 1
...
2, 3 ϭ 1
...
7, 8 ϭ 1
...
9
...
, 9 and
system of nine equations and nine unknowns:

( ) ϭ 4 2, and
(10)

ϭ 1
...
5, 6 ϭ 1
...
15

2

Ϫ 1
...
15

3

Ϫ 1
...
85

1

ϭ 0
...
15

4

Ϫ 1
...
85

2

ϭ 0
...
15

5

Ϫ 1
...
85

3

ϭ 0
...
8016

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...
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...
Editorial review has
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...


●

1
...
98

5

ϩ 0
...
0900

1
...
98

6

ϩ 0
...
1024

1
...
98

7

ϩ 0
...
1156

1
...
98

8

ϩ 0
...
1296

Ϫ 1
...
85

8

ϭ Ϫ6
...


We can solve this large system using Gaussian elimination or, with relative ease,
by means of a computer algebra system
...
4047,
2 ϭ 3
...
2010, 4 ϭ 4
...
1359, 6 ϭ 5
...
6117,
8 ϭ 5
...
8855
...


(11)

The number 1 in (11) is simply a guess for the unknown slope of the solution curve at
the known point ( , ( ))
...
If
b1 agrees with the given value ( ) ϭ b to some preassigned tolerance, we stop; otherwise, the calculations are repeated, starting with a different guess Ј( ) ϭ 2 to obtain
a second approximation b2 for ( )
...
can be adjusted in some systematic way;
linear interpolation is particularly successful when the differential equation in (11) is
linear
...
See Problem 14 in Exercises 9
...

Of course, underlying the use of these numerical methods is the assumption,
which we know is not always warranted, that a solution of the boundary-value problem exists
...
See Problem 13
in Exercises 9
...


In Problems 1 – 10 use the finite difference method and the
indicated value of to approximate the solution of the given
boundary-value problem
...


The electrostatic potential between two concentric
spheres of radius ϭ 1 and ϭ 4 is determined from
2

(2) ϭ Ϫ1;

(1) ϭ 5, (2) ϭ 0;

2

ϩ

2

ϭ 0,

(1) ϭ 50,

(4) ϭ 100
...
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...
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...


Use ϭ 5 and the system of equations found in
parts (a) and (b) to approximate the solution of the
original boundary-value problem
...

Find the difference equation corresponding to the
differential equation
...
,
Ϫ 1 the difference equation yields equations in
ϩ 1 unknows Ϫ1, 0 , 1 , 2 ,
...
Here Ϫ1
and 0 are unknowns, since Ϫ1 represents an
approximation to at the exterior point ϭ Ϫ and
0 is not specified at ϭ 0
...


Consider the boundary-value problem Љ ϭ Ј Ϫ sin ( ),
(0) ϭ 1, (1) ϭ 1
...
Use the shooting method to approximate the solution of this problem
...
1; or, even better, if you have
access to a CAS such as
or
the
function can be used
...
Compute to four rounded decimal places
...
1 and then use ϭ 0
...


Use the Adams-Bashforth-Moulton method to approximate (0
...
Use ϭ 0
...


Ј ϭ 2 ln , (1) ϭ 2;
(1
...
2), (1
...
4), (1
...
5) ϭ 0
...
6), (0
...
8), (0
...
0)

Ј ϭ sin 2 ϩ cos 2, (0) ϭ 0;
(0
...
2), (0
...
4), (0
...
1), (1
...
3), (1
...
5)
Use Euler’s method to approximate (0
...
First use
one step with ϭ 0
...
1
...
1 to approximate (0
...
2), where ( ), ( ) is the solution of the initialvalue problem
Јϭ
Јϭ
(0) ϭ 1,

ϩ
Ϫ
(0) ϭ 2
...
55(1 ϩ ) ϭ 1, (0) ϭ 0, (1) ϭ 0
...
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...
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...
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...


Ϫ1 Ϫ

(1)

0

Convergence of the integral requires that
relation

Ϫ 1 Ͼ Ϫ1 or

Ͼ 0
...
4, can be obtained from (1) with integration by parts
...
In this manner, it is seen that when is a positive integer, ⌫( ϩ 1) ϭ !
For this reason, the gamma function is often called the
Although the integral form (1) does not converge for Ͻ 0, it can be shown
by means of alternative definitions that the gamma function is defined for all real and
complex numbers
ϭ Ϫ , ϭ 0, 1, 2,
...
1
...

In Problems 31 and 32 of Exercises 6
...
This
2
result can be derived from (1) by setting ϭ 1:
2

()

( 1) ϭ
2

⌫

Graph of ⌫( ) for
neither 0 nor a negative integer

When we let
͵0 Ϫ 2 ϭ ͵0

ϭ
Ϫ

2

Ϫ

2

Switching to polar coordinates
double integral:

0

Hence


...
But


...

2
Ϫ

2

␪ ϭ ␲
...
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...

2
In view of (2) and (4), it follows that, with ϭ Ϫ1,
2
⌫

( )

( 1) ϭ Ϫ1 ⌫ (Ϫ1)
...

2

I
Evaluate
...
[

( )
that ⌫ (6) ϭ 0
...

to evaluate


...
89
3

to evaluate


...

A definition of the gamma function due to Carl
Friedrich Gauss that is valid for all real numbers, except
ϭ 0, Ϫ1, Ϫ2,
...
[

to show that ⌫( )

ϩ

0

Use (1) and the fact that ⌫

Ϫ1 Ϫ

0

⌫ Ϫ5
2

Use (1) and the fact

͵

͵

1

Use the fact that ⌫( ) Ͼ

: Let ϭ Ϫln ]

!

...
( ϩ )

Use this definition to show that ⌫( ϩ 1) ϭ ⌫( )
...
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...


...



...


...


...


)


...
An ϫ matrix is called a
of order
The entry in the th row and th column of an ϫ matrix is written

...

A 1 ϫ 1 matrix is sximply one constant or function
...


...


ϭ(

1) ϫ1
...

11
12

...


...


...


...


2

)

ϭ(

)

ϫ

,

where is a constant or a function
...
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...
For

Ϫ3

Ϫ3

Ϫ3

Ϫ3

The

of two

matrices

and

ϩ

ϫ

ϭ(


...


In other words, when adding two matrices of the same size, we add the corresponding entries
...

Ϫ3

0
Ϫ2
ϩ 7 ϩ
0
5
0

3
ϭ 1
0

2

0
ϩ 7
5

Ϫ2
ϩ
0
0


...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Let be a matrix having rows and
rows and columns
...


...



...


...


...


columns and be a matrix having
to be the ϫ matrix

11

12

21

22


...


1

2


...


...


ϩ
...
ϩ
21 ϩ

...


...
ϩ
21 ϩ

11 11

12 21

1

1

21

22

2

1

1 11

ϩ

(͚ )
ϭ1

2

1

1
2


...


...


...


ϩ
...
ϩ
2 ϩ

...


...
ϩ
2 ϩ

12 2

1

22

2

2

)


...
6 that the product
ϭ is defined only when
the number of columns in the matrix is the same as the number of rows in
...


q

Also, you might recognize that the entries in, say, the th row of the final matrix
are formed by using the component definition of the inner, or dot, product of the th
row of with each of the columns of
...

34

΃

Ϫ3
,
0

΃ ΂

5 ؒ (Ϫ3) ϩ 8 ؒ 0
Ϫ4
1 ؒ (Ϫ3) ϩ 0 ؒ 0 ϭ Ϫ4
2 ؒ (Ϫ3) ϩ 7 ؒ 0
6

΃

Ϫ15
Ϫ3
...

30 53
, whereas in part (b)
Observe in part (a) of Example 4 that
ϭ
48 82
the product
is not defined, since Definition II
...

We are particularly interested in the product of a square matrix and a column
vector
...
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...
0
0 1 0
...

ϭ
...


...


...
1
is called the
any ϫ matrix
...
6 that for
ϭ

Also, it is readily verified that, if

is an

ϭ


...


A matrix consisting of all zero entries is called a
is denoted by
...
If

and

are

ϫ

ϭ

0
0 ,
0

ϩ

ϭ


...


ϫ

If all products are defined, multiplication is
over addition:
(

)ϭ

ϩ

ϭ

΂

det

3
2
Ϫ1

6
5
2

3
ϭ p 2
Ϫ1

(

ϩ

)

ϭ


...


is a number called the

For

and

ϩ

΃

2
1 we expand det
4
6
5
2

2
5
1p ϭ 3
2
4

by cofactors of the first row

͉ ͉ ͉

1
2
Ϫ6
4
Ϫ1

͉ ͉

1
2
ϩ2
4
Ϫ1

5
2

͉

ϭ 3(20 Ϫ 2) Ϫ 6(8 ϩ 1) ϩ 2(4 ϩ 5) ϭ 18
...
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...
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...
If det has a row (or a column) containing many zero entries,
then wisdom dictates that we expand the determinant by that row (or column)
...


...


12

22


...


1

2


...


...


΂

΃

2
1 is
4

3
ϭ 6
2

)


...


΃

Ϫ1
2
...


matrix
...


be an ϫ matrix
...


An ϫ matrix
nonsingular
...


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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Let be an ϫ nonsingular matrix and let
ϭ (Ϫ1) ϩ
, where
is the
determinant of the ( Ϫ 1) ϫ ( Ϫ 1) matrix obtained by deleting the th row and
th column from
...


Each
in Theorem II
...
Note that the transpose is utilized in formula (2)
...


ϭ

Thus

΂Ϫ

1
det

22

Ϫ

21

΃
...
Carrying out the transposition gives

Ϫ1

ϭ

1
det

΂

11

21

31

12

22

32

13

23

33

΃


...

10

Since det ϭ 10 Ϫ 8 ϭ 2 0,
Theorem II
...
From (3) we fin
Ϫ1

ϭ

΂

1 10
2 Ϫ2

is nonsingular
...


Ϫ2
1
2

Not every square matrix has a multiplicative inverse
...
Hence Ϫ1 does not exist
...

1

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...
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...


●

Since det ϭ 12 0, the given matrix is nonsingular
...

1

23

ϭϪ

33

0
ϭ Ϫ2
1

͉

1
ϭ Ϫ3
0

ϭ

If follows from (4) that
1
Ϫ1
ϭ
12

΂

1
5
Ϫ3
Ϫ1

You are urged to verify that

΃ ΂

Ϫ2
2
6

1
2
12
5
Ϫ2 ϭ 12
6
Ϫ1
4

ϭ

Ϫ1

Ϫ1
6

1
6
1
2

1
6
1
Ϫ6
1
2

΃


...


Formula (2) presents obvious difficulties for nonsingular matrices larger than
3 ϫ 3
...
* In the case of a large matrix there are more efficien
ways of finding Ϫ1
...

Since our goal is to apply the concept of a matrix to systems of linear first-orde
differential equations, we need the following definitions

If ( ) ϭ ( ( )) ϫ is a matrix whose entries are functions differentiable on a
common interval, then
ϭ

΂ ΃


...

ϫ

To differentiate (integrate) a matrix of functions, we simply differentiate
(integrate) each entry
...


΂ ΃
sin 2

If

()ϭ

3

8 Ϫ1

,

΂ ΃΂
sin 2

then

Ј( ) ϭ

3

ϭ

2 cos 2
3 3
8

΃

(8 Ϫ 1)

*

Strictly speaking, a determinant is a number, but it is sometimes convenient to refer to a determinant as
if it were an array
...
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...
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...

3
3
4 2Ϫ

Matrices are an invaluable aid in solving algebraic systems of
variables or unknowns,
ϩ
1 ϩ

ϩиииϩ
2 ϩ и и и ϩ

12 2
22

2


...
However, that rule requires a
herculean effort if is larger than 3 ϫ 3
...


The

of the system (5) is the

(

11

12

21

22


...


2


...


...


1

If is the column matrix of the
is denoted by ( ͉ )
...


...


, ϭ 1, 2,
...
These operations on equations in a system are, in turn,
equivalent to
on an augmented matrix:
( ) Multiply a row by a nonzero constant
...

( ) Add a nonzero constant multiple of one row to any other row
...

( ) Rows consisting of all 0’s are at the bottom of the matrix
...
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...
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...


The augmented matrices

΂

1
0
0

5
1
0

0
0
0

p

2
Ϫ1
0

p

7
Ϫ1
0

΃

΂0
0

and

0
0

1
0

Ϫ6
0

2
1

͉ 2΃
4

are in row-echelon form
...
Note that the remaining entries in the columns
containing a leading entry 1 are all 0’s
...
In other words, by using different sequences of row
operations we may arrive at different row-echelon forms
...
In Gauss-Jordan elimination we stop when we have
obtained
augmented matrix in reduced row-echelon form
...

This method does not require back-substitution; the solution of the system will be
apparent by inspection of the final matrix
...
The process is repeated on the other variables
...


We can always interchange equations so that the first equation contains the variable

1
...
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...
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...

0
1
5

The last matrix is in row-echelon form and represents the system
1

ϩ2

2

Ϫ
3
2

9
2

3

ϭ

3

2

ϩ

ϭ Ϫ1

3

ϭ 5
...
Substituting both
these values back into the first equation finally yield 1 ϭ 10
...
Since the first entries in the second and third
rows are 1’s, we must, in turn, make the remaining entries in the second and
third columns 0’s:

(

Ϳ

1 2 Ϫ1 Ϫ1
3
9
_
_
0 1
2
2
0 0
1
5

)

Ϫ2

2

ϩ

(

1

Ϳ

1 0 Ϫ4 Ϫ10
3
9
_
_
0 1
2
2
0 0
1
5

)

4
3
Ϫ_
2

3
3

ϩ
ϩ

1
2

(

Ϳ

)

1 0 0 10
0 1 0 Ϫ3
...
Because of what the matrix
means in terms of equations, it is evident that the solution of the system is 1 ϭ 10,
2 ϭ Ϫ3, 3 ϭ 5
...

We solve the system using Gauss-Jordan elimination:

1
__
Ϫ 11
1
__
Ϫ 11

2
3

(
(

Ϳ
Ϳ

1
3 Ϫ2 Ϫ7
4
1
3
5
2 Ϫ5
7 19
1
0
0

3 Ϫ2 Ϫ7
1 Ϫ1 Ϫ3
1 Ϫ1 Ϫ3

)
)

Ϫ4
Ϫ2

1
1

ϩ
ϩ

Ϫ3 2 ϩ
Ϫ 2ϩ

2
3

1
3

(
(

Ϳ
Ϳ

1
3 Ϫ2 Ϫ7
0 Ϫ11 11 33
0 Ϫ11 11 33
1
0
0

)
)

0
1
1
1 Ϫ1 Ϫ3
...
Since only is common to both equations (the nonzero rows), we

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

can assign its values arbitrarily
...

Because of the number of determinants that must be evaluated, formula (2) in Theorem II
...
In the case of 3 ϫ 3 or larger matrices the method
described in the next theorem is a particularly efficient means for findin Ϫ1
...
The same
sequence of operations that transforms into the identity will also transform into Ϫ1
...


...



...


...


...


and simultaneously
with the identity as

)

1 0
...
0

...


...


...


...
This
means that is nonsingular
...


By simultaneously applying
the same row operations
to , we get Ϫ1
...

6

We shall use the same notation as we did when we reduced an
augmented matrix to reduced row-echelon form:

(

Ϳ

2 0 1 1 0 0
Ϫ2 3 4 0 1 0
Ϫ5 5 6 0 0 1

) (
1
_
2

1

Ϳ

_
_
1 0 1 1 0 0
2
2
Ϫ2 3 4 0 1 0
Ϫ5 5 6 0 0 1

) (
2
5

ϩ
1ϩ
1

2
3

Ϳ

_
_
1 0 1 1 0 0
2
2
0 3 5 1 1 0
__
_
0 5 17 5 0 1
2
2

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...
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...


)

●

1
_
3
1
_
5

30

2
3

3

(
(

1 0
0 1
0 1

1
_
2
5
_
3
17
__
10
1

Ϳ
Ϳ

1
_
2
1
_
3
1
_
2

0 0
1
_
0
3
_
0 1
5

) (
) (
Ϫ

_
1 0 _ 1
0 0
2
2
1
1
5
_
_
_
0 1 3 3
0
3
0 0 1 5 Ϫ10 6

2

ϩ

_
Ϫ1
3
_
Ϫ5
3

1 0
0 1
0 0

3

3
3

ϩ
ϩ

1
2

1
_
2
5
_
3
1
__
30

Ϳ

1
_
2
1
_
3
1
_
6

0 0
1
_
0
3
_
_
Ϫ1 1
3
5

)
)

Ϳ

1 0 0 Ϫ2
5 Ϫ3
0 1 0 Ϫ8
17 Ϫ10
...

6

If row reduction of ( ͉ ) leads to the situation
( Ϳ )

row
operations

( Ϳ ),

where the matrix contains a row of zeros, then necessarily is singular
...


Gauss-Jordan elimination can be used to find the eigenvectors of a square matrix
...
A number l is said to be an
exists a
solution vector of the linear system
ϭ␭
...


of

if there
(6)

is said to be an

corresponding to the

The word
is a combination of German and English terms adapted
from the German word
which, translated literally, is “proper value
...


Verify that

΂΃

1
ϭ Ϫ1 is an eigenvector of the matrix
1

΂

0
2
ϭ
Ϫ2

Ϫ1
3
1

΃

Ϫ3
3
...
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...
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...

Ϫ2
1
1
1
Ϫ2
1

We see from the preceding line and Definition II
...

Using properties of matrix algebra, we can write (6) in the alternative form
(

Ϫ␭ )

ϭ ,

(7)

where is the multiplicative identity
...
ϩ

...


...


...


...

ϭ 0
...
, ϭ 0, we are seeking
only nontrivial solutions
...
, in (5)) has a nontrivial solution
if and only if the determinant of the coefficient matrix is equal to zero
...


Inspection of (8) shows that the expansion of det( Ϫ l ) by cofactors results in an
th-degree polynomial in l
...
Thus
To fin
an eigenvector corresponding to an eigenvalue l, we simply solve the system of
equations ( Ϫ l ) ϭ by applying Gauss-Jordan elimination to the augmented
matrix ( Ϫ ␭ ͉ )
...

Ϫ1

To expand the determinant in the characteristic equation, we use the
cofactors of the second row:
det(

1Ϫ␭
Ϫ␭ )ϭ p 6
Ϫ1

2
Ϫ1 Ϫ ␭
Ϫ2

1
0
p ϭ Ϫ␭3 Ϫ ␭2 ϩ 12 ␭ ϭ 0
...
To find the eigenvectors, we must now reduce
( Ϫ ␭ ͉ ) three times corresponding to the three distinct eigenvalues
...
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...
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...


3

(

Ϳ)
Ϳ)

1
2
1 0
0 Ϫ13 Ϫ6 0
0
0
0 0

ϩ

1

Choosing

(

3

1
__
1 0 13 0
6
__
0 1 13 0
...

Ϫ13

For l 2 ϭ Ϫ4,

(

Ϳ)

5
2 1 0
Ϳ )ϭ 6
( ϩ4
3 0 0
Ϫ1 Ϫ2 3 0

Ϫ6
Ϫ5

ϩ
1ϩ
1

2
3

(

Ϳ)

1
2 Ϫ3 0
0 Ϫ9 18 0
0 Ϫ8 16 0

implies that
eigenvector

1

ϭϪ

3

and

_
Ϫ1
9
_
Ϫ1
8

2

2
3

(

Ϫ

3
31

Ϳ)

Ϳ)

1 2 Ϫ3 0
0 1 Ϫ2 0
0 1 Ϫ2 0

ϭ 2 3
...
The choice of
2

3

1
3

(

Ϳ)

1 0
1 0
0 1 Ϫ2 0
0 0
0 0

΂΃

(

ϭϪ

ϩ
ϩ

Ϫ1
2
...

Ϫ2

When an ϫ matrix possesses distinct eigenvalues l 1, l 2 ,
...
,
can be
found
...


*
Of course, 3 could be chosen as any nonzero number
...

†
Linear independence of column vectors is defined in exactly the same manner as for functions

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

Find the eigenvalues and eigenvectors of

3
΂Ϫ1 4΃
...
In the case of a 2 ϫ 2
matrix there is no need to use Gauss-Jordan elimination
...


It is apparent from this system that
single eigenvector

1

ϭ 2 2
...

1

΂

9
ϭ 1
1

Find the eigenvalues and eigenvectors of

The characteristic equation

2

1
9
1

΃

1
1
...

For l 1 ϭ 11 Gauss-Jordan elimination gives

( Ϫ 11 Ϳ ) ϭ

Hence

1

ϭ

3

and

2

ϭ

(

Ϳ)

Ϫ2
1
1 0
1 Ϫ2
1 0
1
1 Ϫ2 0

3
...

0 0
0 0

ϭ 1, then

1

΂΃

1
ϭ 1
...

0 0 0 0

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...
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...
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deemed that any suppressed content does not materially affect the overall learning experience
...


●

In the equation 1 ϩ 2 ϩ 3 ϭ 0, we are free to select two of the variables arbitrarily
...

3 ϭ
1

΂1 2΃ and
2 4

II

If

ϭ

4
΂Ϫ6 5΃ and
9

ϩ

ϭ

΂ ΃

Ϫ
If

ϭ

΂

0
1 and
3

3
ϭ
0
Ϫ4

΃

Ϫ3
and
4

ϭ
2

΂ ΃

If

1
ϭ 5
8

If

ϭ

4
10 and
12

1
΂Ϫ2

΃

Ϫ1
2 , fin
Ϫ2

If

΂Ϫ4
1

΂6 3΃, and
2 1

6
Ϫ3

ϭ

If

ϭ (5

΂

1
ϭ 0
3

Ϫ6
2
1
2

)

7),

(

ϭ

΃

΂ ΃

ϭ

5
΂Ϫ4 9΃ and
6

΂2
1

ϩ

΃

11
, fin
2

( ϩ

΂

Ϫ3
5
Ϫ4

)

1
2
0

΃΂Ϫ2΃ Ϫ ΂Ϫ1 6΃΂Ϫ7΃
5
Ϫ2 3
2

΃΂ ΃ ΂ ΃ ΂ ΃

4
Ϫ1
Ϫ2

2 Ϫ1 ϩ
Ϫ

Ϫ
2
1 Ϫ
8
4
Ϫ6

In Problems 15–22 determine whether the given matrix is
singular or nonsingular
...
2
...


fin
(

΂3 4΃ and
8 1

ϩ

2

ϭ

ϭ

΃

ϭ
(

΂Ϫ1 6΃, fin
3 2

ϭ

Ϫ2
,
4

If

ϭ
2

ϩ

2( ϩ )

Ϫ

2
΂Ϫ5

΃

6
, fin
Ϫ10
2 ϩ3

Ϫ

Ϫ2
ϭ
4
7

If

΂Ϫ2
8

ϭ

4
΂Ϫ3

8
Ϫ5

΂

1
2
2

)

2
ϭ Ϫ1
1

ϭ
ϭ

΃

΃

0
1
1

΂2 5΃
1 4
΂7
2

ϭ

΂

3
4
Ϫ2

΃

10
2

2
1
5

΃

1
0
Ϫ1

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...
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...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

΂

2
ϭ 1
3

΃

1
Ϫ2
2

΂

1
Ϫ3
4

4
ϭ
6
Ϫ2

΃

1
2
Ϫ1

Ϫ1
Ϫ3
2

In Problems 23 and 24 show that the given matrix is
nonsingular for every real value of Find Ϫ1( ) using
Theorem II
...

()ϭ

΂2
4

()ϭ

΂2

4

Ϫ

4

3

Ϫ

sin
cos

΃
Ϫ2 cos
sin

΃

Ϫ
Ϫ

ϭ

Ϫ

΂ ΃

1
ϭ2
Ϫ1

΂΃

2
ϩ4
1

2

΂2

4

Let ( ) ϭ

΂Ϫ3 sin 2
sin 2
1
2

Ϫ3

ϭ

΃

cos

...
Find
4

2
0
1
1

΂

0
0
0
1

0
1
0
0

1
Ϫ1
ϭ
2
1

΃
΃

ϭ1
ϭ0
4ϭ6
ϭ9

3

2

΂
΂

( )

6
and ( ) ϭ
1>

͵

Ϫ
Ϫ
1ϩ2 2Ϫ2
4 1ϩ7 2Ϫ7
1

In Problems 41 – 46 use Theorem II
...


2

()

Ϫ
ϩ
3Ϫ
3ϩ
3

2

΃

0

3

Ϫ
ϩ
2ϩ
2Ϫ2
2

1

4
ϭ
2
Ϫ1

͵

2

2

ϩ
ϩ
1Ϫ
4 1ϩ
1

ϩ2 ϩ4 ϭ2
2 ϩ4 ϩ3 ϭ1
ϩ2 Ϫ ϭ7

΂ ΃

5
ϭ
2
Ϫ7

ϩ
2 ϭ8
ϩ2 Ϫ2 ϭ4
2 ϩ5 Ϫ6 ϭ6

In Problems 39 and 40 use Gauss-Jordan elimination to
demonstrate that the given system of equations has no
solution
...


( ) ( )

1

΂Ϫ1 2΃
Ϫ7 8
In Problems 31–38 solve the given system of equations by
either Gaussian elimination or Gauss-Jordan elimination
...
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...
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...
Find the eigenvectors of the matrix
...
[

΂

12

21

11 ,

12 ,

21 ,

ϭ

, show that

are nonsingular, show that (

and

be

ϫ

Ϫ1

)

ϭ

ϭ
...


matrices
...
The multiplicative identity matrix
is an example of a diagonal matrix
...
Solve for
ϭ
...


΃

11

is nonsingular and

(

: Find a matrix
ϭ

for which
show that

( ) Ј( ) ϩ

If
Let

If ( ) is a 2 ϫ 2 matrix of differentiable functions and
( ) is a 2 ϫ 1 column matrix of differentiable functions, prove the product rule
[ ( ) ( )] ϭ

If

and

22
...

11
Find the inverse of a 3 ϫ 3 diagonal matrix
whose main diagonal entries are all nonzero
...
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...
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...
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...
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...
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...
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...
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...
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...
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...
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)
...
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...
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...
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...
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...

ϭ 1͞(1 Ϫ ) on (Ϫϱ, 1);
ϭ Ϫ1͞( ϩ 1) on (Ϫ1, ϱ);
ϭ 0 on (Ϫϱ, ϱ)
ϭ 3sin 2
ϭ0
no solution

ϩ ;

ϭ ()

ϩ

ϭ 10

cos ϩ 1 sin
4

ϭ

ϭϪ

Ϫ

ϭ 1 and ϭ 2 , 1 and 2 constants
Јϭ 2ϩ 2
The domain is the set of all real numbers
...


ΆϪ, ,
2

ϭ

2

Ͻ0
0

ϭ1
2

(0, ϱ)
ϭ
0

3 3 Ϫ3
2

ϭ Ϫ3,

(Ϫϱ, ϱ)

ϩ
1

9 Ϫ ϩ1
2

3

Ϫ1
2

Ϫ

Ϫ2

Ϫ2
...

2 is semi-stable
...

Ϫ1 is asymptotically stable (attractor); 0 is unstable
(repeller)
...
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...
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...

2

ϭ ͵4

2

-

2

ϭ [Ϫ1 ϩ (1 ϩ 2 )]2
1
(10 )

ϭ tan Ϫ sec ϩ
ϭ 221

Ϫ

( ) ϭ (4 ͞ 2 )

5

ϭ
ϭ

1
4
1
3

Ϫ

2

3

ϭ1
2

Ϫ

Ϫ
3

(

is transient

Ϫ2

ϩ

, (0, ϱ);

Ϫ2 Ϫ

, (0, ϱ);

Ϫ4

ϭ
ϭ

Ϫ1

ϩ

5

ϩ (2 Ϫ )

΂

0

Ϫ

΃

Ϫ2 Ϫ

; (Ϫϱ, ϱ)
; (0, ϱ)

Ϫ1

Ϫ >

Ͻ1
1

2

,

ϩ 1 1␲
2

2

Ϫ2

0Յ Յ1
Ͼ1

) 2,

(erf( ) Ϫ erf(1))

Ϫ( Ϫ4)/

0

ϩ3
2

2

5 2
2

ϩ7 ϭ

Ϫ3 ϩ4 ϭ
ϩ 2 cos Ϫ 1
2

2

ϩ

2

ϩ

; (Ϫϱ, ϱ)

2

ϩ 4

Ϫ2

4

ϭ

not exact
2

ϭ

Ϫ

ϭ4
3

Ϫ 1
ϩ 1

ϩ 4) ϭ 20
)ϭϪ

2

2(

)ϭϪ

2

( )ϭ8

is transient
is transient

ϭ 2 6 ϩ 4, (0, ϱ)
ϭ sin ϩ cos , (Ϫp͞2, p͞2)
( ϩ 1)
ϭ 2 ϩ , (Ϫ1, ϱ); solution is transient
(sec u ϩ tan u) ϭ u Ϫ cos u ϩ , (Ϫp͞2, p͞2)
ϭ Ϫ3 ϩ Ϫ1 Ϫ3 , (0, ϱ); solution is transient
1
ϭ Ϫ1 Ϫ 25 ϩ 76
5
25

Ϫ1

1(

is transient

Ϫ4

ϩ

)

0Յ

Ϫ

2

ln ϩ Ϫ1, (0, ϱ); solution is transient
Ϫ cos , (0, ϱ)
1
5

ϩ

,

ϩ 2Ϫ5 ϩ3 2Ϫ ϭ8
sin Ϫ 3 Ϫ 2 ϩ ln Ϫ ϭ 0
2 2
ϭ 10
cos ϭ
2 2
3
ϩ ϭ
3 2 3ϩ 4ϭ
Ϫ2 3 ϩ 10 3 ϩ ϭ
3

, (Ϫϱ, ϱ);

, (Ϫϱ, ϱ);

3
2

not exact
ϩ2 Ϫ2 3ϭ
Ϫ2
3 3
Ϫ1
Ϫ tan 3 ϭ
Ϫln͉ cos ͉ ϩ cos sin ϭ
4
Ϫ5 3Ϫ ϩ 3ϭ

2

Ϫ

2

Ϫ

,

2 2

4

ϩ

3

2

Ϫ

Ϫ1

ϭ

Ϫ

ϩ

ϭ ϩ

1 3
7

2

ϩ4

1
2

()ϭ

1 3
3

, (Ϫϱ, ϱ)

3

ϭ
ϭ

1

ϩ3
2

0Յ Յ3
Ͼ3

2 Ϫ 1 ϩ 4 Ϫ2 ,
4 2 ln ϩ (1 ϩ 4

ϭ

); (Ϫϱ, ln2)
3 Ϫ 4 Ϫ1
ϭ 2, ϭ Ϫ2, ϭ 2
3 ϩ 4 Ϫ1
ϭ Ϫ1 and ϭ 1 are singular solutions of Problem 21;
ϭ 0 of Problem 22
ϭ1

1

1
2

Ϫ(1ϩ1/ )

)

ϭ Ϫln(2 Ϫ

1
ϭ 1 ϩ 10 tan

Ά

1
Ϫ2
),
2 (1 Ϫ
1 6
Ϫ2
Ϫ 1)
2(

1ϩ

ϭ sin

Ϫ 1; ΂Ϫϱ, Ϫ

ϩ

Ά(

ϭ

4 cos ϭ 2 ϩ sin 2 ϩ
( ϩ 1) Ϫ2 ϩ 2( ϩ 1) Ϫ1 ϭ

Ά

ϭ

ϩ 2 ϩ ln ͉ ͉ ϩ

B3

Ϫ

4

Ϫ

3

ϩ4

4

Ϫ

3

ϩ4

9

12
...
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...
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...
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...
38 lb
( ) ϭ 3 Ϫ 3 Ϫ500 ; : 3 as : ϱ
5
5
5
1
100

()ϭ
()ϭ

ϭ 2
...
1151
ϭ 2
...
6533; ϭ
5 ϭ 0
...
4124
5 ϭ 0
...
5565
5 ϭ 1
...
2696
Euler: 10 ϭ 3
...
9363
RK4: 10 ϭ 42
...
0132
2

10

Ά

1
Ϫ 100

3

()ϭ

true
ϭ

΂

ϩ

0

΃

Ϫ

Ϫ /

as : ϱ

()ϭ

Ϫ50

0 Յ Յ 20
Ͼ 20

Ϫ

:

()ϭ

ϭ sin

3

; ()ϭ1
2

60 Ϫ 60 Ϫ /10,
60( 2 Ϫ 1) Ϫ /10,

΂

ϩ
Ϫ ͞ , a repeller for Ͼ 0, an attractor for Ͻ 0
true

Ϫ50

4

΂

΂
0

0

Ϫ /

΃

Ϫ

ϩ

΃

Ϫ

΃

0

΂ ΃

3

Ϫ

0

4

0

ϩ

0

331 seconds
3

( ) ϭ 0 ( 1Ϫ 2 )
As : ϱ, ( ) : >
...
988 ft

1

ϩ (sin ) ϭ
ϭ ( Ϫ 1) 2 ( Ϫ 3)2
semi-stable for even and unstable for odd;
semi-stable for even and asymptotically stable
for odd
...
9 yr; 10 yr
760; approximately 11 persons/yr
11 h
136
...
00098 0 or approximately 0
...
67° F; approximately 3
...
1 s; approximately 145
...
6 hours prior to the discovery of the body
( ) ϭ 200 Ϫ 170 Ϫ /50

ϭ 2000
2000
()ϭ
; (10) ϭ 1834
1999 ϩ
1,000,000; 5
...

3
0 Ϫ 4

5
13
13
2 0Ϫ5
ϩ
tan Ϫ
ϩ tanϪ1
2
2
2
13
time of extinction is
2
5
2 0Ϫ5
ϭ
tanϪ1
ϩ tanϪ1
13
13
13

΂

΄

()ϭ

()ϭ 1

΂

΄

is 0 Յ Յ 1

576 110 s or 30
...
65 s or 14
...
05 min

29
...
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...
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...
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...
3 ϫ 109 years

)

Ϫ␭1

Ϫ

΂1 Ϫ ␭ ␭ ␭
Ϫ

1

2

1

1

a

(1ϩ )

0 Յ Ͻ 10
10

l1
()ϭ
l1 ϩ l2
l2
()ϭ
l1 ϩ l2

΃

5, 20, 147 days
...


1

a

2

Ϫ␭1

0

()ϭ

4 Ϫ 1 2,
5
20,

()ϭ

ϭ 0 and ϭ 2
( ) ϭ 2 sech2 ( Ϫ 1)
( ) ϭ 2 sech2
1
()ϭ
(Ϫ0
...
01)100
approximately 724 months
approximately 12,839 and 28,630,966

2

,

ϭϪ ϩ1ϩ

Ϫ

()ϭ

΃

1

2

Ά

where r is the weight density of water
Ϫ

΂10 ϩ 1100 Ϫ ΃ Ϫ 1100 Ϫ

͞ ϭ 0
...
99 billion

΃

ϩ

Ϫ

΃

0

2 ϭ Ϫ( ͞ )ln cosh

where

1

1

ϩ

1 2

2) 2

ϩ(

1

ϩ

(0) ϭ Ϫ

0,

1 3

ϭ ()

3) 3

ϩ

ϭ ()

(0) ϭ 0

47
...

The functions satisfy the DE and are linearly independent
on the interval since ( cos 2 , sin 2 ) ϭ 2 2
0;
ϭ 1 cos 2 ϩ 2 sin 2
The functions satisfy the DE and are linearly
independent on the interval since ( 3, 4 ) ϭ 6 0;
ϭ 1 3 ϩ 2 4
...
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...
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...
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...
All Rights Reserved
...
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)
...
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...
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...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...

16

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


2

●

the 20-kg mass
the 20-kg mass; the 50-kg mass
ϭ p, ϭ 0, 1, 2,
...

( ) ϭ 1 cos 2 ϩ 3 sin 2
2
4
213
4
213
4

()ϭ
()ϭ

sin(2 ϩ 0
...


()ϭ

()ϭ
ϭ 0, 1, 2,
...
234799

B

Ϫ

0

1
1

0
2

ϭ 1, 2, 3,
...
51933,

ϩ

1

1
1

sin
Ϫ

24

( )ϭϪ

0Ͻ␤Ͻ

1

( )ϭ

)

247
2

␤ϭ

Ϫ /2

)

0

1 Ϫ ␥2

0 ␥
1 Ϫ ␥2

Ϫ

Ϫ 1 sin 4
2

Ϫ

cos

sin 4 ϩ 4
...
294 s
5
2

0

0

΃ cos 1

(cos 10 ϩ sin 10 ) ϩ 3 ; 3 C
2 2

Ϫ10

ϩ 1

΂

ϭ 0, 1, 2,
...
14 ft below the equilibrium position
...
568 C; 0
...
432 C
ϭ 100 sin ϩ 150 cos
13
13

␲
␲
0
...
3545 ϩ ,
5
5
23
12

or

ϩ ␻ 2 ϭ ␻ 2 ( ),

ϩ 2␭

2

0

ϭ 5 sin (10 Ϫ 0
...
3545 ϩ ,
5

ϭϪ ( Ϫ )Ϫ␤

2

( ) ϭ Ϫcos 2 Ϫ 1 sin 2 ϩ 3 sin 2 ϩ 5 cos 2
8
4
4

cos(2 Ϫ 0
...
721 s
(2 ϩ 1)␲
ϩ 0
...
597 ft
Ј(3) ϭ Ϫ5
...
702 ft/s2
Ϯ8 1 ft/s
3

2

΃

sinh
cosh

B

B

ϭ sin

2

(2 Ϫ 1) ␲
, ϭ 1, 2, 3,
...
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...
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...
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...
;

␲2
,
25

1
1
( ) ϭ Ϫ150 sin 100 ϩ 75 sin 50

ϭ cos

2

␭ ϭ

ϭ 1, 2, 3,
...
;
ϭ sin( ln )
4 4
l ϭ p ,
ϭ 1, 2, 3,
...
;

Ϫ
Ϫ

1

΃

1

ϩ

Ϫ
Ϫ

ϭ sin

΂΃

΄

1ϩ

1
2 1ϩ

Ϫ

ϭ0

ϩ

2

sgn( Ј) ϩ

Љϩ

␲

0

ϭ0

[Ϫ1, 1), ϭ 1
2 2
2
[0, 2 ], ϭ 1
3
3
ϱ
(Ϫ1)
͚ !2
ϭ0
ϱ
Ϫ1

(Ϫ1,1], ϭ 1
(Ϫ5, 15), ϭ 10

ϭ0

( )ϭ

ϭ 0, 1, 2,
...
5 ft
8
Љ ϭ 11 ϩ ( Ј)2
...

When
1,
ϩ

␲
,
50

ϭ

2

΂΃ ΅
1Ϫ

1
1Ϫ

(Ϫ75, 75 ),
32 32

ϭ 75
32

(Ϫ1)

ϱ

͚ 2 ϩ1
ϭ0

͚
ϭ1

(Ϫ1)
( Ϫ 2p) 2
ϩ 1)!

ϱ

͚ (2
ϭ0

ϩ1

Ϫ2
3

3

2
ϩ 15

5

4
Ϫ 315

7

ϩ
...
, (Ϫp>2, p>2)

ϱ

͚(
ϭ3

ϱ

Ϫ 2)

͚ [(
ϭ0

Ϫ2

ϩ 1)

ϩ1

Ϫ

ϱ

ϩ

1Ϫ
When ϭ 1,
( )ϭ

2

2

1

ϩ

͚ [2(
ϭ1

ϩ 1)

ϩ1

ϩ6

Ϫ1]

ϱ

΄

1 1
(
2 2

2

Ϫ

sin
B
B
use at umax, sin 1 > ϭ 1

1
) ϩ ln

2

0

΅

ϩ2

2

ϩ

͚ [(
ϭ1

ϩ 2)( ϩ 1)

1
(5 )
0͚
ϭ0 !
ϱ

ϭ

The paths intersect when Ͻ 1
...
΅
ϩ
...
΅
΄1 ϩ 3 1ؒ 2 ϩ 6 ؒ 5 1ؒ 3 ؒ 2
1
ϩ
ϩ и и и΅
9ؒ8ؒ6ؒ5ؒ3ؒ2
΄ ϩ 4 1ؒ 3 ϩ 7 ؒ 6 1ؒ 4 ؒ 3
2

3

4

3

6

9

2(
8
) ϩ 17

1
΄1 Ϫ 2!
1
΄ Ϫ 3!

0

2(

5
8 ft
4 m
False; there could be an impressed force driving the
system
...
4 lb

0

2(

21,797 cm/s

)ϭ

Ϫ

)ϭ

1

4

ϩ

1
10 ؒ 9 ؒ 7 ؒ 6 ؒ 4 ؒ 3

7

10

΅

ϩиии

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


]

●

1(

)ϭ

0

2( ) ϭ

1

1( ) ϭ

0

΄
1
΄ ϩ 3!
1
΄1 Ϫ 3!
2
΄ Ϫ 4!
1
2!

1Ϫ

2

)ϭ

1

ϩ

3

42
ϩ
6!

2

2

ϱ

1(
2(

)ϭ

0;

2(

)ϭ

[1 ϩ
1
1[ ϩ 2

)ϭ

0

)ϭ

1

1 2
2

͚
ϭ1

21
6!

6

45
7!

7

΅
ϩ и и и΅
9

ϩ

1 3
2

)ϭ

2

1(

)ϭ

2(

)ϭ

2

7

1

ϩ

1 4
6

ϩ

1 4
4

ϭ 1,
3

( )ϭ

]
ϩ и и и]

1

5

7

3

4

ϩ
1

ϭ 5,
2

4

]
6
ϩ и и и]

1
ϩ 120

5

1
ϩ 180

1

ϭ 2,
3

3/2

2

2
΄1 Ϫ 5

΄

1ϩ

2

ϩ

2
7ؒ5ؒ2

23
9 ؒ 7 ؒ 5 ؒ 3!

1ϩ2 Ϫ2

2

ϩ

3

ϭ 0,

ϩ

1
2

ϩ

ϭ 1,

( )ϭ

2

( )ϭ

2/3

1
5ؒ2

2

2

1
6

3

΅

΅

ϩиии

2

2

΅

ϩиии

Ϫ

1
6

1

1
ϩ 1)!

͚ (2
ϭ0

2

3

΅

Ϫиии

Ϫ1

1
͚ (2 ϩ 1)!
ϭ0

1

2

1 sinh

ϩ

2

1

ϱ

ϩ

Ϫ1

2

2 ϩ1

͚ (2
ϭ0

2

)!
1

ϱ

ϩ

Ϫ1

2

͚ (2
ϭ0

2

)!

cosh ]

ϭ0
ϩ

1

2

[

ln Ϫ 1 ϩ 1
2
3

1
ϩ 72

4

2

]

ϩиии

ϭ0
1

( )ϩ

΄

2

ϱ

1(

)ϭ

͚
ϭ0

1
!

1(

) ln ϩ

Ϫ
where

3

ϭ Ϫ1

2

΅

Ϫиии

1
8ؒ5ؒ2

ϩ

5
1
[1 Ϫ 1 ϩ 28 2 Ϫ 21 3 ϩ и и и]
2
7
1/3
[1 Ϫ 1 ϩ 1 2 Ϫ 120 3 ϩ и и и]
2
5

ϩиии
3

΅

ϩиии

ϭ1
3
1

Ϫ

1
ϩ 12

ϭ

2

2ؒ2
22 ؒ 3
ϩ
7
9ؒ7

1
΄1 ϩ 3

2

1
ϭ [

2

23
3 ؒ 3!

3

ϱ

ϭ

΅

ϩиии

2

23 ؒ 4
11 ؒ 9 ؒ 7

ϱ

1

Ϫ
ϩ

΄

5/2

2

( )ϭ

ϭ0
1

1

ϩ
1

ϩиии

ϭ0

ϩ

ϩиии

3

΅

3

2

1
3 ؒ2

ϩ

1
33 ؒ 3!

1
΄1 ϩ 2

4

( Ϫ 1)2
for ϭ 1: ( ) ϭ 5, ( ) ϭ
ϩ1
5( ϩ 1)
for ϭ Ϫ1: ( ) ϭ
, ( )ϭ
Ϫ1
1
1 ϭ 3 , 2 ϭ Ϫ1

( )ϭ

2

2

( )ϭ

23
17 ؒ 9 ؒ 3!

1
΄1 ϩ 3
ϩ

ϭ 0, irregular singular point
ϭ Ϫ3, regular singular point;
ϭ 3, irregular singular point
ϭ 0, 2 , Ϫ2 , regular singular points
ϭ Ϫ3, 2, regular singular points
ϭ 0, irregular singular point;
ϭ Ϫ5, 5, 2, regular singular points

ϭ

1/3

ϩиии

6

22
9ؒ2

2

ϭ0

2

( )ϭ

1

΄

1Ϫ2 ϩ
Ϫ

4

ϩ4

[1 Ϫ
1
1 [ Ϫ 12

3
2, 2

2

23
31 ؒ 23 ؒ 15 ؒ 3!

ϩ
1 3
6

0

Ϫ

΅

΅

2

ϭ8 Ϫ2
( ) ϭ 3 Ϫ 12

΄

2
22
ϩ
15
23 ؒ 15 ؒ 2

1Ϫ

ϩиии

3

1

7/8

ϩ

2

2

0

1

ϭ0

ϩиии

1
΄1 ϩ 4 Ϫ 4 ؒ74! ϩ 23ؒ ؒ6!7 Ϫ и и и΅
8
1
14
34 ؒ 14
( )ϭ ΄ Ϫ
ϩ
Ϫ
Ϫ и и и΅
6
2 ؒ 5!
4 ؒ 7!
1
1
1
( ) ϭ Ϫ2 ΄1 ϩ
ϩ
ϩ
ϩ и и и΅ ϩ 6
2!
3!
4!

1(

2

( )ϭ

1

1 3
6

ϭ 7,
8

1

Ϫиии

72 ؒ 42
6
Ϫ
9!

10

ϩ

2

ϩ

5 ؒ2
ϩ
7!

8 ؒ5 ؒ2
Ϫ
10!
1(

5

Ϫ

2

4

2

4

5
5!

3

2

2(

3
4!

Ϫ

1
3 ؒ 3!

3

1(

ϩ

΂

) Ϫ ϩ
1
4 ؒ 4!

4

1
4

2

΃΅

Ϫиии

ϭ

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


(Ϫ1)
1␭
ϩ 1)!

ϱ

1(

)ϭ

(Ϫ1)
͚ (2 )! 1␭
ϭ0

(

͚ (2
ϭ0

΂ 1␭ ΃ ϩ

ϱ

2(

)ϭ

ϭ

Ϫ1

sin

1

(

)2
)

2

ϭ

cos 1␭

(

)

΂ 1␭ ΃

ϭ

cos

2

sin 1␭
1␭

●

(

ϭ 0 is an ordinary point
( )ϭ

)

1
΄1 Ϫ 3 ϩ 3 1ؒ 2!
1
1
ϩ ΄ Ϫ
ϩ
4
4ؒ7
3

0

4

1

1
4 ؒ 7 ؒ 10

Ϫ

1
32 ؒ 2!

ϩ
ϭ 1 1/3( ) ϩ 2 Ϫ1/3( )
ϭ 1 5/2 ( ) ϩ 2 Ϫ5/2 ( )
ϭ 1 0( ) ϩ 2 0( )
ϭ 1 2 (3 ) ϩ 2 2 (3 )
ϭ 1 2/3(5 ) ϩ 2 Ϫ2/3 (5 )
ϭ 1 Ϫ1/2 1/2(a ) ϩ 2 Ϫ1/2 Ϫ1/2(a )
ϭ Ϫ1/2 [ 1 1 (4 1/2) ϩ 2 1 (4 1/2)]
ϭ [ 1 1( ) ϩ 2 1( )]
ϭ 1/2 [ 1 3/2( ) ϩ 2 3/2( )
ϭ Ϫ1 1 1/2 1 2 ϩ 2 Ϫ1/2 1 2
2
2

[

ϭ
ϭ

ϭ

Ϫ1/2(

)]

[ 1 1/2(1 2) ϩ 2 Ϫ1/2(1 2)]
8
8
1 2
Ϫ3/2
Ϫ3/2
sin(8 ) ϩ 2
cos(1 2)
1
8
2
1/2
3/2
) ϩ 2 1/2 Ϫ1/3(32 a 3/2)
1
1/3( 3 a

ϩ 16
5

5

1

[

2(

)ϭ

1(

)ϭ

2(

)ϭ

[1 Ϫ
3
0 [1 ϩ 2
1
1[ ϩ 2
2

ϩ1
6

2

2

ϩ1
2

3

1
4

ϩ

2(

[

)ϭ

2

[

[1 ϩ

ϩ

2

ϩ1
3

( )ϭ31Ϫ
Ϫ2
1
6␲

[

Ϫ1
2

3

1 2
2
4

ϩ1
8

]

ϩиии

]
3
ϩ 5 4 ϩ и и и]
8
4
ϩ и и и]
1
Ϫ 90

ϭ 3, 2 ϭ 0
1
3
1 ϩ 1 ϩ 20
1( ) ϭ 1
4

1

Ϫ

2

3

ϩ

]

1
Ϫ 15
5

6

1
Ϫ 48

Ϫ

4

1

9

1

2

Ϫ

1
3

3

΅

ϩиии

Ϫ

2

1

ϩ

Ϫ

2

7

Ϫ1
1
ϩ2 ϩ2

2

48
5

Ϫ1
( ϩ 1)2
4
10
Ϫ
2
6

΅

ϩиии

Ϫ

ϩ

6
3

ϩ

2
3
2

ϩ

1

Ϫ
2

ϭ

6

Ϫ

2

3

1
Ϫ4
15
8
Ϫ 2
3
ϩ9

1

2
1
ϩ
Ϫ2
Ϫ4

ϩ

ϩ

3

ϩ

Ϫ

and linearity to show that
}ϭ

1
1
Ϫ
2( Ϫ 2) 2
1 3
630

Ϫ

2

ᏸ {sinh

2

1
33 ؒ 3!

1

Use sinh

False
[Ϫ1, 1]
2 2
2
( Ϫ 1) Љ ϩ Ј ϩ ϭ 0
1
1 ϭ 2, 2 ϭ 0
1
1/2
1 Ϫ 1 ϩ 30
1( ) ϭ 1
3

Ϫ

2

l1 ϭ 2, l 2 ϭ 12, l 3 ϭ 30
3

ϩиии ϩ

1

Ϫ

9

3

7

2

ϭ
2( ), 3( ), 4( ), and 5( ) are given in the text,
1
6
Ϫ 315 4 ϩ 105 2 Ϫ 5),
6 ( ) ϭ 16 (231
1
7
Ϫ 693 5 ϩ 315 3 Ϫ 35 )
7 ( ) ϭ 16 (429
ϭ Ϫ4

6

1
3 ؒ 3!

Ϫ

΅ ΄5
2

10

1
( Ϫ 4)2

Ϫ1/2

[ 1 1/2( ) ϩ 2
1 sin ϩ 2 cos

Ϫ

1 ϩ Ϫ␲
2
ϩ1
1
1
1
Ϫ 2ϩ 2

( )]

1/2

ϭ

( )

2

6

2

4 cos 5 ϩ (sin 5)
2
ϩ 16


...
3
1
5

2

0
...
6

Ϫ0
...
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...
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...
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...
7,
0

Ϫ

0

60

ᐁ

Ϫ0
...
1 at

max

3

sin
Ϫ
Ϫ

ᐁ ( Ϫ 3)

΂ Ϫ 32␲΃
10
3␲
3␲
ϩ
cos Ϫ ΃ ᐁ΂ Ϫ ΃
101 ΂
2
2
1
3␲
3␲
ϩ
sin΂ Ϫ ΃ ᐁ΂ Ϫ ΃
101
2
2
Ϫ

1
( Ϫ 10)2

Ϫ10

Ϫ5( Ϫ3)

ϩ

0

4

24

΃ ᐁ΂
4

Ϫ

2

0

Ϫ

4
...
5 Ϫ (230 Ϫ 57
...
All Rights Reserved
...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...
All Rights Reserved
...
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s)
...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...
Note
that 3 ϭ 2 1 ϩ 2
...
54 s
ϭ

Јϭ

΂

΃

3
4

Ϫ5
8

΂
΂

Ϫ3
Јϭ
6
10

ϭ

΃

4
Ϫ1
4

1 Ϫ1
Јϭ 2
1
1
1
where

, where

ϭ

΂΃

ϭ3

΂΃
΃ ΂ ΃ ΂΃ ΂΃

Ϫ9
0
3
1
Ϫ1
1

, where

0
ϩ Ϫ3 2 ϩ
2

1

ϭ

0 ϩ
Ϫ

Ϫ1
0 ,
2

ϭ

΂΃

ϭ

1

1

1

1

ϭ4 ϩ2 ϩ
ϭϪ ϩ3 Ϫ
ϭ

Ϫ

ϩ2 ϩ

ϩ
Ϫ

ϭ3 Ϫ4 ϩ ϩ2

Ϫ3
Ϫ

ϭ Ϫ2 ϩ 5 ϩ 6 ϩ 2

ϭ

ϩ
Ϫ

1

Ϫ

Yes; ( 1, 2 ) ϭ Ϫ2 Ϫ8
0 implies that
are linearly independent on (Ϫϱ, ϱ)
...
All Rights Reserved
...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...

3
1
0

ϭ

1

ϩ

1
1
0

2

1
΂Ϫ1΃

ϩ2

2

Ϫ

2

΂΃

/2

΂2΃
1

ϩ

ϩ

΂΃
4
1

3

2

ϩ

1
΂Ϫ1΃

΂ ΃
10
3

2

΂1΃
1
2

2

1

ϭ

1

ϩ

΂Ϫ1΃
1

Ϫ

1
΂Ϫ1΃

Ϫ2

ϩ

ϩ

2

΂Ϫ3΃
1
2

΂1΃
1

΂ ΃ ϩ ΂Ϫ2΃
1
4
Ϫ1
4

3
4

ϩ
4

΂Ϫ ΃
1
4
3
4

5

3
2
7
2

Ϫ

4

2

΂Ϫ9΃
6

΂11΃ Ϫ ΂15΃
11
10
Ϫ

΂΃

/2

΂3΃
3

ϩ

΂4΃
2

3 /2

Ϫ3

13
2
13
4

Ϫ

΂΃
15
2
9
4

ϩ

΂ ΃ ΂΃

ϩ

΂Ϫ2΃

Ϫ12
0

Ϫ

1
2

4
3
4
3

Ϫ

sin
΂cos ΃ ϩ ΂Ϫcos ΃ ϩ ΂cos ΃
sin
sin
2

΂Ϫsin ΃ ln͉ cos
cos
΂cos ΃
sin

ϩ

2

͉

sin
΂Ϫcos ΃

ϩ

΂cos ΃
sin

cos
sin
cos
΂Ϫsin ΃ ϩ ΂cos ΃ ϩ ΂Ϫsin ΃
2

Ϫsin
sin
΂sin tan ΃ Ϫ ΂cos ΃ ln͉ cos

͉

sin
΂2cos ΃

sin
΂3 cos ΃

ϩ

΂Ϫcos ΃
sin

2

΂2 cos ΃
Ϫsin

ln͉ sin ͉ ϩ

1
2

ϩ

΂2 cos ΃
Ϫsin

΂ ΃ ΂΃ ΂΃
΂
΃
1
Ϫ1 ϩ
0

1
1
0

2

2

3
2

0
0
1

3

΂Ϫ2΃
2

4

ln͉ cos ͉

ϩ

3

2

1 2 3
2

΂Ϫ1΃
3

ϩ

ϩ

2 1
2

Ϫ1 2 ϩ 1 2
4
2
ϩ Ϫ ϩ1 2 ϩ1
4
2
ϭ

2

΂Ϫ΃

ϩ

΂΃ ΂΃
1
2
2

3

ϩ

΂ ΃
Ϫ2
1

΂ ΃
55
36
Ϫ19
4

ϩ

ϩ

΂Ϫ4΃
6

΂1΃ ϩ ΂3΃
1
2
2
1

7

ϭ
2

΂2΃
2

2

ϩ

΂ ΃ ϭ 2 ΂1΃
3
1

Ϫ2

΂Ϫ1΃
1
ϩ

2

ϩ

2

ϩ

΂ ΃

6
3
29 Ϫ1

Ϫ12

Ϫ

ϩ

΂2΃
0

4

΂ ΃

4 19
cos
29 42

΂ ΃

4 83
sin
29 69

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


/2

●

ϭ

΂0 ΃;

ϭ

΂

ϭ

0

2

΃

ϩ1
Ϫ2

Ϫ2

΂΃

ϩ

Ϫ2 ϩ 1

΂΃
0
1

2

΃

ϩ

ϩ1 ϩ
Ϫ2

2

Ϫ2

3

ϭ

1

΂1΃
0

ϭ

ϩ

΂0΃
1

4

΃

2

3

Ϫ2 ϩ 1

΂ ΃ ΂ ΃ ΂

ϭ

ϭ

ϭ

ϭ

΂

3 2
2

Ϫ1
2
Ϫ 2 ϩ

΂

Ϫ2

3 2
2

1

Ϫ1
2
Ϫ 2 ϩ

΂ ΃

3

3
Ϫ2

΂

2

3 2
4
Ϫ1 2
2

2

Ϫ2
Ϫ2

΃

ϩ

4

2

ϩ3

΂1 ϩ 3 ΃

1

ϭ

3 3
2
1 3 3
2

ϭ

3

΂

Ϫ
Ϫ

΂1΃
1

3

2

1 5
2
3 5
2

ϩ

2

Ϫ3
4
ϩ3
2

΃

΃

4

2

΂1΃
3

΂

΂4΃
1

4

ϩ

ϩ

3

΂ ΃

Ϫ3

16
11
΂Ϫ4΃ ϩ ΂Ϫ1΃

2

΂sin sin cos ΃ ln͉ csc
ϩ

΃;
΃

or

1

1

Ϫ cot ͉

΂ ΃ ΂ ΃ ΂΃
Ϫ1
1 ϩ
0

Ϫ1
0 ϩ
1

2

3

1
1
1

3

ϭ 1
...
1)2
ϭ 0
...
02

0
...
0244
Actual value is (0
...
2214
...
0214
...
05, 2 ϭ 1
...

Error with ϭ 0
...
0214
...
05
is 0
...


΃;

΂1Ϫ93 ΃
Ϫ

ϩ

2

΂΃

7
12
Ϫ16

΂cos cos sin ΃ ϩ ΂sin sin cos ΃ Ϫ ΂1΃
Ϫ
ϩ
1

Љ( )

΂ ΃

2

Ϫ2

΂

1
Ϫ2

ϩ

Ϫ2

3 2
Ϫ 3 Ϫ2
4
4
Ϫ1 2 ϩ 3 Ϫ2
2
2

Ϫ9 2
2
Ϫ3

2

ϭ

ϩ

2

4

for ϭ 0
...
0801; for ϭ 0
...
0592
for ϭ 0
...
5470; for ϭ 0
...
5465
for ϭ 0
...
4053; for ϭ 0
...
4054
for ϭ 0
...
5503; for ϭ 0
...
5495
for ϭ 0
...
3260; for ϭ 0
...
3315
for ϭ 0
...
8254; for ϭ 0
...
8840;
at ϭ 0
...
5) ϭ 3
...
8
2

or

Љ( )

2

ϭ5

Ϫ2

(0
...
025
2

Ϫ2

Յ 0
...
1
...
1) ϭ 0
...
Error is 0
...

If ϭ 0
...
8125
...
1 is 0
...
Error with ϭ 0
...
0109
...


2

ϭ

1

ϭ

1

΂ ΃

΄΂ ΃ ΂ ΃ ΅
cos
sin
΂Ϫsin22 ΃ ϩ ΂cos 22 ΃
1
Ϫ1

ϩ

2

1
Ϫ1
2

0
ϩ
1

Յ 19(0
...
19
2
If ϭ 0
...
8207
...
05, 10 ϭ 1
...

Error with ϭ 0
...
2325
...
1109
...
05

Copyright 2012 Cengage Learning
...
May not be copied, scanned, or duplicated, in whole or in part
...
Editorial review has
deemed that any suppressed content does not materially affect the overall learning experience
...


●

2
1

...
9123, 1 ϭ 4
...
9123, 2 ϭ 4
...
4179, 1 ϭ Ϫ2
...
4173, 2 ϭ Ϫ2
...
1)2
ϭ 0
...
1, 5 ϭ 0
...
If ϭ 0
...
4124
...
1 is 0
...
Error with ϭ 0
...
0069
...
9078; actual value is (0
...
9082
ϭ 2
...
5463
5 ϭ 0
...
5493
5 ϭ 1
...
7130
5
5

tanh
; (5) ϭ 35
...
1, 4 ϭ 903
...
05, 8 ϭ 1
...
82341667
()ϭ

5

(5)

( )

5!

5

ϭ 40

Ϫ2

5!

Յ 40

2(0)

(0
...
333 ϫ 10Ϫ6
Actual value is (0
...
8234134413
...
225 ϫ 10Ϫ6 Յ 3
...

If ϭ 0
...
82341363
...
1 is 3
...
Error with
ϭ 0
...
854 ϫ 10Ϫ7
...
6774, 2 ϭ Ϫ2
...
3226
1 ϭ Ϫ0
...
3356, 3 ϭ Ϫ0
...
2167
1 ϭ 3
...
6306, 3 ϭ 3
...
2355,
5 ϭ 2
...
8842, 2 ϭ 2
...
2064, 4 ϭ 1
...
0681, 6 ϭ 0
...
2913
1 ϭ 0
...
5097, 3 ϭ 0
...
9471,
5 ϭ 1
...
3353, 7 ϭ 1
...
6855,
9 ϭ 1
...
3492, 2 ϭ 0
...
1363, 4 ϭ 1
...
2118, 6 ϭ 2
...
8490
0 ϭ Ϫ2
...
0755, 2 ϭ Ϫ1
...
6126, 4 ϭ Ϫ1
...
1)5
24
Յ 24
ϭ 2
...
1, 5 ϭ 0
...

From calculation with ϭ 0
...
40546511
...
2) ϭ 1
...
4) ϭ 1
...
6) ϭ 1
...
8) ϭ 1
...

4 ϭ 0
...
2, 5 ϭ 1
...
1, 10 ϭ 1
...
2, 5 ϭ 0
...
1, 10 ϭ 0
...
2) ϭ Ϫ1
...
6800
1 ϭ Ϫ1
...
4919
1 ϭ 1
...
4640
1 ϭ 8
...
4199;
2 ϭ 8
...
4199

Comparison of numerical methods with

1
...
20
1
...
40
1
...
1386
2
...
5136
2
...
0201

2
...
3439
2
...
8246
3
...
1556
2
...
5695
2
...
1197

Comparison of numerical methods with

1
...
20
1
...
40
1
...
1469
2
...
5409
2
...
0690

2
...
3450
2
...
8269
3
...
6000
0
...
8283
0
...
0921

0
...
7191
0
...
9752
1
...
05:

2
...
3454
2
...
8278
3
...
60
0
...
80
0
...
00

ϭ 0
...
1:

0
...
7194
0
...
9757
1
...
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...
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...
Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it
...
05:
nonsingular;

0
...
70
0
...
90
1
...
6024
0
...
8356
0
...
1044

0
...
7193
0
...
9755
1
...
6049
0
...
8431
0
...
1169

ϭ 0
...
2) 3
...
1: (0
...
2) 1
...
2) 1
...
297

11
΂2 Ϫ1΃
2
2
28
΂12 Ϫ12΃

΂Ϫ6
14

΂Ϫ11
17
19
΂Ϫ30

27
΂Ϫ32 Ϫ1΃
Ϫ4
6
΂19 22΃
3

΃
Ϫ18
31΃
6
Ϫ22

΂
΂Ϫ4
8

΂ ΃
΂0 0΃
0 0
9
3

24
8

3
Ϫ6

΂

4
8
10

180

8
16
20

7
΂10 38΃
75
΂Ϫ14΃
1
΂Ϫ38΃
Ϫ2

΃

1
Ϫ19

ϭ4

΂
΂

΃

7
΂10 38΃
75

4

2

Ϫ 12

Ϫ␲ sin ␲
6

2

΂

΃

1 8
4

Ϫ1
4
4

΃

0
6

΃

Ϫ1
4

(1/␲) sin ␲
3
Ϫ
ϭ 3, ϭ 1, ϭ Ϫ5
ϭ 2 ϩ 4 , ϭ Ϫ5 Ϫ , ϭ
ϭ Ϫ1, ϭ 3, ϭ 7
2
2
2
1 ϭ 1, 2 ϭ 0, 3 ϭ 2, 4 ϭ 0
2

΂
΂

1
3

2
3
Ϫ1
3
Ϫ2
3

΃

1
3
Ϫ2
3

0

5
ϭ
2
Ϫ1

΃

6
2
Ϫ1
Ϫ2
3

Ϫ3
Ϫ1
1
Ϫ1
6

1
3
Ϫ1
3

7
6
Ϫ4
3
1
3
1
2

1
3
Ϫ1
3
1
2

1

␭1 ϭ 6, ␭2 ϭ 1,

1

␭1 ϭ ␭2 ϭ Ϫ4,

1

΃

΂2΃,
7
1
ϭ΂ ΃
Ϫ4
ϭ

2

ϭ

΂1΃
1

␭1 ϭ 0, ␭2 ϭ 4, ␭3 ϭ Ϫ4,

1

΂ ΃ ΂΃ ΂΃

9
ϭ 45 ,
25

2

1
ϭ 1 ,
1

3

1
ϭ 9
1

␭1 ϭ ␭2 ϭ ␭3 ϭ Ϫ2,

singular
nonsingular;

Ϫ3

Ϫ

΂

΂΃

΃

΂2΃
1

Ϫ1
7
Ϫ5

Ϫ

Ϫ1
2
1
Ϫ1
ϭ
0
Ϫ1
2
6
12
Ϫ5

Ϫ 4
2 Ϫ

΃

Ϫ2
5
Ϫ1

Ϫ

1
΂Ϫ1΃

1 4
4

Ϫ1

΃
Ϫ5
10΃

10
20
25

΂

0
Ϫ1
ϭ 0

8
Ϫ16

΂

Ϫ2
1
Ϫ13
9
8

΂ ΃

4

81␲
Ϫ
15

ϭϪ

1
3 4
2 3 Ϫ4 Ϫ

Ϫ5
ϭ Ϫ2
7

3
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