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Ordinary Differential Equations
A differential equation is an equation for a function that relates the values of the function
to the values of its derivatives
...
These equations may include
the unknown function as well as known functions of the same variable
...
Generally
speaking, a differential equation is a representation of a physical phenomenon, where
the derivatives correspond to the "rates of change" of the unknown function with respect
to the variable
...
g
...
For example:

d2 f
df
+
+  2 f = Ae it
2
dt
dt
dy
− ky = 0
dx

On the other hand, partial differential equations (PDEs) contain functions of multiple
variables and partial derivatives, for example:
𝜕𝑓(𝑥, 𝑦)
= 2𝑥𝑦 + 𝑥 2
𝜕𝑥𝜕𝑦
We will cover only ODEs in this unit
...

Solving an ODE means finding an equation with no derivatives that satisfies the given
ODE
...
It
is important to be able to identify the type of ODE we are dealing with before we attempt
to solve it
...
An equation containing only first derivatives is a first-order
differential equation; an equation containing the second derivative is a second-order
differential equation, and so on
...
The simplest ODE
𝑑𝑦

can be integrated directly by finding antiderivatives, for example 𝑑𝑥 = 2𝑥
...


First-order ODEs
The standard form for first-order DEs can be put into the form:
dy
= f ( x, y )
dx

Which is equivalent to 𝑑𝑦 = 𝑓(𝑥, 𝑦)𝑑𝑥 which gives 𝑑𝑦 − 𝑓(𝑥, 𝑦)𝑑𝑥 = 0
M ( x, y )dx + N ( x, y )dy = 0

The general solution of a first order DE contains one arbitrary constant
...
We will consider only few most practically
important special cases
...
+ a1 ( x) + a0 ( x) y = F ( x)
dx
dx
dx
That means all the derivatives (including the zeroth one, i
...
y) are present only to the
first power and there no terms with mixed order (e
...


𝑑2 𝑦
𝑑𝑥 2

 𝑦 2)
...
But the one below is linear of order 2
...
Keeping the above notation, the equation is
homogeneous if 𝑓(𝑥) = 0
...
+ a1 ( x) + a0 ( x) y = 0
n −1
n
n −1
dx
dx
dx

is homogeneous
...

The Solution of a differential equation is the function, e
...
f (t ), y ( x), ( x) which satisfies
the equation identically
...
To assign values to these constants, n additional conditions (initial
conditions, boundary conditions) are needed
...


Types of solutions of DEs
1
...


Example 1
Solve the 𝑦 ′ − 𝑘𝑦 = 0
...


Solution
This is a nasty looking non-linear differential equation, but it is actually separable:

(

) dy
= 2 x + xy
dx

(

) dy
= x(2 + y )
dx

y 1 + x2
y 1 + x2

2

2

y
x
dy =
dx
2
(2 + y )
1 + x2

(

)

Integrating both sides:
1
1
ln(2 + y 2 ) = ln(1 + x 2 ) + C
2
2

and rearranging:
1
2 + 𝑦2
ln (
)=𝐶
2
1 + 𝑥2
ln (

2 + 𝑦2
) = 2𝐶
1 + 𝑥2

2 + 𝑦2
= 𝑒 2𝑐 = 𝑘
1 + 𝑥2
2 + 𝑦 2 = 𝑘(1 + 𝑥 2 )
𝑦 2 = 𝑘(1 + 𝑥 2 ) − 2
𝑦 = ±√𝑘(1 + 𝑥 2 ) − 2
2 + 𝑦 2 − 𝑘(1 + 𝑥 2 ) = 0

is the general solution
...
What happened to the one on the left? The answer is
quite straightforward
...


Example 3
Solve the following differential equation
...


Plug this into the general solution and then solve to get an explicit solution
...
Solve
6

3𝑥 2 + 4𝑥 − 4
𝑦 =
, 𝑦(1) = 3
2𝑦 − 4
′

(answer: 𝑦 = 2 + √𝑥 3 + 2𝑥 2 − 4𝑥 + 2)

2
...

𝑑𝑥
𝑑𝑦
= 5𝑥 − 5𝑥 2
𝑑𝑥
𝑑𝑦 = 5𝑥 − 5𝑥 2 𝑑𝑥
3
...

3= 0+𝑐
C=3
(Answer: 𝑦 =

−7
2

𝑥 2 + 3)
...
Solve the ODE:
𝑦 ′ = 𝑒 −𝑦 (2𝑥 − 4),

𝑦(5) = 0

(ANSWER: 𝑦(𝑥) = 𝑙𝑛(𝑥 2 − 4𝑥 − 4)

5
...

(Answer: 𝑦 = 3𝑥 2 − 2𝑥 + 3)

6
...


2
...

𝑑𝑦

An equation of the form 𝑃 𝑑𝑥 = 𝑄, where P and Q are functions of both x and y of the same degree
throughout, is said to be homogeneous in y and x
...
However, 𝑓 (𝑥, 𝑦) =

𝑥2 − 𝑦
2𝑥 2 +𝑦2

homogeneous since the term in y in the numerator is of degree 1 and the other three terms are of
degree 2
...
Then 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
Substituting
𝑣+𝑥
𝑣+𝑥

𝑑𝑣 3𝑥(𝑣𝑥) + (𝑣𝑥)2
=
𝑑𝑥
𝑥2

𝑑𝑣 3𝑣𝑥 2 + 𝑣 2 𝑥 2
=
= 3𝑣 + 𝑣 2
𝑑𝑥
𝑥2
𝑣+𝑥
𝑥

𝑑𝑣
= 3𝑣 + 𝑣 2
𝑑𝑥

𝑑𝑣
= 3𝑣 + 𝑣 2 − 𝑣
𝑑𝑥
𝑥

𝑣2

𝑑𝑣
= 𝑣 2 + 2𝑣
𝑑𝑥
1
1
𝑑𝑣 = 𝑑𝑥
+ 2𝑣
𝑥

Partial fractions
1
1
𝑑𝑣 = 𝑑𝑥
𝑣(𝑣 + 2)
𝑥
Integrating both sides
𝑣
= 𝑘 2𝑥2
𝑣+2
But 𝑣 =

𝑦
𝑥

𝑦
𝑥 = 𝑘 2𝑥2
𝑦
+
𝑥 2
𝑦
𝑦
= 𝑘 2 𝑥 2 ( + 2)
𝑥
𝑥

8

is not

Procedure

Example
Solve the DE: 𝑦 − 𝑥 = 𝑥

𝑑𝑦
,
𝑑𝑥

given 𝑦(1) = 2
...
Solve 𝑥

𝑑𝑦
𝑑𝑥

=

𝑥 2 +𝑦 2
𝑦

at 𝑦(1) = 4
...

2𝑦−𝑥 𝑑𝑦

2
...

Answer 𝑥 2 + 𝑥𝑦 − 𝑦 2 = 1

Task

10

3
...
These DEs are non-separable or non-homogenous, for example,
𝑑𝑦
+ 𝑦 = 2𝑥 2
𝑑𝑥
𝑑𝑦 1
+
...

(ii) Determine the integrating factor 𝑒 ∫ 𝑃𝑑𝑥 and simplify
...
F
...

Given boundary conditions, the particular solution may be determined
...


Solution
𝑑𝑦
𝑑𝑥

1

2

+ (𝑥) 𝑦 = 𝑥

The integrating factor is
1

𝑒 ∫𝑥𝑑𝑥 = 𝑒 ln 𝑥 = 𝑥
...
F
...

𝑑𝑥

𝑑𝑦

𝑑

𝑑𝑦

NB: 𝑑𝑥 (𝑥𝑦) = 𝑥 𝑑𝑥 + 𝑦
...

𝑥

Notice that this differential equation is not separable because it’s impossible to factor
the expression for 𝑦’ as a function of 𝑥 times a function of 𝑦
...

It turns out that every first-order linear differential equation can be solved in a similar
fashion by multiplying both sides by a suitable function 𝜆 called an integrating factor
...

3

3

∫ 2 × 3𝑥 2 𝑒 𝑥 𝑑𝑥 = 2 ∫ 𝑒 𝑢 𝑑𝑢 = 2𝑒 𝑢 + 𝑐 = 2𝑒 𝑥 + 𝑐
Example
Find the solution of the initial-value problem 𝑥 2 𝑦’ + 𝑥𝑦 = 1, 𝑥 > 0 and 𝑦(1) = 2
...
𝑥 𝑑𝑥 + 𝑦 = 1
...
𝑑𝑥 − 2(𝑥 + 1)3 = 𝑥+1 𝑦 at 𝑦(0) = 0
...


Multiplying by 𝜆 we get
(

1
𝑑𝑦
1
4
1
)
−(
)
𝑦 = 2(
)(𝑥 + 1)3
4
4
(𝑥 + 1) 𝑑𝑥
(𝑥 + 1) 𝑥 + 1
(𝑥 + 1)4
𝑑
𝑦
2
(
)=
4
𝑑𝑥 (𝑥 + 1)
𝑥+1

Integrating both sides (LHS is 𝜆𝑦) ALWAYS)
𝑦
2
=
∫
𝑑𝑥
(𝑥 + 1)4
𝑥+1
𝑦
= 2 ln|𝑥 + 1| + 𝑐
(𝑥 + 1)4
14

Applying initial conditions
0=0+c
C=0
𝑦
= 2 ln|𝑥 + 1|
(𝑥 + 1)4
𝑦 = 2 ln|𝑥 + 1| (𝑥 + 1)4
...


Applications of First-Order Differential Equations
Exponential Growth and decay problems
Let N(t) denote the amount of substance {or population) that is either growing or
decaying
...
Then for growth
implies that

𝑑𝑁
𝑑𝑡

𝑑𝑁
𝑑𝑡

= 𝑘𝑁, where k is the constant of proportionality
...

Example
The population of a certain country is known to increase at a rate proportional to the
number of people presently living in the country

𝑑𝑁
𝑑𝑡


...

Solution

15

𝑑𝑁
= 𝑘𝑁
𝑑𝑡
To solve this, we have to integrate both sides
∫

𝑑𝑁
= 𝑘 ∫ 1 𝑑𝑡
𝑁

ln 𝑁 = 𝑘
...
𝑒 𝑐
𝑁(𝑡) = 𝐴𝑒 𝑘𝑡 where A=𝑒 𝑐
When t=0, 𝑁(0) = 𝐴𝑒 𝑘
...

ln (2) = ln (𝑒 2𝑘 )
ln 2 = 2𝑘
𝑘=

1
ln 2
2

16

Modelling Continuous Compound Interest
When interest on an account is compounded continuously, the account grows at a rate that is directly
proportional to the size of the account
...
This can be described by the differential equation

where A is the balance in the account after t years and r is the annual interest rate in decimal form
...
𝐴 = 𝑃(1 + 𝑖)𝑛 where 𝑖 interest rate per annum
1
...


1
...
Solve the initial-condition problem

, A(0) = 1000
...
07t
...
2
...

The investment will grow to approximately $8,166 in 30 years
...
07)30 = $7 612
...
035)60 = $7878
...

2
...
5%?

𝑁(𝑡) = 𝑁0 𝑒 𝑘𝑡 = 5000𝑒 0
...

3
...
5%
...
045𝑡
It will take about 6
...


4
...
Consider the population of bacteria
...
02𝑡 , where 𝑡 is measured in minutes
...
The population of Goth island at the end of the year 2000 was 500
...
What will the population be at the
end of the year 2050?
Solution
𝑁(𝑡) = 𝑁0 𝑒 𝑘𝑡 = 500𝑒 0
...
58 ≈ 74 207
3
...
How much does
the student need to invest today to have $1 million when she retires at age 65?
4
...
Assuming no additional deposits or withdrawals, how much will be in
the account after seven years if the interest rate is a constant 8
...
25 percent for the last three years?

Answers
1
...


3
...


19

Second-order Linear ODEs
Linear ordinary differential equation of order 𝑛, in the dependent variable 𝑦 and the
independent variable 𝑥, is an equation that is in, or can be expressed in, the form

where 𝑎0 ≠ 0
...
This is a hyperbola
...
𝑦 = 2 sin 𝑥 +
2
...

(3) that no transcendental (trigonometric, logarithmic, exponential) functions of 𝑦 and/or
its derivatives occur
...
In each case y
is the dependent variable
...

𝑑𝑦
𝑑𝑥

= 𝑘𝑥 is a first-order DE and first degree DE
...

𝑑2 𝑦

2

(𝑑𝑥 2 ) = 3𝑥
...


A nonlinear ordinary differential equation is an ordinary differential equation that is not
linear
...
Equation 2 owes its nonlinearity to the presence of the term
𝑑𝑦

5(𝑑𝑥 )3, which involves the third power/degree of the first derivative
...


21

Linear ordinary differential equations are further classified according to the nature of the
coefficients of the dependent variables and their derivatives
...

Linear 2nd order DEs with constant coefficients
What we have stated so far applies generally to any linear ODE
...
That means that all
the functions an ( x), an−1 ( x),
...
a1 , a0
...
+ a1
+ a0 y = F ( x )
n −1
n
n −1
dx
dx
dx

Complementary (homogeneous) solution
...
+ a1
+ a0 y = 0
n −1
n
n−1
dx
dx
dx

is always found following a simple general rule:
Use a trial solution of the form
y = e x

where 𝜆 is some constant, which we do not know, but will find by substituting this trial
solution to the above homogeneous equation
...
+ a1 + a0 = 0

which in general will have n roots
...
If the roots 1 , 2 ,
...
+ Cn en x

where C1 , C2 ,
...


22

2
...
g
...
you
must multiply the corresponding exponential with a polynomial of the order k1 − 1 , k 2 − 1
etc
...
The solution then looks like this:

(

)

(

)

y = C1 + C2 x +
...
Dk2 x k2 −1 e2 x +
...
e
...

At this level, we restrict our attention to second-order linear homogeneous differential
equations with constant coefficients only
...

In this section we study the case where 𝐺(𝑥) = 0, for all x, in the ODE
...
Or we can rewrite:

d2y
dy
a 2 + b + cy = 0
dx
dx

(basic form)

where a, b, and c are constants
...

Thus, the general solution of the 2nd – order linear differential equation

23

d2y
dy
a 2 + b + cy = 0
dx
dx

(where a, b, and c are constants)

depends on the roots of the auxiliary quadratic equation

am2 + bm + c = 0

such that

i) if 𝑏 2 − 4𝑎𝑐 ≥ 0 (2 distinct real roots m1 and m2)
then

y = c1e m1 x + c2 e m2 x

ii) if 𝑏 2 − 4𝑎𝑐 = 0
then

(1 real repeated root m1=m2=m)

y = c1emx + c2 xemx

iii) if 𝑏 2 − 4𝑎𝑐 < 0

(2 complex roots

m1 =  + i and m2 =  − i )

x
y
=
e
(c1 cos x + c2 sin x)
then

Note:
Each of these solutions contains two arbitrary constants c1 and c2 since solution of 2nd
– order differential involves two integrations
...


d2y
dy
2 2 + 5 − 3y = 0
dx
dx
Auxiliary eqn
...


Auxiliary eqn
...
The general solution will be:
y = (C1 + C2 x )e 4 x

Example

25

Task
We could verify that this is indeed a solution by differentiating and substituting into the
differential equation
...

Solution

Case III: 𝒃𝟐 − 𝟒𝒂𝒄 < 𝟎
In this case the roots and of the auxiliary equation are complex numbers
...
The problem 𝑦" + 2𝑦′ =
𝑒𝑥; 𝑦(0) = 1, 𝑦(1) = 1 is a boundary-value problem, because the two subsidiary
conditions are given at the different values x = 0 and x = 1
...

NB: If all of the associated supplementary conditions relate to one 𝑥 value, the problem
is called an initial-value problem (or one-point boundary-value problem)
...


Example
Solve the boundary-value problem 𝑦” + 2𝑦’ + 𝑦 = 0, 𝑦(0) = 1 and 𝑦(1) = 3
...


THE METHOD OF UNDETERMINED COEFFICIENTS
𝐺(𝑥) is a polynomial and it is reasonable to guess that there is a particular solution yp
that is a polynomial of the same degree as G because if y is a polynomial, then 𝑎𝑦" +
29

𝑏𝑦’ + 𝑐𝑦 is also a polynomial
...

Example

Example
Solve 𝑦’’ − 𝑦’ − 2𝑦 = 4𝑥 2
Solution

30

The end

31



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