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Title: Ordinary Differential Equations (ODE)
Description: Well comprehensive notes on Ordinary Differential Equations (ODE)
Description: Well comprehensive notes on Ordinary Differential Equations (ODE)
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Lecture Notes for Math250:
Ordinary Differential Equations
Wen Shen
2011
NB! These notes are used by myself
...
They can not substitute the textbook
...
Introduction
Definition: A differential equation is an equation which contains derivatives of the unknown
...
)
Two classes of differential equations:
• O
...
E
...
D
...
(partial differential equations)
...
• order of the equation: the highest order of derivatives
...
Then,
a0 (t)y (n) + a1 (t)y (n−1) + · · · + an (t)y = g(t),
(∗)
is a linear equations
...
Two things you must know: identify the linearity and order of an equation
...
Let y(t) be the unknown
...
(a)
...
3y ′ + (t + 4)y = t2 + y ′′,
(c)
...
y (4) + ty ′′′ + cos t = ey
...
Problem
(a)
...
3y ′ + (t + 4)y = t2 + y ′′
(c)
...
y (4) + ty ′′′ + cos t = ey
order
1
2
3
4
linear?
No
Yes
No
No
What is a solution? Solution is a function that satisfied the equation and
the derivatives exist
...
Verify that y(t) = eat is a solution of the IVP (initial value
problem)
y ′ = ay,
y(0) = 1
...
Answer
...
They are both OK
...
Example 3
...
2
Answer
...
OK
...
Example 4
...
This means there are
infinitely many solutions
...
(meaning: y = 1 when
t = 0) Then
t2
− t + ln |t + 1| + 1
...
y(0) = 0 + ln |1| + c = c = 1,
Review on integration:
Z
Z
Z
Z
xn dx =
so
y(t) =
1
xn+1 + c,
n+1
1
dx = ln |x| + c
x
sin x dx = − cos x + c
cos x dx = sin x + c
Z
ex dx = ex + c
Z
ax
+c
ax dx =
ln a
3
(n 6= 1)
Integration by parts:
Chain rule:
Z
u dv = uv −
Z
v du
d
(f (g(t)) = f ′ (g(t)) · g ′ (t)
dt
Directional field: for first order equations y ′ = f (t, y)
...
Example 5
...
We know the following:
2
• If y = 3, then y ′ = 0, flat slope,
• If y > 3, then y ′ < 0, down slope,
• If y < 3, then y ′ > 0, up slope
...
5
4
3
...
5
2
1
...
5
0
0
1
2
3
As t → ∞, we have y → 3
...
y ′ = t + y
• We have y ′ = 0 when y = −t,
4
4
5
6
• We have y ′ > 0 when y > −t,
• We have y ′ < 0 when y < −t
...
• If y(0) > −1, then y → ∞ as t → ±∞
...
• If y(0) = −1, the y(t) = −t − 1
...
nonlinear;
• modeling;
• autonomous equations
...
1: Linear equations; Method of integrating
factors
The function f (t, y) is a linear function in y, i
...
So we will study the equation
y ′ + p(t)y = g(t)
...
In particular, we require:
µ(t)y ′ + µ(t)p(t)y = (µ(t)y)′,
µ(t)y ′ + µ(t)p(t)y = µ(t)y ′ + µ′ (t)y
⇒
6
which requires
µ′ (t) =
dµ
= µ(t)p(t),
dt
Integrating both sides
ln µ(t) =
which gives a formula to compute µ
µ(t) = exp
Z
Z
⇒
dµ
= p(t) dt
µ
p(t) dt
p(t) dy
...
Putting back into equation
(A), we get
Z
d
(µ(t)y) = µ(t)g(t), µ(t)y = µ(t)g(t) dt + c
dt
which give the formula for the solution
Z
Z
1
µ(t)g(t) dt + c , where µ(t) = exp
p(t) dt
...
Solve y ′ + ay = b (a 6= 0)
...
We have p(t) = a and g(t) = b
...
b at
e +c
a
Example 2
...
Answer
...
So
Z
µ(t) = exp( 1 dt) = et
7
=
b
+ ce−at ,
a
and
−t
y(t) = e
Z
t 2t
t
e e dt = e
Z
3t
−t
e dt = e
1 3t
e +c
3
1
= e2t + ce−t
...
Solve
(1 + t2 )y ′ + 4ty = (1 + t2 )−2 ,
y(0) = 1
...
First, let’s rewrite the equation into the normal form
y′ +
so
p(t) =
4t
y = (1 + t2 )−3 ,
2
1+t
4t
,
1 + t2
g(t) = (1 + t2 )−3
...
Z
Z
p(t) dt
= exp
Then
2 −2
y = (1+t )
Z
2 2
2 −3
2 −2
(1+t ) (1+t ) dt = (1+t )
Z
(1+t2 )−1 dt =
By the IC y(0) = 1:
y(0) =
0+c
= c = 1,
1
⇒
y(t) =
arctan t + 1
...
Solve ty ′ − y = t2 e−t , (t > 0)
...
Rewrite it into normal form
1
y ′ − y = te−t
t
so
g(t) = te−t
...
(1 + t2 )2
We have
µ(t) = exp(
and
y(t) = t
Z
1 −t
te dt = t
t
Z
(−1/t)dt) = exp(− ln t) =
Z
1
t
e−t dt = t(−e−t + c) = −te−t + ct
...
Solve y − 31 y = e−t , with y(0) = a, and discussion how the
behavior of y as t → ∞ depends on the initial value a
...
Let’s solve it first
...
4
3
y(0) = e0 (− + c) = a,
4
so
c=a+
3
4
3
3
3 −4t
3
= − e−t + (a + )et/3
...
The
first term e−t goes to 0 as t grows
...
So we have
• If a +
3
4
= 0, i
...
, if a = − 43 , we have y → 0 as t → ∞;
• If a +
3
4
> 0, i
...
, if a > − 43 , we have y → ∞ as t → ∞;
• If a +
3
4
< 0, i
...
, if a < − 43 , we have y → −∞ as t → ∞;
Example 6
...
Answer
...
We have
Z
µ(t) = exp( 2/t dt) = exp(2 ln t) = t2
and
−2
y(t) = t
By IC y(1) = 2,
Z
4t · t2 dy = t−2 (t4 + c)
y(1) = 1 + c = 2,
c=1
we get the solution:
1
, t > 0
...
y(t) = t2 +
In the graph below we plot several solutions in the t − y plan, depending on
initial data
...
10
5
0
−5
−4
−3
−2
−1
0
10
1
2
3
4
2
...
Then we have
Z
Z
N(y) dy = M(x) dx,
⇒
N(y) dy = M(x) dx
and we get implicitly defined solutions of y(x)
...
Consider
dy
sin x
=
...
3
If one has IC as y(π) = 2, then
2−
1 3
· 2 = − cos π + c,
3
⇒
5
c=− ,
3
so the solution y(x) is implicitly given as
1
5
y − y 3 + cos x + = 0
...
Find the solution in explicit form for the equation
3x2 + 4x + 2
dy
=
,
dx
2(y + 1)
Answer
...
⇒
(y − 1)2 = x3 + 2x2 + 2x + c
Set in the IC y(0) = −1, i
...
, y = −1 when x = 0, we get
(−1 − 1)2 = 0 + c,
(y − 1)2 = x3 + 2x2 + 2x + 4
...
To determine which sign is the correct one, we check again by the initial
condition:
√
y(0) = 1 ± 4 = 1 ± 2 = −1
We see we must choose the ‘-’ sign
...
On which interval will this solution be defined?
x3 + 2x2 + 2x + 4 ≥ 0,
⇒
x2 (x + 2) + 2(x + 2) ≥ 0
⇒
(x2 + 2)(x + 2) ≥ 0,
⇒
x ≥ −2
...
At this point |dy/dx| →
∞, therefore solution can not be defined at this point
...
We also include the solution with the ‘+’ sign, using dotted line
...
Solve y ′ = 3x2 + 3x2 y 2, y(0) = 0, and find the interval where
the solution is defined
...
Let’s first separate the variables
...
Set in the IC:
arctan 0 = 0 + c,
⇒
c=0
we get the solution
arctan y = x3 ,
y = tan(x3 )
...
e
...
Solve
y′ =
⇒
−
h π i1/3
1 + 3x2
,
3y 2 − 6y
2
h π i1/3
2
...
Answer
...
Set in the IC: x = 0, y = 1, we get
1 − 3 = c,
⇒
c = −2,
Then,
y 3 − 3y 2 = x3 − x − 2
...
To find the valid interval of this solution, we note that y ′ is not defined is
3y 2 − 6y = 0, i
...
, when y = 0 or y = 2
...
To find the corresponding
values of x, we use the solution expression:
y=0:
x3 + x − 2 = 0,
(x2 + x + 2)(x − 1) = 0,
⇒
⇒
x=1
and
x3 + x − 2 = −4,
y=2:
⇒
⇒
(x2 − x + 2)(x + 1) = 0,
x3 + x + 2 = 0,
⇒
x = −1
(Note that we used the facts x2 + x + 2 6= 0 and x2 − x + 2 6= 0 for all x
...
14
-
2
x
2
...
2
...
For a linear equation
y ′ + p(t)y = g(t),
y(t0 ) = y0 ,
we have the following existence and uniqueness theorem
...
If p(t) and g(t) are continuous and bounded on an open interval
containing t0 , then it has an unique solution on that interval
...
Find the largest interval where the solution can be defined for
the following problems
...
ty ′ + y = t3 , y(−1) = 3
...
Rewrite: y ′ + 1t y = t2 , so t 6= 0
...
(B)
...
Answer
...
t0 = 1, the interval is t > 0
...
(t − 3)y ′ + (ln t)y = 2t, y(1) = 2
ln t
2t
Answer
...
Since t0 = 1, the interval is then 0 < t < 3
...
y ′ + (tan t)y = sin t, y(π) = 100
...
Since t0 = π, and for tan t to be defined we must have t 6=
k = ±1, ±2, · · ·
...
2
2k+1
π,
2
For non-linear equation
y ′ = f (t, y),
y(t0 ) = y0 ,
we have the following theorem:
Theorem
...
15
We note that the statement of this theorem is not as strong as the one for
linear equation
...
Example 1
...
Consider
dy
t
= f (t, y) = − ,
dy
y
y(−2) = 0
...
So the conditions of the Theorem are not satisfied, and we
expect something to go wrong
...
Both
are solutions
...
Example 2
...
Consider a simple non-linear equation:
y′ = y2,
y(0) = 1
...
But, due to the
non-linearity of f , solution can not be defined for all t
...
Z
Z
1
1
−1
dy = dt,
− = t + c, y(t) =
...
t−1
We see that the solution blows up as t → 1, and can not be defined beyond
that point
...
16
2
...
Model I: Exponential growth/decay
...
We can write
dQ
(t) = r · Q(t),
dt
r : rate of growth/decay
If r > 0: exponential growth
If r < 0: exponential decay
Differential equation:
Q′ = rQ,
Solve it: separable equation
...
ln Q = rt + c,
⇒
Q(t) = ert+c = cert
Here r is called the growth rate
...
The solution
is
Q(t) = Q0 ert
...
Q(TD ) = Q0 erTD = 2Q0 ,
erTD = 2,
rTD = ln 2,
TD =
ln 2
...
2 0
1
Q(TH ) = Q0 erTH = Q0 ,
2
1
erTD = ,
2
Note here that TH > 0 since r < 0
...
−r
NB! TD , TH do not depend on Q0
...
Example 1
...
Answer
...
08, we have TD =
ln 2
...
08
Example 2
...
Find
its half life
...
Model:
Q(t) = Q0 ert
...
We have the solution
1
Q(10) = Q0 ,
3
1
Q0 e10r = Q0 ,
3
r=
− ln 3
...
r
− ln 3
ln 3
Model II: Interest rate/mortgage problems
...
Start an IRA account at age 25
...
Interest rate 8% annually, but assume
compounded continuously
...
Answer
...
08S + 2000,
dt
S(0) = 2000
...
Solve it by integrating factor
S ′ − 0
...
08t
Z
e−0
...
08t
0
...
08t
2000
2000 · e
dt = e
S(t) = e
+c =
+ ce0
...
08
−0
...
08
C = 2000(1 +
18
1
) = 27000,
0
...
08t − 25000
...
2 − 25000 ≈ 637, 378
...
Example 4: A home-buyer can pay $800 per month on mortgage payment
...
Determine maximum amount this buyer can afford to borrow
...
Set up the model: Let Q(t) be the amount borrowed (principle)
after t years
dQ
= 0
...
We must find Q(0)
...
09Q = −9600,
µ = e−0
...
09t
−0
...
09t
0
...
09t
−9600
(−9600)e
dt = e
Q(t) = e
−0
...
09
By terminal condition
Q(20) =
9600
+ ce0
...
09
so we get
c=−
9600
0
...
8
9600 0
...
0
...
09 · e1
...
8 ) ≈ 89, 034
...
1
...
09
0
...
09
Model III: Mixing Problem
...
At t = 0, a tank contains Q0 lb of salt dissolved in 100 gal of
water
...
At the same time, the well-mixed mixture is
draining from the tank at the same rate
...
Find the amount of salt in the tank at any time t ≥ 0
...
When t → ∞, meaning after a long time, what is the limit amount
QL ?
Answer
...
Q(0) = Q0
...
Solve the equation
Q′ +
−(r/100)t
Q(t) = e
By IC
Z
r
r
Q= ,
100
4
µ = e(r/100)t
...
4
4
r
Q(0) = 25 + c = Q0 ,
c = Q0 − 25,
we get
Q(t) = 25 + (Q0 − 25)e−(r/100)t
...
As t → ∞, the exponential term goes to 0, and we have
QL = lim Q(t) = 25lb
...
Tank contains 50 lb of salt dissolved in 100 gal of water
...
From t = 0, 1/4 lb of salt/gal is entering at a rate of 4
gal/min, and the well-mixed mixture is drained at 2 gal/min
...
20
Answer
...
Since the inflow rate 4 gal/min is larger than the outflow rate
2 gal/min, the tank will be filled up at tf :
tf =
400 − 100
= 150min
...
Let Q(t) be the amount of salt at t min
...
50 + t
(3)
...
400
400(50 + 150)
64
Model IV: Air resistance
Example 7
...
5 kg is thrown upward with initial velocity
10 m/sec from the roof of a building 30 meter high
...
Find the max height above ground the ball reaches
...
Let S(t) be the position (m) of the ball at time t sec
...
Let upward be
the positive direction
...
8 is the gravity, and m = 0
...
We have an equation
for v:
dv
1
= − v − 9
...
1(v + 98),
dt
10
so
Z
Z
1
dv = (−0
...
1t + c
v + 98
which gives
v + 98 = c¯e−0
...
1t
...
1t
...
1t )dt = −98t + 108e−0
...
1) + c
By IC for S,
S(0) = −1080 + c = 30,
S(t) = −98t − 1080e−0
...
c = 1110,
At the maximum height, we have v = 0
...
v(T ) = 0,
−98 + 108e−0
...
1T = ln(98/108),
98 = 108e−0
...
1T = 98/108,
T = −10 ln(98/108) = ln(108/98)
...
1 ln(108/98) + 1110
98
108
− 1080(98/108) + 1110 ≈ 34
...
98
Other possible questions:
• Find the time when the ball hit the ground
...
22
• Find the speed when the ball hit the ground
...
• Find the total distance traveled by the ball when it hits the ground
...
23
2
...
Simplest example: y ′ = ry, exponential growth/decay, where solution is
y = y0 ert
...
Why? Because if f (y0) = 0, then y(t) = y0 is a constant solution
...
Question: Is an equilibrium stable or unstable?
Example 1
...
We have two critical points: y1 = 0, y2 = 2
...
5
3
2
2
y
f
1
...
5
+
1
+
0
0
−1
_
−0
...
5
0
0
...
5
2
2
...
5
1
1
...
Example 2
...
5
3
5
4
3
2
f
1
0
−1
−2
−3
−4
−5
0
1
2
3
y
4
5
6
• (A)
...
Are they stable or unstable?
• (C) Sketch the solutions in the t − y plan, and describe the behavior of
y as t → ∞ (as it depends on the initial value y(0)
...
(A)
...
(B)
...
25
(C)
...
1
0
...
3
0
...
5
t
0
...
7
0
...
9
1
Asymptotic behavior for y as t → ∞ depends on the initial value of y:
• If y(0) < 1, then y(t) → 1,
• If y(0) = 1, then y(t) = 1;
• If 1 < y(0) < 3, then y(t) → 1;
• If y(0) = 3, then y(t) = 3;
• If 3 < y(0) < 5, then y(t) → 5;
• If y(0) = 5, then y(t) = 5;
• if y(t) > 5, then y(t) → 5
...
Example 3
...
For y < 0,
we have y ′ > 0, and for y > 0 we also have y ′ > 0
...
So on the interval y < 0, solution approaches y = 0 as t
grows, so it is stable
...
This type of critical point is called semi-stable
...
Example 4
...
5
f
1
0
...
5
−1
−1
−0
...
5
1
1
...
5
3
3
...
Identify equilibrium points;
• (B)
...
Sketch solution in y − t plan;
• (D)
...
Answer
...
y = 0, y = 1, y = 2, y = 3 are the critical points
...
y = 0 is stable, y = 1 is semi-stable, y = 2 is unstable, and y = 3 is
stable
...
The Sketch is given in the plot:
27
4
3
...
5
y
2
1
...
5
0
−0
...
2
0
...
6
0
...
2
1
...
6
1
...
The asymptotic behavior as t → ∞ depends on the initial data
...
Application in population dynamics: let y(t) be the population of a species
...
dt
the logistic equation
dy
y
= r(1 − ))y,
dt
k
r=intrinsic growth rate,
k=environmental carrying capacity
...
Here y = 0 is unstable, and y = k is stable
...
28
Chapter 3: Second Order Linear Equations
General form of the equation:
a2 (t)y ′′ + a1 (t)y ′ + a0 (t)y = b(t),
where
a2 (t) 6= 0,
y(t0) = y0 , y ′ (t0 ) = y¯0
...
Otherwise, it is called non-homogeneous
...
1: Homogeneous equations with constant coefficients
This is the simplest case: a2 , a1 , a0 are all constants, and g = 0
...
Example 1
...
Answer
...
Check: y ′′ = et , so y ′′ − y = et − et = 0, ok
...
Check: y ′ = −e−t , so y ′′ = e−t , so y ′′ − y = et − et = 0, ok
...
) is also a solution
...
Actually this is a general property
...
Theorem Let y1 (t) and y2 (t) be solutions of
a2 (t)y ′′ + a1 (t)y ′ + a0 (t)y = 0
29
Then, y = c1 y1 + c2 y2 for any constants c1 , c2 is also a solution
...
(I)
If y2 solves the equation, then
a2 (t)y2′′ + a1 (t)y2′ + a0 (t)y2 = 0
...
Let y = c1 y1 + c2 y2 , we have
a2 (t)y ′′ + a1 (t)y ′ + a0 (t)y = 0
therefore y is also a solution to the equation
...
Find r
...
Conclusion: If r is a root of the characteristic equation, then y = ert is a
solution
...
Example 2
...
• (a)
...
30
• (b)
...
• (c) What happens when t → ∞?
Answer
...
The characteristic equation is: r 2 −5r+6 =, so (r−2)(r−3) =
0, two roots: r1 = 2, r2 = 3
...
(b)
...
y ′(0) = 5: we have y ′ = 2c1 e2t + 3c2 e3t , so y ′(0) = 2c1 + 3c2 = 5
...
Then c2 = 7
...
(c)
...
So y → ∞
...
Find the solution for 2y ′′ + y ′ − y = 0, with initial conditions
y(1) = 0, y ′(1) = 3
...
Characteristic equation:
2r 2 + r − 1 = 0,
⇒
(2r − 1)(r + 1) = 0,
General solution is:
⇒
1
r1 = , r2 = −1
...
The ICs give
y ′ (1) = 3 :
(A)+(B) gives
1
y(1) = 0 : c1 e 2 + c2 e−1 = 0
...
2
2
3 1
c1 e 2 = 3,
2
1
c1 = 2e− 2
...
31
(A)
(B)
The solution is
1 1
1
y(t) = 2e− 2 e t − 2ee−t = 2e 2 (t−1) − 2e−t+1 ,
2
and as t → ∞ we have y → ∞
...
Write the characteristic equation;
2
...
Write the general solution;
4
...
Example 4
...
• (a)
...
• (b)
...
(a)
...
General solution is
y(t) = c1 e−
√
5t
√
+ c2 e
5t
...
If y(t) remains bounded as t → ∞, then the term e √5t must vanish,
5t
which means we must have c2 = 0
...
If √y(0) = 1,
− 5t
′
then y(0) = c1 =
1, so y(t) = e
...
Example 5
...
The characteristic equation is
2r 2 + 3r = 0,
⇒
r(2r + 3) = 0,
32
⇒
3
r1 = − , r2 = 0
2
The general solutions is
3
3
y(t) = c1 e− 2 t + c2 e0t = c1 e− 2 t + c2
...
Example 6
...
Answer
...
The characteristic equation could be (r − 3)(r + 1) = 0, or
this equation multiplied by any non-zero constant
...
NB! This answer is not unique
...
33
3
...
Theorem
...
If p(t), q(t) and g(t) are continuous and bounded on an open interval I containing t0 , then there exists exactly one solution y(t) of this equation, valid
on I
...
Given the equation
(t2 − 3t)y ′′ + ty ′ − (t + 3)y = et ,
y ′(1) = 1
...
Answer
...
t(t − 3)
We see that we must have t 6= 0 and t 6= 3
...
See the figure below
...
Given two functions f (t), g(t), the Wronskian is defined as
W (f, g)(t) =
˙ f g ′ − f ′ g
...
f g
Main property of the Wronskian:
• If W (f, g) ≡ 0, then f anf g are linearly dependent
...
Example 2
...
(a)
...
Answer
...
(b)
...
Answer
...
(c)
...
Answer
...
(In fact, we have g(t) = 4 · f (t)
...
f (t) = 2t, g(t) = |t|
...
Note that g ′(t) = sign(t) where sign is the sign function
...
So they are linearly dependent
...
Suppose y1 (t), y2 (t) are two solutions of
y ′′ + p(t)y ′ + q(t)y = 0
...
They are
also called to form a fundamental set of solutions
...
The next Theorem is probably the most important one in this chapter
...
Then, the Wronskian
W (y1, y2 ) on I is given by
Z
W (y1 , y2 )(t) = C · exp( −p(t) dt),
for some constant C depending on y1 , y2, but independent on t in I
...
We skip this part
...
Example 3
...
36
Find W (y1 , y2 ) without solving the equation
...
We first find the p(t)
p(t) = −
t+2
t
which is valid for t 6= 0
...
t
NB! The solutions are defined on either (0, ∞) or (−∞, 0), depending on t0
...
Example 4
...
Answer
...
By Abel’s Theorem we have
Z
2
dt = C · e− ln t = Ct−2
...
So we have
W (y1 , y2 )(5) = 25−2 =
2
...
If W (f, g) = 3e4t , and f = e2t , find g
...
By definition of the Wronskian, we have
W (f, g) = f g ′ − f ′ g = e2t g ′ − 2e2t g = 3e4t ,
which gives a 1st order equation for g:
g ′ − 2g = 3e2t
...
We can choose c = 0, and get g(t) = 3te2t
...
Example 6
...
Find the general
solution
...
The characteristic equation is r 2 + 2r + 1 = 0, which given double
roots r1 = r2 = −1
...
How can we
find another solution y2 that’s linearly independent?
By Abel’s Theorem, we have
W (y1, y2 ) = C exp
Z
−2 dt = Ce−2t ,
and we can choose C = 1 and get W (y1, y2 ) = e−2t
...
These two computation must have the same answer, so
e−t (y2′ + y2 ) = e−2t ,
y2′ + y2 = e−t
...
Solve it:
Z
t
−t
et e−t dt = e−t (t + c)
...
The general solution is
y(t) = c1 y1 + c2 y2 = c1 e−t + c2 te−t
...
We will study it more later
in chapter 3
...
38
3
...
Consider
the equation
ay ′′ + by ′ + cy = 0,
The two roots are
→
ar 2 + br + c = 0
...
2a
If b2 − 4ac < 0, the root are complex, i
...
, a pair of complex conjugate
numbers
...
There are two solutions:
r1,2 =
−b ±
y1 = e(λ+iµ)t = eλt eiµt ,
y2 = y1 = e(λ−iµ)t = eλt e−iµt
...
A couple of Examples to practice this formula:
i 56 π
e
√
5
3
5
1
= cos π + i sin π = −
+i
...
ea+ib = ea eib = ea (cos b + i sin b)
...
But these solutions are complex valued
...
If y1 , y2 are two solutions,
then 21 (y1 + y2 ), 2i1 (y1 − y2 ) are also solutions
...
2i
To make sure they are linearly independent, we can check the Wronskian,
W (˜
y1 , y˜2) = µe2λt 6= 0
...
39
So y1 , y2 are linearly independent, and we have the general solution
y(t) = c1 eλt cos µt + c2 eλt sin µt = eλt (c1 cos µt + c2 sin µt)
...
(Perfect Oscillation: Simple harmonic motion
...
6
Answer
...
The general solution is
y(t) = c1 cos 2t + c2 sin 2t
...
y ′( ) = 1 : −2c1 sin + 2c2 cos = −2c1
6
3
3
2
2
Solve these two equations, we get c1 = −
y(t) = −
√
3
4
and c2 = 14
...
This is also called perfect oscillation or simple
harmonic motion
...
(Decaying oscillation
...
Answer
...
So the general solution is
y(t) = e−t (c1 cos 10t + c2 sin 10t),
so
y ′ (t) = −e−t (c1 cos t + c2 sin t) + e−t (−10c1 sin t + 10c2 cos t)
Fit in the ICs:
y(0) = 1 :
y ′(0) = 0 :
y(0) = e0 (c1 + 0) = c1 = 1,
y ′ (0) = −1 + 10c2 = 0,
c2 = 0
...
Solution is
y(t) = e−t (cos t + 0
...
The graph is given below:
1
0
...
6
0
...
2
0
−0
...
4
−0
...
8
−1
0
0
...
5
2
2
...
5
4
We see it is a decaying oscillation
...
As t → ∞, we have y → 0
...
(Growing oscillation) Find the general solution of y ′′ − y ′ +
81
...
Answer
...
25 = 0,
⇒
r = 0
...
5,
µ = 2
...
5t (c1 cos 9t + c2 sin 9t)
...
In the limit when
t → ∞, y oscillates between −∞ and +∞
...
We note that since λ =
−b
,
2a
so the sign of λ follows the sign of −b
...
4: Repeated roots; reduction of order
For the characteristic equation ar 2 + br + c = 0, if b2 = 4ac, we will have two
repeated roots
b
r1 = r2 = r = −
...
How can we find the second solution which
is linearly independent of y1 ?
Example 1
...
We have r 2 +4r+4 = 0,
and r1 = r2 = r = −2
...
What is y2 ?
Method 1
...
By Abel’s Theorem we
have
Z
W (y1 , y2) = c exp(− 4 dt) = ce−4t = e−4t , (let c = 1)
...
They must equal to each other:
e−2t (y2′ + 2y2 ) = e−4t ,
y2′ + 2y2 = e−2t
...
Method 2
...
We guess a solution of the form
y2 = v(t)y1 = v(t)e−2t , and try to find the function v(t)
...
Put them in the equation
e−2t (v ′′ − 4v ′ + 4v) + 4e−2t (v ′ − 2v) + 4v(t)e−2t = 0
...
So
y2 (t) = vy1 = (c1 t + c2 )e−2t = c1 te−2t + c2 e−2t
...
Therefore we can choose
c1 = 1, c2 = 0, and get y2 = te−2t , which gives the same general solution as
Method 1
...
A typical solution graph is included below:
2
...
5
1
0
...
5
1
1
...
5
3
3
...
5
5
We see if c2 > 0, y increases for small t
...
One can show that in general if one has repeated roots r1 = r2 = r, then
y1 = ert and y2 = tert , and the general solution is
y = c1 ert + c2 tr rt = ert (c1 + c2 t)
...
Solve the IVP
y ′′ − 2y ′ + y = 0,
y(0) = 2,
45
y ′ (0) = 1
...
This follows easily now
r 2 − 2r + 1 = 0,
⇒
r1 = r2 = 1,
⇒
y(t) = (c1 + c2 t)et
...
The ICs give
y(0) = 2 :
y ′(t) = (c1 + c2 t)et + c2 et ,
y ′ (0) = c1 + c2 = 1,
So the solution is y(t) = (2 − t)et
...
Summary: For ay ′′ + by ′ + cy = 0, and ar 2 + br + c = 0 has two roots r1 , r2 ,
we have
y(t) = c1 er1 t + c2 er2 t ;
• If r1 6= r2 (real):
• If r1 = r2 = r (real):
• If r1,2 = λ ± iµ complex:
y(t) = (c1 + c2 t)ert ;
y(t) = eλt (c1 cos µt + c2 sin µt)
...
Example 3
...
Answer
...
By Abel’s
Theorem, and choose C = 1, we have
Z
Z
3
3t
dt = exp − ln t = t−3/2
...
t
t
46
Solve this for y2 :
Z
Z
1
1 2 3
3
1
µ = exp(
t · t− 2 dt = ( t 2 + C)
...
Since 32 is a constant multiplication, we can
√
drop it and choose y2 = t
...
We saw in the previous example
that this method is inferior to Method 1, therefore we will not focus on it at
all
...
Let’s introduce another method that combines the ideas from Method 1 and
Method 2
...
We will use Abel’s Theorem, and at the same time we will seek
a solution of the form y1 = vy1
...
Now,
seek y2 = vy1
...
Note that this is a general formula
...
3
Drop the constant 32 , we get
3
y2 = vy1 = t 2
1
1
= t2
...
We
will focus on this method from now on
...
Consider the equation
t2 y ′′ − t(t + 2)y ′ + (t + 2)y = 0,
47
t > 0
...
Answer
...
2
t
t
t
Let y2 be the second solution
...
t
p(t) = −
Let y2 = vy1, the W (y1 , y2) = v ′ y12 = t2 v ′
...
(A cheap trick to double check your solution y2 would be: plug it back into
the equation and see if it satisfies it
...
We observe here that Method 3 is very efficient
...
Given the equation
√
y1 = t(1/4) e2 t , find y2
...
We will always use method 3
...
By Abel’s
Theorem, setting c = 1, we have
Z
W (y1 , y2) = exp( 0dt) = 1
...
Then, W (y1, y2 ) = y12 v ′ = t 2 e4 t v ′
...
√
1
Let u = −4 t, so du = −2t− 2 dt, we have
Z
1
1 √
1
v = − eu du = − eu = − e−4 t
...
The general solution is
1
√
y(t) = c1 y1 + c2 y2 = t 4 (c1 e2
48
t
√
+ c2 e−2 t )
...
6: Non-homogeneous equations; method of
undetermined coefficients
Want to solve the non-homogeneous equation
y ′′ + p(t)y ′ + q(t)y = g(t),
(N)
Steps:
1
...
e
...
2
...
The general solution for (N) is then
y = yH + Y = c1 y1 + c2 y2 + Y
...
Key step: step 2
...
Main focus: constant coefficient case, i
...
,
ay ′′ + by ′ + cy = g(t)
...
Find the general solution for
y ′′ − 3y ′ + 4y = 3e2t
...
Step 1: Find yH
...
Step 2: Find Y
...
Y ′ = 2Ae2t ,
Y ′′ = 4Ae2t
...
2
So Y = − 12 e2t
...
The general solution to the non-homogeneous solution is
1
y(t) = yH + Y = c1 e−t + c2 e4t − e2t
...
But this is not always true
...
Find general solution for
y ′′ − 3y ′ + 4y = 2e−t
...
The homogeneous solution is the same as Example 1: yH = c1 e−t +
c2 e4t
...
e
...
So Y ′ = −Ae−t , Y ′′ = Ae−t
...
So it doesn’t work
...
It will never work
...
So
Y ′ = Ae−t − Ate−t ,
Y ′′ = −Ae−t − Ae−t + Ate−t = −2Ae−t + Ate−t
...
Summary 1
...
case
form of the particular solution Y
r1 6= α and r2 6= α
Y = Aeαt
r1 = α or r2 = α, but r1 6= r2
Y = Ateαt
r1 = r2 = α
Y = At2 eαt
Example 3
...
Answer
...
Note that g(t) is a polynomial of degree 2
...
51
Compare the coefficient, we get three equations for the three coefficients
A, B, C:
−4A = 3
→ A=−
3
4
9
8
55
1
2A − 3B − 4C = 2, → C = (2A − 3B − 2) = −
4
32
−(6A + 4B) = 0, → B =
So we get
3
9
55
Y (t) = − t2 + t −
...
Example 4
...
Answer
...
New try: multiply by a t
...
Then
Y ′ = 3At2 + 2Bt + C, Y ′′ = 6At + 2B
...
Compare the coefficient, we get three equations for the three coefficients
A, B, C:
−9A = 3
→ A=−
1
3
(6A − 6B) = 0, → B = A = −
1
3
1
8
2B − 3C = 2, → C = (2B − 2) = −
3
9
So Y = t(− 13 t2 − 13 t − 98 )
...
If g(t) is a polynomial of degree n, i
...
,
g(t) = αn tn + · · · + α1 t + α0
the particular solution for
ay ′′ + by ′ + cy = g(t)
(where a 6= 0) depends on b, c:
case
c 6= 0
c = 0 but b 6= 0
form of the particular solution Y
Y = Pn (t) = An tn + · · · + A1 t + A0
Y = tPn (t) = t(An tn + · · · + A1 t + A0 )
c = 0 and b = 0 Y = t2 Pn (t) = t2 (An tn + · · · + A1 t + A0 )
Example 5
...
Answer
...
Note that (sin t)′ =
cos t, so we must have the cos t term as well
...
Then
Y ′ = A cos t − B sin t,
Y ′′ = −A sin t − B cos t
...
So we must have
−5A + 3B = 1,
−3A − 5B = 0,
So we get
Y (t) = −
→
A=
5
3
sin t +
cos t
...
34
But this guess won’t work if the form is a solution to the homogeneous
equation
...
Find a general solution for y ′′ + y = sin t
...
Let’s first find yH
...
For the particular solution Y : We see that the form Y = A sin t + B cos t
won’t work because it solves the homogeneous equation
...
Then
Y ′ = (A sin t + B cos t) + t(A cos t + B sin t),
Y ′′ = (−2B − At) sin t + (2A − Bt) cos t
...
2
The general solution is
1
y(t) = yH + Y = c1 cos t + c2 sin t − t cos t
...
If g(t) = a sin αt + b cos αt, the form of the particular solution
depends on the roots r1 , r2
...
54
Example 7
...
Answer
...
Also we see r1 = −1, r2 = 4, so r1 6= a and r2 6= a
...
e
...
So
Y ′ = Aet + (At + b)et = (A + b)et + Atet ,
Y ′′ = · · · = (2A + B)et + Atet
...
We must have −6At − A − 6B = t, i
...
,
−6A = 1,
−A−6B = 0,
⇒
1
1
A = − ,B = ,
6
36
⇒
1
1
Y = (− t+ )et
...
We must multiply it by t in that case
...
Find a particular solution of
y ′′ − 3y ′ − 4y = te−t
...
Since a = −1 = r1 , so the form we used in Example 7 won’t work
here
...
Then
Y ′ = · · · = [−At2 + (2A − B)t + B]e−t ,
Y ′′ = · · · = [At2 + (B − 4A)t + 2A − 2B]e−t
...
55
So we must have −10At + 2A − 5B = t, which means
−10A = 1,
2A − 5B = 0,
Then
Y =
⇒
A=−
1
1
, B=−
...
10
25
Summary 4
...
case
form of the particular solution Y
r1 6= a and r2 6= a
Y = P˜n (t)eat = (An tn + · · · + A1 t + A0 )eat
r1 = a or r2 = a but r1 6= r2
Y = tP˜n (t)eat = t(An tn + · · · + A1 t + A0 )eat
r1 = r2 = a
Y = t2 P˜n (t)eat = t2 (An tn + · · · + A1 t + A0 )eat
Other cases of g are treated in a similar way: Check if the form of g is a
solution to the homogeneous equation
...
If yes, then multiply it by t or t2
...
Summary 5
...
Then
case
form of the particular solution Y
r1,2 6= α ± iβ
Y = eαt (A cos βt + B sin βt)
r1,2 = α ± iβ
Y = t · eαt (A cos βt + B sin βt)
Summary 6
...
Then
56
r1,2
case
6= α ± iβ
form of the particular solution Y
Y = e [(An t + · · · + A0 ) cos βt + (Bn tn + · · · + B0 ) sin βt]
αt
n
Y = t · eαt [(An tn + · · · + A0 ) cos βt + (Bn tn + · · · + B0 ) sin βt]
r1,2 = α ± iβ
If the source g(t) has several terms, we treat each separately and add up
later
...
This claim follows from the principle of superposition
...
Example 9
...
Answer
...
Example 10
...
Answer
...
We also note that the terms sin 4t and −4 cos 4t are of the same type, and
we must multiply it by t
...
Example 11
...
Answer
...
Then,
for the term et cos t we must multiply by t
...
57
3
...
The spring-mass system: See figure below
...
Figure (B): we put a mass m on the spring, and the spring is stretched
...
Force diagram at equilibrium position: mg = F s
...
58
So: we have mg = kL which give
k=
mg
L
which gives a way to obtain k by experiment: hang a mass m and measure
the elongation L
...
Total elongation: L + u
Total spring force: Fs = −k(L + u)
Other forces:
* damping/resistent force: Fd (t) = −γv = −γu′ (t), where γ is the damping
constant, and v is the velocity
* External force applied on the mass: F (t), given function of t
P
Total force on the mass:
f = mg + Fs + Fd + F
...
Since mg = kL, by rearranging the terms, we get
mu′′ + γu′ + ku = F
where m ia the mass, γ is the damping constant, k is the spring constant,
and F is the external force
...
Case 1: Undamped free vibration (simple harmonic motion)
...
So the equation becomes
mu′′ + ku = 0
...
m
General solution
u(t) = c1 cos ω0 t + c2 sin ω0 t
...
We can write
!
q
c
c
2
1
u(t) = c21 + c22 p 2
cos ω0 t + p 2
sin ω0 t
...
So amplitude is R =
p
c21 + c22 and phase is δ = arctan
A few words on units:
force (f ) weight (mg)
lb
lb
newton
newton
c2
...
8 m/sec2
Example 1
...
If the mass is
displaced an additional 2 in, and is then set in motion with initial upward
velocity of 1 ft/sec, determine the position, frequency, period, amplitude and
phase of the motion
...
We see this is free harmonic oscillation
...
g
32
16
And the elongation is L = 2in = 16 ft
...
Let u(t) be the
position from equilibrium, we get the equation
mu′′ + ku = 0,
5 ′′
u + 60u = 0,
16
therefore
1
u′′ + 192u = 0,
u(0) = , u′ (0) = −1
...
ω0
192
(Note that c1 = u(0) and c2 = u′ (0)/ω0
...
u(t) = cos ω0 t − √
6
192
The four terms of the motion are
ω0 =
and
√
192,
π
2π
=√ ,
T =
ω0
48
r
q
19
2
2
≈ 0
...
= − arctan
c1
4
192
Case II: Damped free vibration
...
mu′′ + γu′ + ku = 0
then
p
γ 2 − 4km
...
−γ ±
2
61
√
• If γ 2 − 4km > 0, (i
...
, γ > 4km) we have two real roots, and the
general solution is u = c1 er1 t + c2 er2 t , with r1 < 0, r2 < 0
...
The mass will simply return to the equilibrium position exponentially
...
√
• If γ 2 −4km = 0, (i
...
, γ = 4km) we have double roots r1 = r2 = r < 0
...
Depending on the sign of c1 , c2 (which is determined by the ICs), the
mass may cross the equilibrium point maximum once
...
√
• If γ 2 − 4km < 0, (i
...
, γ < 4km) we have complex roots
p
4km − γ 2
γ
, µ=
...
This motion is damped oscillation
...
c1
Here the term e−λt R is the amplitude, and µ is called the quasi fre
...
Summary: For all cases, since the real part of the roots are always negative,
u will go to zero as t grow
...
Example 2
...
8 kg is hanging on a spring with k = 1
...
The mass is then further stretched for another 2ft, then
released from rest
...
Answer
...
So the equation for u is
1
u′′ + u′ + u = 0,
8
mu′′ + γu′ + ku = 0,
u(0) = 2,
u′ (0) = 0
...
By ICs, we have u(0) = c1 = 2, and
u′ (t) = −
1
1
u(t) + e− 16 t (−ω0 c1 sin ω0 t + ω0 c2 cos ω0 t),
16
u′ (0) = −
1
u(0) + ω0 c2 = 0,
16
c2 = √
2
...
255
3
...
(The case
where F (t) = F0 sin ωt is totally similar
...
mu′′ + γu′ + ku = F0 cos ωt
...
From discussion is the previous chapter, we know that uH → 0 as t → ∞
for systems with damping
...
The appearance of U is due to the force term F
...
The form is U = R cos(ωt − δ)
...
As time t → ∞, we have u → U
...
Case 2: Without damping
...
We have two
cases
...
The particular solution should be
U = A cos wt + B cos wt
But there is no u′ term, so we only need U = A cos wt
...
Plug in the equation
m(−w 2 A cos wt) + kA cos wt = F0 cos wt,
64
(k − mw 2 )A = F0 ,
A=
General solution
F0
F0
=
...
Find c1 , c2
...
2
2
sin a+b
...
One particular situation: if w0 6= w but wo ≈ w, then |w0 − w| << |w0 + w|
...
5
1
0
...
5
−1
−1
...
(One observes it by hitting two nearby keys on a piano,
for example
...
The particular solution is
U = At cos w0 t + Bt sin w0 t
A typical plot looks like:
5
4
3
2
1
0
−1
−2
−3
−4
−5
0
0
...
5
2
2
...
5
4
4
...
If the frequency of the source term ω equals to the
frequency of the system ω0 , then, small source term could make the solution
grow very large!
Summary:
• With damping: Transient solution plus the forced response term,
• Without damping:
if w = w0 : resonance
...
66
Chapter 6
...
6
...
Definition: Given a function f (t), t ≥ 0, its Laplace transform F (s) =
L{f (t)} is defined as
Z A
Z ∞
...
−st
e−st f (t) dt
F (s) = L{f (t)} =
e f (t) dt = lim
A→∞
0
0
We say the transform converges if the limit exists, and diverges if not
...
Example 1
...
Answer
...
f (t) = et
...
F (s) = L{f (t)} = lim
A→∞
=
lim −
A→∞
Z
A
−st at
e
e dt = lim
A→∞
0
1
1
e−(s−a)A − 1 =
,
s−a
s−a
67
Z
A
−(s−a)t
e
0
(s > a)
A
1 −(s−a)t
dt = lim −
e
A→∞
s−a
0
Example 3
...
Answer
...
s
So we get a recursive relation
L{tn } =
n
L{tn−1 },
s
∀n,
which means
L{tn−1 } =
n−1
L{tn−2 },
s
L{tn−2 } =
n−2
L{tn−3 }, · · ·
s
By induction, we get
n (n − 1)
n (n − 1) (n − 2)
n
L{tn−1 } =
L{tn−2 } =
L{tn−3 }
s
s
s
s
s
s
1
n! 1
n!
n (n − 1) (n − 2)
· · · L{1} = n = n+1 , (s > 0)
= ··· =
s
s
s
s
s s
s
L{tn } =
Example 4
...
Answer
...
Compute by definition, with integration-by-parts,
twice
...
)
Method 2
...
By Example 2 we have
L{eiat } =
1
1(s + ia)
s + ia
s
a
=
= 2
= 2
+i 2
...
Remark: Now we will use
sion
...
Example 5
...
We do this by definition:
Z ∞
Z 2
Z ∞
−st
−st
F (s) =
e f (t) dt =
e dt +
(t − 2)e−st dt
0
0
2
∞ 2 Z A
1 −st
1
1 −st
=
e + (t − 2) e−st −
e dt
−s
−s
2 −s
t=0
t=2
∞
1 1 −st
1
1 −2s
1 −2s
(e − 1) + (0 − 0) +
e
(e − 1) + 2 e−2s
=
=
−s
s −s
−s
s
t=2
69
6
...
Properties of Laplace transform:
1
...
2
...
3
...
4
...
5
...
This also implies L{tf (t)} =
−F ′ (s)
...
L{eat f (t)} = F (s − a) where F (s) = L{f (t)}
...
Remarks:
• Note property 2 and 3 are useful in differential equations
...
• Property 5 is the counter part for Property 2
...
• Property 6 is also known as the Shift Theorem
...
3
...
This follows by definition
...
By definition
Z
′
L{f (t)} =
∞
−st ′
e
−st
f (t)dt = e
0
∞ Z
f (t) −
0
∞
0
(−s)e−st f (t)dt = −f (0)+sL{f (t)}
...
This one follows from Property 2
...
4
...
5
...
ds 0
∂s
0
0
6
...
0
0
By using these properties, we could find more easily Laplace transforms of
many other functions
...
From
L{tn } =
n!
n!
...
From
L{sin bt} =
s2
71
b
...
From
L{cos bt} =
s
,
s2 + b2
L{eat cos bt} =
we get
s−a
...
L{t3 + 5t − 2} = L{t3 } + 5L{t} − 2L{1} =
3!
1
1
+5 2 −2
...
L{e2t (t3 + 5t − 2)} =
1
1
3!
+
5
−
2
...
L{(t2 + 4)e2t − e−t cos t} =
because
L{t2 + 4} =
2
4
+ ,
3
s
s
2
4
s+1
+
−
,
3
(s − 2)
s − 2 (s + 1)2 + 1
L{(t2 + 4)e2t } =
⇒
2
4
+
...
Example 7
...
b
L{t sin bt} = − 2
s + b2
′
=
−2bs
...
L{t cos bt} = −
s
2
s + b2
72
′
= ··· =
s2 − b2
...
Definition:
L−1 {F (s)} = f (t),
if
F (s) = L{f (t)}
...
Some simple examples:
Example 10
...
s2 + 4
2 s2 + 2 2
2
s2 + 2 2
2
Example 11
...
·
4
4
4
4
(s + 5)
3 (s + 5)
3
(s + 5)
3
s
3
Example 12
...
L
2
2
2
s +4
s +4
2
s +4
2
Example 13
...
s2 − 4
(s − 2)(s + 2)
s−2 s+2
4
4
Here we used partial fraction to find out:
s+1
A
B
=
+
,
(s − 2)(s + 2)
s−2 s+2
73
A = 3/4,
B = 1/4
...
We will go through one example first
...
(Two distinct real roots
...
Answer
...
Take Laplace transform on both sides: Let L{y(t)} =
Y (s), and then
L{y ′(t)} = sY (s)−y(0) = sY −1,
L{y ′′(t)} = s2 Y (s)−sy(0)−y ′(0) = s2 Y −s−2
...
s
⇒
Now we get an algebraic equation for Y (s)
...
s(s − 5)(s + 2)
Step 3: Take inverse Laplace transform to get y(t) = L−1 {Y (s)}
...
Y (s) =
A
B
C
A(s − 5)(s + 2) + Bs(s + 2) + Cs(s − 5)
s2 − s + 2
= +
+
=
...
The previous equation holds for all values of s
...
Set s = 5: we get 35B = 22, so B = 22
...
Now, Y (s) is written into sum of terms which we can find the inverse transform:
1
1
1
1 22
4
y(t) = AL−1 { } + BL−1 {
} + CL−1 {
} = − + e5t + e−2t
...
You will get an algebraic equation for Y
...
• Take inverse transform to get y(t) = L−1 {Y }
...
(Distinct real roots, but one matches the source term
...
Answer
...
s−2
Solve it for Y :
(s2 −s−2)Y (s) =
1
s−1
+1 =
,
s−2
s−2
⇒
Y (s) =
s−1
s−1
=
...
2
(s − 2) (s + 1)
s + 1 s − 2 (s − 2)2
Compare the numerators:
s − 1 = A(s − 2)2 + B(s + 1)(s − 2) + C(s + 1)
Set s = −1, we get A = − 92
...
Set s = 0 (any convenient values of s can be used in this step), we get B = 29
...
9
9
3
Compare this to the method of undetermined coefficient: general solution
of the equation should be y = yH + Y , where yH is the general solution to
the homogeneous equation and Y is a particular solution
...
Since 2 is a root, so the form of the particular solution
is Y = Ate2t
...
This fits well with our result
...
(Complex roots
...
y(0) = 0,
Answer
...
r 2 − 2r + 2 = 0,
(r − 1)1 + 1 = 0,
yH = c1 et cos t + c2 et sin t,
r1,2 = 1 ± i,
Y = Ae−t ,
so the solution should have the form:
y = yH + Y = c1 et cos t + c2 et sin t + Ae−t
...
2
(s − 1)2 + 1
(s − 1)2 + 1
s+1
(s − 1)2 + 1
s+1
This gives us some idea on which terms to look for in partial fraction
...
76
1
s+2
+1 =
s+1
s+1
s+2
s+2
A
B(s − 1) + C
Y (s) =
=
=
+
2
2
(s + 1)(s − 2s + 2)
(s + 1)((s − 1) + 1)
s + 1 (s − 1)2 + 1
s2 Y −1−2sY +2Y =
1
,
s+1
(s2 −2s+2)Y (s) =
⇒
Compare the numerators:
s + 2 = A((s − 1)2 + 1) + (B(s − 1) + C)(s + 1)
...
Compare coefficients of s2 -term: A + B = 0, B = −A = − 15
...
Y (s) =
1
9
s−1
1
1 1
−
+
2
5 s + 1 5 (s − 1) + 1 5 (s − 1)2 + 1
1
1
9
y(t) = e−t − et cos t + et sin t
...
Example 17
...
) Solve
y ′′ + y = cos 2t,
y(0) = 2,
y ′ (0) = 1
...
Again, let’s first predict the terms in the solution:
r 2 + 1 = 0,
r1,2 = ±i,
yH = c1 cos t + c2 sin t,
so
y = yH + Y = c1 cos t + c2 sin t + A cos 2t,
and the Laplace transform would be
Y (s) = c1
s1
1
s
s
+ c2 2
+A 2
...
2
2
(s + 4)(s + 1)
s +1
s +4
(s2 + 1)Y (s) =
Comparing numerators, we get
2s3 + s2 + 9s + 4 = (As + B)(s2 + 4) + (Cs + D)(s2 + 1)
...
Let’s try by setting s into complex numbers
...
3
Set now s = 2i:
−16i − 4 + 18i + 4 = (2Ci + D)(−3),
then
0 + 2i = −3D − 6Ci,
So
Y (s) =
⇒
1
D = 0, C = −
...
3
3
A very brief review on partial fraction, targeted towards inverse
Laplace transform
...
We assume n < m
...
First, fact out Pm (s), write it into product of terms
of (i) s − a, (ii) s2 + a2 , (iii) (sa )2 + b2
...
term in PM (s)
from where?
real root, or
s−a
g(t) = eat
term in partial fraction
inverse L
...
A
s−a
Aeat
A
B
+
s − a (s − a)2
Aeat + Bteat
double roots,
(s − a)2
or r = a and g(t) = eat
double roots,
(s − a)3
and g(t) = eat
A
B
C
+
+
s − a (s − a)2 (s − a)3
Aeat + Bteat +
C 2 at
te
2
imaginary roots or
s2 + µ 2
g(t) = cos µt or sin µt
As + B
s2 + µ 2
A cos µt + B sin µt
A(s − λ) + B
(s − λ)2 + µ2
eλt (A cos µt + B sin µt)
complex roots, or
(s − λ)2 + µ2
g(t) = eλt cos µt(or sin µt)
In summary, this table can be written
Pn (s)
(s − a)(s − b)2 (s − c)3 ((s − λ)2 + µ2 )
B1
B2
C1
C2
C3
D1 (s − λ) + D2
A
+
+
+
+
+
+
...
3: Step functions
Topics:
• Definition and basic application of unit step (Heaviside) function,
• Laplace transform of step functions and functions involving step functions (piecewise continuous functions),
• Inverse transform involving step functions
...
Unit step function(Heaviside function):
0, 0 ≤ t < c,
uc t =
1, c ≤ t
...
A plot of uc (t) is below:
6uc
1
-
c
0
For a given function f (t), if it is multiplied with uc (t), then
0,
0 < t < c,
uc tf (t) =
f (t), c ≤ t
...
Example 1
...
A plot of this is given below
80
t
61 − uc
1
-
c
0
t
We see that this function picks up the interval [0, c)
...
Rectangular pulse
...
We see it can be expressed as
ua (t) − ub (t)
and it picks up the interval [a, b)
...
For the function
g(t) =
f (t), a ≤ t < b
0,
otherwise
We can rewrite it in terms of the unit step function as
g(t) = f (t) · ua (t) − ub (t)
...
For the function
sin t, 0 ≤ t < 1,
et ,
1 ≤ t < 5,
ft =
2
t
5 ≤ t,
81
t
we can rewrite it in terms of the unit step function as we did in Example 3,
treat each interval separately
f (t) = sin t · u0 (t) − u1 (t) + et · u1 (t) − u5 (t) + t2 · u5 (t)
...
Shift of a function: Given f (t), t > 0, then
f (t − c), c ≤ t,
g(t) =
0,
0 ≤ t < c,
is the shift of f by c units
...
f
g
6
6
-
0
t
-
c
0
Let F (s) = L{f (t)} be the Laplace transform of f (t)
...
0
c
Let y = t − c, so t = y + c, and dt = dy, and we continue
Z ∞
Z ∞
−s(y+c)
−sc
L{g(t)} =
e
f (y) dy = e
e−sy f (y) dy = e−cs F (s)
...
Note now we are only considering the domain t ≥ 0
...
In following examples we will compute Laplace transform of piecewise continuous functions with the help of the unit step function
...
Given
f (t) =
sin t,
sin t + cos(t − π ),
4
π
0≤t< ,
4
π
≤ t
...
4
(Or, if we write out each intervals
π
π
f (t) = sin t(1 − u π4 (t)) + (sin t + cos(t − ))u π4 (t) = sin t + u π4 (t) · cos(t − )
...
)
And the Laplace transform of f is
F (s) = L{sin t} + L{u π4 (t) · cos(t −
π
1
s
π
)} = 2
+ e− 4 s 2
...
Given
f (t) =
t, 0 ≤ t < 1,
1, 1 ≤ t
...
83
The Laplace transform is
L{f (t)} = L{t} − L{(t − 1)u1 (t)} =
1
1
− e−s 2
...
Given
f (t) =
0,
0 ≤ t < 2,
t + 3, 2 ≤ t
...
The Laplace transform is
L{f (t)} = L{(t − 2)u2 (t)} + 5L{u2(t)} = e−2s
1
1
+ 5e−2s
...
Given
g(t) =
1, 0 ≤ t < 2,
t2 , 2 ≤ t
...
Observe that
t2 − 1 = (t − 2 + 2)2 − 1 = (t − 2)2 + 4(t − 2) + 4 − 1 = (t − 2)2 + 4(t − 2) + 3 ,
we have
g(t) = 1 + (t − 2)2 + 4(t − 2) + 3 u2 (t)
...
Given
2
4
3
+
+
s3 s2 s
0, 0 ≤ t < 3,
et , 3 ≤ t < 4,
f (t) =
0, 4 ≤ t
...
We can rewrite it in terms of the unit step function as
f (t) = et (u3 (t) − u4 (t)) = u3 (t)et−3 e3 − u4 (t)et−4 e4
...
− e4 e−4s
=
s−1
s−1
s−1
Inverse transform: We use two properties:
1
L{uc (t)} = e−cs ,
s
and
L{uc (t)f (t − c)} = e−cs · L{f (t)}
...
Example 10
...
3
s
s
s
1 2
−1 1
We know that L { s3 } = 2 t , so we have
1 2
t,
0 ≤ t < 2,
1
1 2
2
2
−1
f (t) = L {F (s)} = t − u2 (t) (t − 2) =
2
2
1 t2 − 1 (t − 2)2 , 2 ≤ t
...
Given
e−3s
1
F (s) = 2
= e−3s
= e−3s
s + s − 12
(s + 4)(s + 3)
B
A
+
s+4 s−3
...
So
1
f (t) = L−1 {F (s)} = u3 (t) Ae−4(t−3) + Be3(t−3) = u3(t) −e−4(t−3) + e3(t−3)
7
which can be written as a p/w continuous function
0 ≤ t < 3,
0,
f (t) =
− 1 e−4(t−3) + 1 e3(t−3) ,
3 ≤ t
...
Given
se−s
s+2−2
s+2−2
−s s + 2 − 2
−s
F (s) = 2
...
6
...
Next we study initial value problems with discontinuous force
...
Example 1
...
y + y + y = g(t),
g(t) =
1, 1 ≤ t,
Answer
...
Also we have L{g(t)} = L{u1 (t)} = e−s 1s
...
2
s(s + s + 1) s + s + 1
Now we need to find the inverse Laplace transform for Y (s)
...
We have
1
A
Bs + C
=
+
...
87
Compare s2 -term: 0 = A + B, so B = −A = −1
...
So
−s
Y (s) = e
We work out some detail
1
s+1
− 2
s s +s+1
+
s2
s+1
...
2
2
3
We conclude
!#
√
3
3
(t − 1) − sin
(t − 1)
y(t) = u1 (t) 1 − e
cos
2
2
"
√
√ #
1
3
1
3
+e− 2 t cos
t + √ sin
t
...
Extra question: What happens when t → ∞?
Answer: We see all the terms with the exponential function will go to zero,
so y → 1 in the limit
...
Since g(t) becomes constant 1 for large t, and the particular
solution (which is also the steady state) with 1 on the right hand side is 1,
which provides the limit for y
...
88
• Actually the solution consists of two part: the forced response and the
homogeneous solution
...
Example 2
...
0, 2π ≤ t,
Rewrite
1
1
L{g} = e−πs − e−2π
...
s
s
e−π
e−2π
s
e−π − e−2π
+
=
−
+ 2
...
1
1 1
}
=
− cos 2t
...
In Example 2, let
0, 0 ≤ t < 4,
et , 4 ≤ 5 < 2π,
g(t) =
0, 5 ≤ t
...
Answer
...
s−1
s−1
Take Laplace transform of the equation, we get
G(s) = L{g(t)} = e4 e−4s
(s2 +4)Y (s) = G(s)+s,
Y (s) = e4 e−4s − e5 e−5s
s
1
+
...
The students may work out the inverse transform as a practice
...
(Undamped system with force, example 2 from the book p
...
Let’s first work on g(t) and its Laplace transform
g(t) =
t−5
1
1
(u5 (t) − u10 (t)) + u10 (t) = u5 (t)(t − 5) − u10 (t)(t − 10),
5
5
5
1
1
1
1
G(s) = L{g} = e−5s 2 − e−10s 2
5
s
5
s
Let Y (s) = L{y}, then
(s2 +4)Y (s) = G(s),
Y (s) =
G(s)
1
1
1
1
= e−5s 2 2
− e−10s 2 2
2
s +4
5
s (s + 4) 5
s (s + 4)
90
Work out the partial fraction:
...
So
1
1 1
1
1
2
1
...
· 2− · 2
2
2
2
s (s + 4)
4 s
8 s +2
4
8
Go back to y(t)
1
1
y(t) = L−1 {Y } = u5 (t)h(t − 5) − u10 (t)h(t − 10)
5
5
1
1
1
1
1
1
=
u5 (t) (t − 5) − sin 2(t − 5) − u10 (t) (t − 10) − sin 2(t − 10)
5
4
8
5
4
8
0,
0 ≤ t < 5,
1
1
(t − 5) − 40
sin 2(t − 5),
5 ≤ 5 < 10,
=
20
1
1
− 40
(sin 2(t − 5) − sin 2(t − 10)), 10 ≤ t
...
1
4
+ R · cos(2t + δ) for some amplitude
The plots of g and y are given in the book
...
During the interval 0 < t < 5,
g = 0 and initial conditions are all 0
...
For large time
t, g = 1
...
Adding the homogeneous solution,
we should have y = 14 + c1 sin 2t + c2 cos 2t for t large
...
91
Chapter 7
...
1: Introduction to systems of differential equations
Given
ay ′′ + by ′ + cy = g(t),
y ′ (0) = β
y(0) = α,
we can do a variable change: let
x2 = x′1 = y ′
x1 = y,
then
(
x′1 = x2
x′2
1
= y = (g(t) − bx2 − cx1 )
a
′′
x1 (0) = α
x2 (0) = β
Observation: For any 2nd order equation, we can rewrite it into a system of
2 first order equations
...
Given
y ′′ + 5y ′ − 10y = sin t,
y(0) = 2,
y ′ (0) = 4
Rewrite it into a system of first order equations: let x1 = y and x2 = y ′ = x′1 ,
then
′
x1 (0) = 2
x1 = x2
I
...
’s:
x2 (0) = 4
x′2 = y ′′ = −5x2 + 10x1 + sin t
We can do the same thing to any high order equations
...
...
Reversely, we can convert a 1st order system into a high order equation
...
Given
′
x1 = 3x1 − 2x2
x′2 = 2x1 − 2x2
x1 (0) = 3
x2 (0) = 21
Eliminate x2 : the first equation gives
3
1
x2 = x1 − x′1
...
This we know how to solve!
Definition of a solution: a set of functions x1 (t), x2 (t), · · · , xn (t) that satisfy
the differential equations and the initial conditions
...
2: Review of matrices
A matrix of size m × n:
a1,1 · · · a1,n
A =
...
We consider only square matrices, i
...
, m = n, in particular for n = 2 and 3
...
• Addition: A + B = (aij ) + (bij ) = (aij + bij )
• Scalar multiple: αA = (α · aij )
• Transpose: AT switch the ai,j with aji
...
• Product: For A · B = C, it means ci,j is the inner product of (ith row
of A) and (jth column of B)
...
Example 1
...
′
x1 = a(t)x1 + b(t)x2 + g1 (t)
x′2 = c(t)x1 + d(t)x2 + g2 (t)
⇒
x1
x2
′
1 −1 3
x1
4
2 0
5
0
· x2
=
0 1 −1
x3
7
=
Some properties:
• Identity I: I = diag(1, 1, · · · , 1), AI = IA = A
...
x y z
• Inverse inv(A) = A−1 : A−1 A = AA−1 = I
...
– (6) All eigenvalues of A are non-zero
...
3: Eigenvalues and eigenvectors
Eigenvalues and eigenvectors of A (A is 2 × 2 or 3 × 3
...
If A~v = λ~v , then (λ, ~v) is the (eigenvalue, eigenvector) of A
...
Remark: If ~v is an eigenvector, then α~v for any α 6= 0 is also an eigenvector,
because
A(α~v) = αA~v = αλ~v = λ(α~v)
...
We see that det(A − λI) is a polynomial of degree 2 (or 3) in λ, and it is also
called the characteristic polynomial of A
...
Example 1: Find the eigenvalues and the eigenvectors of A where
1 1
A=
...
Let’s first find the eigenvalues
...
Now, let’s find the eigenvector ~v1 for λ1 = −1: let ~v1 = (a, b)T
0
a
1 − (−1)
1
,
=
·
(A − λ1 I)~v1 = 0, ⇒
0
b
4
1 − (−1)
0
a
2 1
,
=
·
⇒
0
b
4 2
so
1
...
2
Example 2
...
2 −9
A=
...
2 − λ −9
= (2 −λ)2 + 36 = 0,
det(A−λI) = det
4
2−λ
⇒
λ1,2 = 2 ±6i
¯ 1 , complex conjugate
...
e
...
So we need to only find one
...
~v1 =
2
i
− 32 i
3
−6iv 1 − 9v 2 = 0,
so
97
7
...
If ~g = 0, it is homogeneous
...
Superposition: If ~x1 (t) and ~x2 (t) are two solutions of the homogeneous
system, then any linear combination c1~x1 + c2~x2 is also a solution
...
If det X(t) 6= 0, then ~x1 (t), ~x2 (t), · · · , ~xn (t) is a set of linearly independent
functions
...
The general solution is the linear combination of these solutions, i
...
~x = c1~x1 (t) + c2~x2 (t) + · · · + cn~xn (t)
...
5: Homogeneous systems of two equations
with constant coefficients
...
C
...
Claim: If (λ, ~v) is an eigen-pair for A, then ~z = eλt~v is a solution to ~x′ = A~x
...
~z′ = (eλt~v )′ = (eλt )′~v = λeλt~v
A~z = A(eλt~v ) = eλt (A~v ) = eλt λ~v
Therefore ~z′ = A~z so ~z is a solution
...
• Step II: Find the corresponding eigenvectors ~v1 , ~v2
...
• Step IV: Check that ~z1 , ~z2 are linearly independent: the Wronskian
W (~z1 , ~z2 ) = det(~z1 , ~z2 ) 6= 0
...
)
• Step V: Form the general solution: ~x = c1~z1 + c2~z2
...
99
We will start with an example
...
Solve
′
~x = A~x,
A=
1 1
4 1
...
By an example in 7
...
x2 (t) = −2c1 e−t + 2c2 e3t
Qualitative property of the solutions:
• What happens when t → ∞?
If c2 > 0, then x1 → ∞, x2 → ∞
...
Asymptotic relation between x1 , x2 : look at
As t → ∞, we have
x1
:
x2
c1 e−t + c2 e3t
x1
=
...
3t
x2
2c2 e
2
This means, x1 → 2x2 asymptotically
...
100
1
2
...
is the trajectories of various solutions in the x2 − x1 plane
...
x1
c2 e3t
1
• If c1 = 0, then
=
=
, so the trajectory is a straight line
x2
2c2 e3t
2
x1 = 2x2
...
Since λ2 = 3 > 0, the trajectory is going away from 0
...
Note that this is exactly the direction of ~v1
...
−t
• For general cases where c1 , c2 are not 0, the trajectories should start
(asymptotically) from line x1 = −2x2 , and goes to line x1 = 2x2 asymptotically as t grows
...
A saddle point is unstable
...
To draw the phase portrait, we follow
these guidelines:
• The general solution is
~x = c1 eλ1 t~v1 + c2 eλ2 t~v2
...
We see that the solution
vector is a scalar multiple of ~v2
...
Since λ2 > 0, solutions |~x| → ∞ along this
line, so the arrows are pointing away from the origin
...
We see that the solution vector is a scalar multiple of ~v1
...
Since λ1 < 0,
solutions approach 0 along this line, so the arrows are pointing toward
the origin
...
We need to draw at
least one trajectory in each region
...
e
...
We need to know the asymptotic behavior
...
This
means, all trajectories come from the direction of ~v1 , and will approach
~v2 as t grows
...
102
x2
6
~v2
U
~v1
K
q
- x1
~v2
i
K
~v1
Example 2
...
,
λ1 = −3,
~v2 =
λ1 = 3,
~v1 =
0
−1
Then the phase portrait looks like this:
x2
6
~v1
I
6
Y
-
j
~v2 x
1
?
R
~v1
103
I
-~
v2
If the two real distinct eigenvalue have the same sign, the situation is quite
different
...
Consider the homogeneous system
−3 2
′
...
Answer
...
(Two eigenvalues are both negative!)
• Find the eigenvector for λ1
...
1
−2 + 1
b
1 −1
b
0
This gives a = b
...
• Find the eigenvector for λ2
...
=
·
=
·
(A−λ2 I)~v1 =
0
d
1 2
d
1
−2 + 4
This gives c + 2d = 0
...
So ~v2 = (−2, 1)T
...
x2 (t) = c1 e−t + c2 e−4t
104
−4t
−2
1
...
Since λ2 < 0, the arrows point
toward the origin
...
Since λ1 < 0, the arrows point
toward the origin
...
See the plot below
...
105
Definition: If λ1 6= λ2 are real with the same sign, the critical point ~x = 0
is called a node
...
If λ1 < 0, λ2 < 0, this node is called a sink
...
Example 4
...
, ~v2 =
λ1 = 3, λ2 = 4, ~v1 =
−3
2
(1) Find the general solution for ~x′ = A~x, (2) Sketch the phase portrait
...
(1) The general solution is simple, just use the formula
1
1
4t
λ1 t
λ2 t
3t
...
As t → −∞, ~x → c1 eλ1 t~v1
...
x2
6
~v1
M
~v2
M
~v1
6
- x1
?
N
~v2
Summary:
(1)
...
If λ1 and λ2 are real and with same sign: the origin is a node
...
6: Complex eigenvalues
If A has two complex eigenvalues, they will be a pair of complex conjugate
numbers, say λ1,2 = α ± iβ, β 6= 0
...
e,
¯ 2
...
They are complex-valued functions, and they also are complex conjugate
...
By the principle of superposition,
1
~y1 = (~z1 + ~z2 ) = Re(~z1 ),
2
~y2 =
1
(~z1 − ~z2 ) = Im(~z1 )
2i
are also two solutions, and they are real-valued
...
The general solution is then ~x = c1 ~y1 + c2~y2
...
We have two eigenval¯ two eigenvectors: ~v and ~v¯, which we can write
ues: λ and λ,
λ = α + iβ,
~v = ~vr + i~vi
...
The general solution is
~x = c1 eαt (cos βt · ~vr − sin βt · ~vi ) + c2 eαt (sin βt · ~vr + cos βt · ~vi )
...
e
...
The ~x is a
harmonic oscillation, which is a periodic function
...
108
Example 1
...
) Find the general solution and
sketch the phase portrait of the system:
0 −4
′
~x = A~x,
A=
...
First find the eigenvalues of A:
det(A − λI) = λ2 + 4 = 0,
λ1,2 = ±2i
...
=
·
(A − λI)~v = 0,
0
b
1 −2i
a − 2ib = 0,
then
~v =
choose b = 1, then a = 2i,
2i
1
=
0
1
+i
2
0
...
+ cos 2t ·
+c2 sin 2t ·
− sin 2t ·
~x = c1 cos 2t ·
0
1
0
1
Write out the components, we get
x1 (t) = −2c1 sin 2t + 2c2 cos 2t
x2 (t) = c1 cos 2t + c2 sin 2t
...
• They do not intersect with each other
...
• They are ellipses
...
109
• The arrows are pointing either clockwise or counter clockwise, determined by A
...
By the differential equations, we get ~x ′ = A~x = (0, 1)T , which is a
vector pointing upward
...
See plot below
...
The origin in this case is called a center
...
If the complex eigenvalues have non-zero real part, the situation is still different
...
Consider the system
′
~x = A~x,
A=
3 −2
4 −1
...
Eigenvectors: need to compute only one ~v = (a, b)T
...
Choosing a = 1, then b = 1 − i, so
1
1
0
~v =
=
+i
...
+ c2 et
= c1 et
sin 2t − cos 2t
cos 2t + sin 2t
Phase portrait
...
If this term
is not present, (i
...
, the eigenvalues would be pure imaginary), then the
solutions are perfect oscillations, whose trajectory would be closed curves
around origin, as the center
...
Since α = 1 > 0, all arrows are pointing away from the origin
...
Consider the point (x1 = 1, x2 = 0), then ~x′ = A~x = (3, 4)T
...
At the point ~x = (0, 1)T , we have ~x′ = (−2, −1)T
...
We don’t stress on the
exact shape of the spirals
...
111
2
...
5
1
x2
0
...
5
−1
−1
...
5
−2
...
5
−1
−0
...
5
1
1
...
5
In this case, the origin (the critical point) is called the spiral point
...
Remark: If α < 0, then all arrows will go towards the origin
...
An example is provided in the text book
...
Example 3
...
The eigenvalues and eigenvectors are:
1
0
1
1
λ1,2 = − ± i,
...
We know that one solution is
λ1 t
~z = e
−( 12 +i)t
~v1 = e
112
1
0
±i
0
01
...
We know the real part and the imaginary
part are both solutions, so work them out:
1
0
1
0
− 12 t
~z = e
cos t
− sin t
+ i sin t
+ i cos t
...
If c2 = 0, we have
1
x21 + x22 = (e− 2 t )2 c21
...
The trajectories will be spirals, with arrows pointing toward the origin
...
0
−1
So the spirals rotate clockwise
...
See the picture below
...
8: Repeated eigenvalues
Here we study the case where the two eigenvalues are the same, say λ1 =
λ2 = λ
...
Example 1
...
so λ1 = λ2 = 2
...
1
1
b
1
...
~z1 = e ~v = e
−1
We need to find a second solution
...
We have
~z′ = eλt~v + λteλt~v = (1 + λt)eλt~v
A~z2 = Ateλt~v = teλt (A~v ) = teλt λ~v = λteλt~v
If ~z2 is a solution, we must have
~z′ = A~z
→
1 + λt = λt
which doesn’t work
...
(here ~η is a constant vector to be
determined later)
...
115
Since ~z2 is a solution, we must have ~z′ = A~z
...
This is what one uses to solve for ~η
...
Back to the original problem, to compute this ~η , we plug in A and λ, and
get
1
η1
−1 −1
, η1 + η2 = −1
...
We can choose η1 = 0, then η2 = −1, and so ~η =
−1
So the general solution is
~x = c1~z1 + c2~z2 = c1 eλt~v + c2 (teλt~v + eλt ~η )
0
1
1
2t
2t
2t
...
• As t → −∞, we have ~x → 0
...
Since λ > 0, the arrows point away from the origin
...
For this solution, as t → ∞, the
dominant term in ~x is teλt~v
...
On the other hand, as t → −∞, the dominant term in ~x
is still teλt~v
...
But,
due to the change of sign of t, the ~x will change direction and point
toward the opposite direction as when t → ∞
...
At ~x = (1, 0), we have ~x′ = (1, 1)T , and at ~x = (0, 1),
we have ~x′ = (−1, 3)T
...
See
figure below
...
As t → ∞, the dominant term in ~x is teλt~v
...
As t → −∞, the dominant term in ~x is
still teλt~v
...
But,
due to the change of sign of t, the ~x will change direction and point
toward the opposite direction as when t → ∞
...
5
4
3
2
x2
1
0
−1
−2
z2
−3
z1
−4
−5
−5
−4
−3
−2
−1
0
x1
1
2
3
4
5
Remark: If λ < 0, the phase portrait looks the same except with reversed
arrows
...
If A has repeated eigenvalues, the origin is called a improper
node
...
′
Example 2
...
5 −4
det(A−λI) = (−2−λ)(−4−λ)+1 = λ2 +6λ+8+1 = (λ+3)2 = 0,
117
~x
...
The corresponding eigenvector ~v = (a, b)T
−2 + 3
2
a
1
2
a
(A − λI)~v =
·
=
·
=0
−0
...
5 −1
b
So we
must have a + 2b = 0
...
To find the generalized eigenvector ~η, we solve
~v =
−1
2
1
2
η1
(A − λI)~η = ~v ,
·
=
...
5 −1
η2
−1
This gives us
onerelation η1 + 2η2 = 2
...
The general solution is
and so ~η =
1
0
2
2
3t
3t
λt
λt
λt
3t
...
5
4
3
2
x2
1
0
−1
z2
−2
z1
−3
−4
−5
−5
−4
−3
−2
−1
0
x1
1
The origin is an improper node which is unstable
Title: Ordinary Differential Equations (ODE)
Description: Well comprehensive notes on Ordinary Differential Equations (ODE)
Description: Well comprehensive notes on Ordinary Differential Equations (ODE)