LECTURE 28: NONHOMOGENEOUS SYSTEMS
MINGFENG ZHAO
November 12, 2014
For the the two dimensional autonomous linear system ~x0 = A~x, where A is a 2 × 2 constant matrix. Assume A has
two distinct nonzero eigenvalues λ1 6= λ2 , the behavior of solutions to the two dimensional linear homogeneous system
~x0 = A~x:
Eigenvalues
Behavior
I.
real and both positive
source(unstable)
II.
real and both negative
sink (stable)
III. real and opposite signs
saddle
IV.
purely imaginary
center point (ellipses)
V.
complex with positive real part
spiral source
VI.
complex with negative real part
spiral sink
Undetermined coefficients
Let A be a 2 × 2 matrix,, consider the system:
~x0 = A~x + f~(t).
Let ~a, ~b be two constant vectors, pn (t) and p̃n (t) be polynomials with degree n. If f~(t) has the form:
~a pn (t)emt cos(kt) + ~b p̃n (t)emt sin(kt),
then a particular solution ~xp (t) to ~x0 = A~x + f~(t) can be take as:
~xp (t) =
n+α
X
~vi ti emt cos(kt) +
i=0
n+α
X
w
~ i ti emt sin(kt),
i=0
where
• α is one of 0, 1 and 2 (α is the multiplicity of m + ki as the eigenvalues to A):
– If m + ki is not an eigenvalue of A, then α = 0.
1
2
MINGFENG ZHAO
– If m + ki is an eigenvalue of Aand A has two different eigenvalues, then α = 1.
– If m + ki is an eigenvalue of A and A has only one eigenvalue, then α = 2.
• ~vi and w
~ i are undetermined constant vectors.




t
−1 0
e
. Find a particular solution of ~x0 = A~x + f~(t), where f~(t) = 
.
Example 1. Let A = 
−2 1
t


λ+1
0
 = (λ + 1)(λ − 1) = 0, then λ1 =
First, let’s find the eigenvalues of A, that is, det (λI2 − A) = det 
2
λ−1
and λ2 = −1. Notice that

 



f~(t) = 
et
1
=
1
0
 et + 
0
 t.
1
So we can let ~xp (t) = ~atet + ~bet + ~ct + d~ be a particular solution to ~x0 = A~x = f~(t), then
~x0p (t)
= ~aet + ~atet + ~bet + ~c
= A~xp (t) + f~(t)

= A~atet + A~bet + A~ct + Ad~ + 
1


 et + 
0
0

 t.
1
Then we have

A~a = ~a,
A~b + 
1


 = ~a + ~b,
A~c + 
0
0

 = 0,
~
and ~c = Ad.
1
Solve ~a, ~b, ~c, m and n, we can take




1
0
 , ~b =  2  ,
~a = 
−1
0

~c = 
0
−1


,
and d~ = 

0
−1
.
So a particular solution to ~x0 = A~x + f~(t) is:

~xp (t) = 
0
−1


 tet + 
1
2
0


 et + 
0
−1


t + 
0
−1


=
1 t
2e
−tet − t − 1

.
Variation of Parameters
Let ~x1 (t) and ~x2 (t) be two linearly independent solutions to ~x0 = A(t)~x, then X(t) = [~x1 ~x2 ] is a fundamental matrix,
that is, X 0 (t) = A(t)X(t). Let’s consider
~x0 = A(t)~x + f~(t).
LECTURE 28: NONHOMOGENEOUS SYSTEMS
3
Let ~xp (t) = X(t)~u(t) for some ~u(t) be a particular solution to ~x0 = A(t)~x + f~(t), then
~x0p = X 0 (t)~u(t) + X(t)~u0 (t) = A(t)~xp + f~(t) = A(t)X(t)~u(t) + f~(t).
Since X 0 (t) = A(t)X(t), then
A(t)X(t)~u(t) + X(t)~u0 (t) = A(t)X(t)~u(t) + f~(t).
That is, X(t)~u0 (t) = f~(t), which implies that
~u0 (t) = X(t)−1 f~(t).
Then
Z
~u(t) =
X(t)−1 f~(t) dt.
That is,
Z
~xp (t) = X(t)

Example 2. Let A = 
−1
0
−2
1
X(t)−1 f~(t) dt.


. Find a particular solution of ~x0 = A~x + f~(t), where f~(t) = 
et

.
t

First, let’s find the eigenvalues of A, that is, det (λI2 − A) = det 
λ+1
0
2
λ−1

 = (λ + 1)(λ − 1) = 0, then
and λ2 = −1.
λ1 = 1,
For λ1 = 1, let solve A~x = ~x, that is,

2

0


2

Then 
x1
x2


 = x2 
0


, tha is, 
1
0
x1
0


=
x2
0

.
0

 is an eigenvalue corresponding to λ = 1, which implies that
1

0



 et = 
1
0
e
t

 is a solution to ~x0 = A~x.
For λ2 = −1, let solve A~x = −~x, that is,

0

2
0
−2

x1

x2


=
0
0

.
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MINGFENG ZHAO

Then 
x1


 = x1 
x2
1


1
, tha is, 
1

 is an eigenvalue corresponding to λ = −2, which implies that
1

1




e−t
 e−t = 
1
e
 is a solution to ~x0 = A~x.
−t
Then we can take the fundamental matrix X(t) as:

X(t) = 
0
e−t
et
e−t

.
Then det X(t) = −1, and

X(t)−1 = − 
e−t
−e−t
−et
0


=
−e−t
e−t
et
0

.
Then
Z
~x(t)
=

=


=
0
e−t

et
e−t

0
e−t

t
−t

e

=
X(t)−1 f~(t) dt
X(t)

e



Z

−e−t
e−t
et
0
0
e−t
et
e−t

−1 + te−t

−te − t − 1 +
et
2t

 dt
t

 dt
−t − te−t − e−t


1 2t
2e

1 t
2e
t


e


=

Z
1 t
2e
.
So a particular solution to ~x0 = A~x + f~(t) is:

~xp (t) = 
1 t
2e
−tet − t − 1 + 12 et

.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca