LECTURE 29: NONHOMOGENEOUS SYSTEMS
MINGFENG ZHAO
November 14, 2014
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) +
n+α
X
w
~ i ti emt sin(kt),
i=0
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.
– 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.
Remark 1. If you do not want to compute eigenvalues of A to determin α, you can just try the following particular
solution:
~xp (t) =
n+2
X
~vi ti emt cos(kt) +
i=0
n+2
X
w
~ i ti emt sin(kt)
i=0

Example 1. Find the general solution to ~x0 = 
1
4
1
−2


 ~x − 
et
e
1
t

.
2
MINGFENG ZHAO

Let A = 
1
4
1
−2


 and f~(t) = − 

et
. First, let’s find the eigenvalues of A, that is,
t
e

det (λI2 − A) = det 
λ−1
−1
−4
λ+2

 = (λ − 1)(λ + 2) − 4λ2 + λ − 6 = 0.
Then
λ1 = −3,
and λ2 = 2.
For λ = −3, let’s solve A~x = −3~x, that is,

−4 −1


Then 
x1


 = x1 
x2
−4
−1

x1



=
x2
0

.
0

1
−4
, tha is,


1

−4
 is an eigenvalue for λ = −3.
For λ = 2, let’s solve A~x = 2~x, that is,



Then 
x1
x2


 = x1 
1
1
−1
−4
4

x1



=
x2
0

.
0

, that is,
1

1


 is an eigenvalue for λ = 2.
1

So the general solution to ~x0 = 
1
1
4
−2

 ~x is:

~x(t) = C1 
1
−4


 e−3t + C2 
1
1

 e2t .
LECTURE 29: NONHOMOGENEOUS SYSTEMS

Since − 
et
e
t


=
−1
−1
3

 et , let ~xp (t) = ~aet be a particular solution to ~x0 = A~x + f~(t), then

~x0p (t) = ~aet = A~aet + 
−1
−1

 et .
So we get

−1
(I2 − A)~a = 
−1

.
That is,



−1
0
a1

−4
3


=
a2
−1
−1

.
Then

~a = 
1

.
1
So the general solution to ~x0 = A~x + f~(t) is:

~x(t) = C1 
1
−4


 e−3t + C2 
1


 e2t + 
1
1


 et = 
1

C1 e−3t + C2 e2t + et
−3t
−4C1 e
2t
t
.
+ C2 e + e
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).
Let ~xp (t) = X(t)~u(t) for some ~u(t) be a particular solution to ~x0 = A(t)~x + f~(t), then 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)
X(t)−1 f~(t) dt.
4
MINGFENG ZHAO

Example 2. Find a particular solution to ~x0 = 
1

1

et
 ~x − 
t

.
4 −2
e
By the computations in Example 1, a fundamental matrix for ~x0 = A~x can be:


e−3t
e2t
.
X(t) = 
−3t
2t
−4e
e
Then
X(t)−1

2t
1  e
= −t
5e
4e−3t
−e2t
e−3t

1
Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 = 
4



e3t
−e3t
1
= 
.
5 4e−2t e−2t



1
et
 ~x − 
, then
t
−2
e

0
X(t)~u (t) = − 

et
.
et
Then

~u0 (t) = −X(t)−1 
et


e3t

−e3t
 = −1 
5 4e−2t
e
t
e

−2t
et


0


~u(t) =

Then a particular solution to ~x0 = 
~u0 (t) dt =
1
1
4
−2



 ~x − 

~xp (t) = X(t)~u(t) = 
et
et
−e−t


 dt = 
−3t
e2t
e
2t

0
Example 3. Solve ~x0 = 

Let A = 
1
1
4
−2
1
1
4
−2


 ~x − 
et
et
e−t

e
−t


.

=
Laplace transform for systems

0
 can be:
e−3t
−4e

0
Z
0

 = −1 
=
.
5 5e−t
e
−e−t
t
Then
Z


 and ~x(0) = 0.

~
 and X(s)
= L[~x(t)], then
~
~
L[~x0 (t)] = sX(s)
− ~x(0) = sX(s).
et
e
t

.
LECTURE 29: NONHOMOGENEOUS SYSTEMS

Apply the Laplace transform on the both sides of ~x0 = 
1
4
1
−2

~
~
sX(s)
= AX(s)
−


et
 ~x − 
t
5

, we get
e
L[et ]
t
L[e ]

.
By looking up the table, we have
L[et ] =
1
.
s−1
Then we get


s−1
−4
−1


~
 X(s)
= −
s+2
1
s−1
1
s−1

.
Then we get

s+2
1
~

X(s)
=−
(s − 1)(s + 2) − 4
4
1
s−1


1
s−1
1
s−1


1
=−

(s + 3)(s − 2)
s+3
s−1
s+3
s−1


 = −
1
(s−1)(s−2)
1
(s−1)(s−2)

.
By partial fractions, we have
1
1
1
=
−
.
(s − 1)(s − 2)
s−2 s−1
By looking the table, we have
1
1
= e2t , and L−1
= et .
s−2
s−1



1
et
 ~x − 
 and ~x(0) = 0 is:
−2
et
L−1

Therefore, the solution to ~x0 = 
1
4

~
~x(t) = L−1 [X(s)]
= −
e2t − et
2t
e −e
t


=
et − e2t
t
e −e
2t

.
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