LECTURE 25: SYSTEMS OF ODES AND EIGENVALUE METHOD
MINGFENG ZHAO
November 09, 2015
In this course, we only study the linear system in the following form:

 x0 = ax1 + bx2 + f1 (t)
1
 x0 = cx + dx + f (t)
1
2

Let ~x(t) = 
x1 (t)

,

A=
x2 (t)
a
b
c
d

2

and f~(t) = 
,
2
f1 (t)

, then we have ~x0 = A~x + f~(t) .
f2 (t)
Theorem 1. Let ~xc (t) be the general solution to the homogeneous system ~x0 = A~x, and ~xp (t) be a particular solution
to ~x0 = A~x + f~(t), then the general solution to ~x0 = A~x + f~(t) is:
~x(t) = ~xc (t) + ~xp (t).
Homogeneous system

Theorem 2. Let A be a 2 × 2 matrix, and ~y (t) = 
y1 (t)


 and ~z(t) = 
y2 (t)
z1 (t)

 be two linearly independent solutions
z2 (t)
to ~x0 = A~x, then the general solution to ~x0 = A~x is
~x(t) = C1 ~y (t) + C2 ~z(t).
~
In this case, the 2 × 2 matrix X(t)
= [~y (t) ~z(t)] is called a fundamental matrix of the system ~x0 = A~x. It’s easy to
~ 0 (t) = AX(t),
~
see that X
where

~ 0 (t) = 
X
y1 (t) z1 (t)
0

 := 
y2 (t) z2 (t)
y10 (t) z10 (t)
y20 (t) z20 (t)

.
Definition 1. Let ~x1 (t), · · · , ~xn (t) be n many vector functions, we say that ~x1 (t), · · · , ~xn (t) are linearly dependent if
there exist some constants C1 , · · · , Cn such that
• not all of Ci ’s are zero, that is, there exists some 1 ≤ i0 ≤ n such that Ci0 6= 0.
1
2
MINGFENG ZHAO
• C1 ~x1 (t) + · · · + Cn ~xn (t) = 0.
If ~x1 (t), · · · , ~xn (t) are not linearly dependent, then we say they are linearly independent.
Remark 1. It’s easy to see that for two vectors ~x(t) and ~y (t) are linearly dependent if and only if either ~x(t) is a
constant multiple of ~y (t), or either ~y (t) is a constant multiple of ~x(t).

Example 1. Let ~x1 (t) = 
1


 et , and ~x2 (t) = 
1

1
−1
 et , it’s easy to see that we can not find a constant C such that
~x1 (t) = C~x2 (t) or ~x2 (t) = C~x1 (t), that is, ~x1 (t), ~x2 (t) are linearly independent .

Example 2. Let ~x1 (t) = 
t2


, ~x2 (t) = 
t

0

, and ~x3 (t) = 
−t2
1+t

, then it’s easy to see that ~x1 (t) =
1
~x2 (t) − ~x3 (t), that is, ~x1 (t), ~x2 (t), ~x3 (t) are linearly dependent .
General Approach: Let C1 , C2 and C3 be such that
C1 ~x1 (t) + C2 ~x2 (t) + C3 ~x3 (t) = 0.
That is,

C1 
t2


 + C2 
t
0


 + C3 
1+t
−t2


=
1
0

.
0
So we get


C1 t2 − C3 t2

=

C1 t + C2 (1 + t) + C3
0

.
0
Then
C1 t2 − C3 t2 = 0,
and C1 t + C2 (1 + t) + C3 = 0.
So we get
C1 = C3 ,
C1 + C2 = 0,
and C2 + C3 = 0.
Then
C1 = C3 = −C2 .
If we take C1 = 1, then C2 = −1 and C3 = 1, that is, ~x1 (t) − ~x2 (t) + ~x3 (t) = 0, which implies that
~x1 (t), ~x2 (t), ~x3 (t) are linearly dependent .
Eigenvalue method for the general homogeneous system
LECTURE 25: SYSTEMS OF ODES AND EIGENVALUE METHOD
3
Definition 2. Let A be a 2 × 2 matrix, suppose that there is a scalar λ and a nonzero vector ~v such that A~v = λ~v . In
this case, λ is called an eigenvalue of A, and ~v is called an eigenvector corresponding to λ.
Theorem 3 (Eigenvalue Method). Let λ be an eigenvalue of A and ~v be an eigenvector corresponding to λ, then
~x(t) = eλt~v is a solution to ~x0 = A~x.
Proof. In fact, we have
~x0 (t) = λeλt~v = eλt A~v = A(eλt~v ) = A~x(t).
Theorem 4. Let A be a 2 × 2 matrix, λ is an eigenvalue of A if and only if det (A − λI2 ) = 0, where I2 is the 2 × 2
identity matrix, that is,

I2 = 

Remark 2. Notice that if A = 
a
b
c
d
1
0
0
1

.

, then

det (A − λI2 )
=
det 
a−λ
c
b
d−λ


=
(a − λ)(d − λ) − bc
=
λ2 − (a + d)λ + ad − bc.
The polynomial χA (λ) := λ2 − (a + d)λ + ad − bc is called the characteristic polynomial of A. Hence the complex
number λ is an eigenvalue of A if and only if λ is a root of the characterisitc polynomial of A.
Remark 3. For eigenvalues λ of a 2 × 2 matrix A, since the characteristic polynomial of A is a quadratic polynomial,
that is, χA (λ) = λ2 − (a + d)λ + ad − bc, then we have three cases:
I. A has two different real roots λ1 and λ2 , in this case, eigenvectors ~v 1 and ~v 2 are real.
II. A hs two different complex roots λ1 and λ2 , in this case, eigenvectors ~v 1 and ~v 2 are complex.
III. A has the same real root λ (the most complicated case), in this case, eigenvector ~v is real.
Eigenvalue method with distinct real eigenvalues
4
MINGFENG ZHAO
Theorem 5. Let A be a 2 × 2 matrix with two distinct real eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then the general solution to ~x0 = A~x is:
~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 .
Remark 4. The non-zero eigenvectors corresponding to different eigenvalues are linearly independent.
Example 3. Find the general solution to the system:

Let ~x(t) = 
x1 (t)


, A = 
x2 (t)

2
1
0
Since
1
A=
2
1
0
1
x01
=
2x1 + x2
x02
=
x2 .

, then we need to solve ~x0 = A~x. First, we need to find the eigenvalues of

 and corresponding eigenvectors.

det (A − λI2 ) = det 
2
1
0
1



 − λI2  = det 
2−λ
1
0
1−λ
then

I. When λ1 = 1, let’s solve 
2
0
λ1 = 1,
 
and λ2 = 2.

1
x1
x1

=
, that is,
1
x2
x2


1
1
0
0


x1



=
x2
0
0
Then we know that

x1

x2

So 
1
−1


 = x1 
1
−1

 is an eigenvector corresponding to λ1 = 1.

.

.

 = (2 − λ)(1 − λ) = 0,
LECTURE 25: SYSTEMS OF ODES AND EIGENVALUE METHOD

2
II. When λ1 = 2, let’s solve 
1


x1
1
x1
 = 2

0

x2

5

, that is,
x2
0

1

−1
0
x1



0
=
x2

.
0
Then we know that


x1

 = x1 

x2

So 
1

1
.
0

 is an eigenvector corresponding to λ1 = 2.
0

In summary, the eigenvalues and corresponding eigenvectors of 

λ1 = 1 with 
2
1
0
1

1
−1

 are:

 and λ2 = 2 with 
1

.
0
Therefore, the general solution to ~x0 = A~x is:

x1 (t)



 = C 1 et 
−1
x2 (t)

Example 4. Solve ~x0 = 
2
2
1
3

1

 + C2 e2t 
1


=
0
C1 et + C2 e2t
−C1 et

.

 ~x.

First, let’s find eigenvalues of A := 
2
2
1
3

det (A − λI2 ) = det 

, we need to solve
2−λ
2
1
3−λ

 = (2 − λ)(3 − λ) − 2 = λ2 − 5λ + 4 = 0.
Then
λ1 = 1,
and λ2 = 4.
For λ1 = 1, let’s find the eigenvalue corresponding to λ1 = 1, we need to solve A~x = ~x, that is,


 

−1 −2
x1
0


=
.
−1 −2
x2
0
6
MINGFENG ZHAO

Then 
x1


 = x2 
x2
−2

. That is,
1

−2


 is an eigenvalue corresponding to λ1 = 1.
1
For λ2 = 4, let’s find the eigenvalue corresponding to λ2 = 4, we need to solve A~x = 4~x, that is,


  
2 −2
x1
0


 =  .
−1 1
x2
0

Then 
x1


 = x1 
x2
1

, that is,
1

1


 is an eigenvalue corresponding to λ2 = 4.
1

Therefore, the general solution to ~x0 = 
2
2
1
3

 ~x is:

~x(t) = C1 et 
−2


 + C2 e4t 
1
1

.
1
Problems you can do:
Lebl’s Book [2]: Exercise 3.4.6, 3.4.7, 3.4.103 on Page 108 and Page 109.
Braun’s Book [1]: All exercises on Page 340 and Page 341. Read all materials in Section 3.8.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
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