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M392:
Linear systems of ordinary differential equations
Definitions:  is an eigenvalue of an m by m matrix A provided
the eigenvector equation Ax =  x with x = the m by 1 matrix
x1 x2 , , , , xm T has nontrivial solutions x . The trivial
solution is x = 0 = (0 0 ,,, 0). The eigenvector equation, as we
will see below, can be rewritten as a linear system of m equations
in m unknowns.
If  is an eigenvalue of A, then every solution x of Ax =  x is
called an eigenvector of A corresponding to  .
Problem:
 2 2

1) Find eigenvalues and eigenvectors of the matrix A = 
1
3


2) Use your answer to solve the system of ordinary differential
 y '  2 y1  2 y2
equations  1
for functions
y
'

y

3
y
 2
1
2
 y1 (0)  2
 y1 (t )
subject
to
initial
conditions


 y2 (t )
 y 2 (0)  1
Solution, Part I. Find eigenvalues and eigenvectors:
Method: To solve Ax =  x, move everything to the left side and
rewrite it as a system of linear equations as follows:

Ax  x  0

Ax  Ix  0

( A  I ) x  0
Recall that a system of m equations in m unknowns has exactly
one solution if the determinant of the coefficient matrix is nonzero,
and either no solutions or many solutions if that determinant is
zero.
Since the system of equations Ax =  x has a nontrivial solution x
by definition of eigenvalues, and also has the trivial solution, the
same is true of the rewritten system ( A  I ) x  0 . It follows that
this system has infinitely many solutions. Therefore its coefficient
matrix ( A  I ) has determinant zero. Calculate as follows.
| A  I | 0
 2 2  0 

  
  0
1
3
0


 

2
2
.
0
1
3
(2   )(3   )  2(1)  0
2  5  4  0
(  1)(  4)  0
and so the eigenvalues are   4 and   1.
For each eigenvalue  , solve the eigenvector equation
x 
( A  I ) x  0 where x   1  . In other words, solve
 x2 
2   x1   0 
2  

     
1
3



  x2   0 
Start with eigenvalue   4 and plug in to get
2   x1   0 
2  4

      which says that
1
3

4

  x2   0 
 2 x1  2 x2  0
 x1  x2  0
Row
ops



0  0
1x1  1x2  0
This is a very simple system with solutions
x2  arbitrary  r
x1  r
 x   r  1
Thus the eigenvalue   4 has eigenvectors  1      r   .
 x2   r  1
 2 2  r   4r 
     =4x
Check: Ax = 
1
3

 r   4r 
Warning: check that the coefficient matrix determinant
2 2
| ( A  4I ) | =
is really zero. If not, go back and look for
1 1
your mistake.
x  s
In the same way x1  2 x2  0 Solution  2
 x1  2s
 x    2s    2 
the eigenvalue   1 has eigenvectors  1   
  s 
x
s
  1 
 2 
y’ = 3y solution is y(t) = c e^(3t)
 y '1  2 y1  2 y 2
Solution, Part II. To solve the system 
let y=
y
'

y

3
y
 2
1
2
and write the system as
y′ = A y . Suppose A is a 1 by 1 matrix, ( a number) and
y = y(t). Then the solution of y′ = A y is y(t )  ce At
 y1 
 
 y2 
To solve the general case, Guess y = et c and plug into y′ = Ay :
t
 y1  t  c1   e c1 
   e     t 
 y2 
 c2   e c2 
e t c = A e t c Divide by e t to obtain
Ac =  c
Therefore,  is an eigenvalue of A with
eigenvector c and a solution is y = e t c .
Corresponding to the two eigenvalues (with corresponding
eigenvectors) for the matrix A we get two solutions to the system
of ODE’s:
 1
y 
For   4 with eigenvectors r   we have yI =  1 
 1
 y2 
 re 
=  4 t 
 re 
------------------------------------------------------------y 
  2
For   1 with eigenvectors s  we have yII =  1 
 1 
 y2 
4t
  2se
= 
 se
1t
1t



We found that the system of differential equations y′ = Ay has two
solutions yI and yII . Thus y ' I  Ay I and y ' II  Ay II
y I ' y II '  Ay I  Ay II
Add these equations:
( y I  y II )'  A( y I  y II )
It follows that yI  yII also solves y′ = Ay .
Thus we can add the above two solutions to get solutions
4t
t
 y1 (t )  re  2se
(*)

4t
t
y
(
t
)

re

se
 2
It’s possible to show that there are no other solutions, and so (*) is
called the general solution to the system of ODE’s.
 y (0)  2
Finally, plug the given initial conditions  1
into the
y
(
0
)

1
 2
2  r  2 s
general solution to get 
1  r  s
r  4 / 3
Solve for r and s to get 
 s  1 / 3
4 t 2 4t

y
(
t
)

e  e
1

3
3
Plug into (*) to get the particular solution 
4
1
 y 2 (t )  e t  e 4 t

3
3
Exercise: Check this solution by plugging into the original system
of ODE’s.
Sample problems
1. a) Find the eigenvalues and eigenvectors of the matrix
 5 2 

A = 
2

2


b) Use your answer from a) to solve the following system of
differential equations for y1 (t ) and y 2 (t ) subject to the initial
conditions y1 (0) = 1 and y 2 (0) = 2:

y1  5 y1  2 y 2

y 2  2 y1  2 y 2
2. Solve the following simultaneous differential equations for y1 (t )
and y 2 (t ) subject to initial conditions y1 (0) = 2 and y 2 (0) = -1 .
You must proceed by first finding the eigenvalues and
eigenvectors of an appropriate matrix. No credit for any other
method!

y1  2 y1  1 y2

y2  2 y1  5 y2