First year: Matrices and differential equations
This is an application of the concept of diagonalisation to the study of
simultaneous differential equations.
Prerequisites:
-
introduction to differential equations
eigenvalues, eigenvectors
matrices: diagonalisation
(I)
Theory and example
Say you have to solve the following system of differential equations:
 u ' ( x)  u ( x)  v( x)
.

v' ( x)  4u ( x)  2v( x)
First, you write it in matrix form:
 u ( x) 
1 1 
 and let A  
 .
Let Y  
 v( x) 
 4  2
 u ' ( x) 
 and the system
Then Y '  
 v' ( x) 
 u ' ( x)  u ( x)  v( x)
becomes Y '  AY.

v' ( x)  4u ( x)  2v( x)
If the matrix A is diagonalisable, then it is actually quite easy to solve the
system of differential equations above.
We give the theory below for any n n diagonalisable matrix A , but keep in
mind that, as far as you are concerned, n  2 or n  3 .
Let A be a diagonalisable n n matrix and let Y be a vector whose
coordinates are differentiable functions, i.e.
 y1 ( x ) 


Y   ... .
 y ( x) 
 n 
To solve the system of differential equations Y '  AY :
(a) Find the matrix P that diagonalises A , i.e. the matrix P such that
D  P 1 AP , where D is a diagonal matrix.
(b) Make the change of variable Y  PU , Y '  PU ' . We then have
Y '  AY

 U '  P 1 APU
PU '  APU
 U '  DU .
(c) Solve the system U '  DU for U .
(d) The solution to the system of differential equation is Y  PU , where
U is the solution to U '  DU .
It looks more complicated than it actually is. Let’s see how it works with our
example:
First we have to find the matrix P , i.e. we have to find the eigenvalues and
1 1 
 .
eigenvectors of the matrix A  
 4  2
A  I 
1 
1
4
2
 2    6  (  2)(  3) ,
so that A has two eigenvalues:   2 and   3 .
 x
Let X   .
 y
AX  2 X
 x  y  0
 

 4x  4 y  0
AX  3 X
4 x  y  0
 

4 x  y  0
x y

y  4 x 
1
X    .
1
 1 
X    .
  4
1 1 
 and the diagonal matrix D is
The matrix P is therefore P  
1  4 
2 0 
 .
D  
 0  3
 u ( x) 
Let U   1 . The matrix equality U '  DU is equivalent to
 u 2 ( x) 
u1 ' ( x)  2u1 ( x)
,

u2 ' ( x)  3u2 ( x)
which is easily solved:
u1 ' ( x)  2u1 ( x)
 u1 ( x)  Ae2 x , u2 ( x)  Be 3 x

u2 ' ( x)  3u2 ( x)
 Ae2 x 
 U   3 x .
 Be 
Y  PU is the solution to Y '  AY and we have
1 1  Ae 2 x   Ae 2 x  Be 3 x 
.
 3 x    2 x
Y  PU  
3 x 
1  4  Be   Ae  4 Be 
If you are also given initial conditions, i.e. conditions of the form
 u (0)   1 
    ,
Y (0)  
 v(0)    1
you can find the value of A and the value of B :
3

A

1
A

B

1

 

5,
Y (0)     
 
2

1
A

4
B


1
 

B 
5

1  3e 2 x  2e 3 x 
.
and finally the solution to the initial value problem is Y   2 x
5  3e  8e 3 x 
(II)
Exercises
(1) Decide whether the following matrices are diagonalisable and, if
they are, find the matrix P that diagonalises them.
  3 2

C  
  2 1
 3 0 0


A   0 2 0 ,
 0 1 2


 2 0  2


B  0 3 0 
0 0 3 


 y1 ( x) 


(2) Solve Y '  AY , where Y   y 2 ( x)  and where
 y ( x) 
 3 
(i)
 3  2 0


A    2 3 0 ,
 0
0 5 

(ii)
0
4
 3


A    2  2  1 .
  2 0  3

