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We will be looking for a solution
 y  x, y  x,
1
2
, yn  x  
T
to the system of linear differential equations with constant
coefficients in a cannonical form
y1'  a11 y1  a12 y2 
 a1n yn  f1  x 
y2'  a21 y1  a22 y2 
 a2 n y n  f 2  x 
yn'  an1 y1  an 2 y2 
 ann yn  f n  x 
Satisfying the initial condition
 y1  x0  , y2  x0  ,
, yn  x0     b1 , b2 ,
T
, bn 
T
As the first step, we will find a general solution to the associated
homogeneous system
y1'  a11 y1  a12 y2 
 a1n yn
y2'  a21 y1  a22 y2 
 a2 n y n
yn'  an1 y1  an 2 y2 
 ann yn
It can be proved that such a system has n independent vector
solutions, which can be displayed as columns in the matrix
 y11
y
Y   21


 yn1
y12
y22
yn 2
y1n 
y2 n 



ynn 
 y11
y
Y   21


 yn1
y12
y22
yn 2
y1n 
y2 n 



ynn 
is called the fundamental solution matrix to the system
y1'  a11 y1  a12 y2 
 a1n yn
y2'  a21 y1  a22 y2 
 a2 n y n
yn'  an1 y1  an 2 y2 
 ann yn
 a11
a
Denoting A   21


 an1
The system
a1n 
a2 n 



ann 
a12
a22
an 2
y1'  a11 y1  a12 y2 
 a1n yn
y2'  a21 y1  a22 y2 
 a2 n y n
yn'  an1 y1  an 2 y2 
 ann yn
can be written as
y ,y ,
'
1
'
2

' T
n
,y
 A  y1 , y2 ,
, y2 
T
The derivative of the fundamental solution matrix
 y12'
 '
y21
'

Y 

 '
 yn1
y12'
'
y22
yn' 2
satisfies the matrix equation
Y '  AY
y1' n 
' 
y2 n 

' 
ynn 
Let us examine the matrix function
2
3
x
x
x
e xA  E  A  A2  A3 
1!
2!
3!
We have
de xA
x 2 x2 3
 A A  A 
dx
1!
2!
 Ae xA
so that e xA satisfies the matrix differential equation Y '  AY
xA
The matrix e is a fundamental solution matrix to the system
y1'  a11 y1  a12 y2 
 a1n yn
y2'  a21 y1  a22 y2 
 a2 n y n
yn'  an1 y1  an 2 y2 
 ann yn
It can be proved that any solution to the above system can be
written as
 y1 
 c1 
where
y 
c 
 2   e xA  2 
c1 , c2 , , cn
 
 
 
 
are constants
y
 n
 cn 
The general solution to the system
y1'  a11 y1  a12 y2 
 a1n yn  f1  x 
y2'  a21 y1  a22 y2 
 a2 n y n  f 2  x 
yn'  an1 y1  an 2 y2 
 ann yn  f n  x 
can be written as
 y1 
 c1   y1 
y 
c   
 2   e xA  2    y2 
 
   
 
   
 yn 
 cn   yn 

where y1 , y2 ,
, yn

T
is any solution to
We need to find a particular solution to the system
y1'  a11 y1  a12 y2 
 a1n yn  f1  x 
y2'  a21 y1  a22 y2 
 a2 n y n  f 2  x 
yn'  an1 y1  an 2 y2 
 ann yn  f n  x 
To this end we will again use the variation-of-constants method
Let Y be a fundamental matrix solution to the system
and
 c1  x  


 c2  x  




 cn  x  
y1'  a11 y1  a12 y2 
 a1n yn
y2'  a21 y1  a22 y2 
 a2 n y n
yn'  an1 y1  an 2 y2 
 ann yn
a vector of differentiable functions
Substituting
 y1 
 c1  x  
 


c2  x  
 y2 

  Y 

 


y 
 cn  x  
 n
into
y1'  a11 y1  a12 y2 
 a1n yn  f1  x 
y2'  a21 y1  a22 y2 
 a2 n y n  f 2  x 
yn'  an1 y1  an 2 y2 
 ann yn  f n  x 
 c1'  x  
 c1  x  
 c1  x    f1  x  
 '




 

c
x
c
x
f
x
c
x
   2  
2    2  

yields Y '  2
Y
 AY






 

 '




 

c
x
c
x
f
x
c
x
 n  
 n    n  
 n  
but, since Y is a solution to the associated homogeneous system,
we have Y '  AY and thus
 c1  x  
 c1  x  




c
x
c
x
2  
'  2  

Y
 AY








c
x
c
x
 n  
 n  
This leaves us with the system
 c1'  x    f1  x  
 '
 

f
x
c
x
2  
2  


Y


 

 '
 

f
x
c
x
 n    n  
This is basically a system of linear algebraic equations with
 c1'  x  
 '

c
x
 2  


 '

c
x
 n  
as the vector of
unknowns
Y
as the system
matrix
 f1  x  


f
x
 2  




f
x
 n  
as the righthand-side
column
If
 h1  x  
 c1'  x    f1  x  


 '
 

h
x


 2
 is a solution to Y  c2  x     f 2  x  



 



 '
 

h
x


 n

 cn  x    f n  x  
 h1  x  dx 
 y1 



 
 h2  x  dx 
 y2 


Y


 


 


y 
  hn  x  dx 
 n


is the particular solution we are looking for
If, in addition, the solution has to satisfy a set of initial conditions
0
0
y
,
y
 1 2,
, yn0  at a point x0, we have to find the coefficients
c1, c2, ..., cn in the general solution
 y1 
 c1   y1 
y 
c   
 2   e xA  2    y2 
 
   
 
   
 yn 
 cn   yn 
This can be done by solving the system of linear algebraic
equations
 y10 
 c1   y1  x0  

 0
c  
 y2   e x0 A  2    y2  x0  

 
  

 0
  
 cn   yn  x0  
 yn 