3.5 Systems of Differential Equations
In this section we will apply the concepts of matrices, eigenvalues, and eigenvectors to solve
systems of differential equations. We consider the linear system
x (t) = A(t) x(t) + f(t)
where x (t) is the derivative vector, A(t) is the coefficient matrix, x(t) is the vector of the
dependent variables, and f(t) is the vector of the forcing function. Equations with nonconstant
coefficients A(t) might be solved numerically or by the power series method [ The power series
method will not be covered in this course]. We will consider a homogenous system with constant
coefficients
x (t) = A x(t)
(3.5-1)
We assume a solution of the form x(t) = uet so that x (t) = uet. Substitute x(t) and x (t) into
Eq. (3.5-1) yields
uet = Auet  Au = u
Therefore the solution of the linear system can be obtained form the eigenvalues and
eigenvectors of the matrix A. The general or complete solution of the system (3.5-1) is then
x(t) = c1u1 e 1t + c2u2 e 2t + ... + cnun e nt
where u1, u2, ... and un are the linear independent eigenvectors. c1, c2, ..., cn are the integration
constants that can be determined from the initial conditions t = 0.
Example 3.5-1. Find a general solution of the system
x1 =
 x2 + x3
 0 1 1 
x 2 = 4x1  x2  4x3 ,  A =  4  1  4


 3  1 4 
x 3 =  4x1  x2 + 4x3
(3.5-2)
Solution
Substituting x(t) = uet into the system and rearranging yields
u1 +
u2  u3 = 0
 4u1 + ( + 1)u2 + 4u3 = 0
3u1 + u2 + ( 4)u3 = 0
(3.5-3)
The corresponding characteristic equation is
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
1
 4  1
3
1
4 = ( + 1) (  1) (  3) = 0
4
1
Therefore 1 = 3, 2 = 1, and 3 =  1 are the eigenvalues of the coefficient matrix A.
For  = 1 = 3, system (3.5-3) becomes the linear system
3u1 + u2  u3 = 0
 4u1 + 4u2 + 4u3 = 0
3u1 + u2  u3 = 0
1
The solution for this homogenous system is u = c1   1
 
 2 
1 
1 


Similarly for 2 = 1, u = c2 0 , and for 3 =  1, u = c3  2  . It follows that
 
 
1
1 
1
  1 e3t,
 
 2 
1 
0 et, and
 
1
1 
 2  e-t are particular solution of (3.5-2).
 
1 
The general solution of the given first-order system is
1
1 
1 




3t
t
x = c1  1 e + c2 0 e + c3  2  e-t.
 
 
 
 2 
1
1 
For the initial conditions: x1(0) =  1, x2(0) = 1, x3(0) = 1
 1 1 1  c1    1
 c1   2 
  1 0 2   =      = 
c
 4.5

 c2   1 
 2 

 2 1 1  c3   1 
 c3   1.5 
The final solution is
x1 = 2e3t  4.5 et + 1.5 e-t
x2 =  2e3t + 3e-t
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x3 = 4e3t  4.5 et + 1.5 e-t
The MATLAB command dsolve can solve this linear system as follows
>> [x1,x2,x3]=dsolve('Dx=-y+z,Dy=4*x-y-4*z,Dz=-3*x-y+4*z','x(0)=-1,y(0)=1,z(0)=1')
x1 =
3/2*exp(-t)-9/2*exp(t)+2*exp(3*t)
x2 =
-2*exp(3*t)+3*exp(-t)
x3 =
-9/2*exp(t)+3/2*exp(-t)+4*exp(3*t)
Example 3.5-2. Find a complete solution of the following simple first order system whose
coefficient matrix has complex eigenvalues.
x1 = 2x1  3x2
x 2 = 3x1 + 2x2
Solution
u 
This system will have a solution of the type  1  et if and only if
u2 
3   u1  0
  2
  3   2 u  = 0

  2  
The characteristic equation is
 2
3
3
= 2  4 + 13 = 0
 2
The eigenvalues are 1 = 2 + 3i, 2 = 2  3i
For 1 = 2 + 3i
(2 + 3i  2)u1 + 3u2 = 0
 3 u1 + (2 + 3i  2)u2 = 0
Dividing the system by 3
iu1 + u2 = 0
 u1 + iu2 = 0
The solution to the homogeneous system is u1 = ki and u2 = k
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1
1
1
u = ki   = c    x = c   e(2 + 3i)t
 i 
 i 
 i 
We can use Euler’s formula [eit = cos(t) + isin(t)] to obtain real solution for this system
 cos( 3t )  i sin( 3t )  2t
 i[cos( 3t )  i sin( 3t )] e


 1  (2 + 3i)t
=
 i  e
 
 e i 3t  2t
e =

i 3t 
 ie 
 1  (2 + 3i)t
=
 i  e
 
cos( 3t ) 2t
 sin( 3t )  2t
 sin( 3t )  e + i  cos(3t ) e




The general solution for the system is then
cos( 3t ) 2t
 sin( 3t )  2t
x = c1 
e + c2 

 e
 sin( 3t ) 
 cos(3t )
For the initial conditions: x1(0) = 1, x2(0) = 1  c1 = 1, c2 = 1
x1 = cos(3t)e2t  sin(3t)e2t
x2 = cos(3t)e2t + sin(3t)e2t
The MATLAB command dsolve gives the same solution as expected.
>> [x1,x2]=dsolve('Dx=2*x-3*y,Dy=3*x+2*y','x(0)=1,y(0)=1')
x1 =
-exp(2*t)*(-cos(3*t)+sin(3*t))
x2 =
exp(2*t)*(sin(3*t)+cos(3*t))
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