Repeated Eigenvalues: Example1

Example. Consider the system
 
 
 
 
Answer. The matrix coefficient of the system is
 
 
In order to find the eigenvalues consider the characteristic polynomial
 
 
Since  
 , we have a repeated
eigenvalue equal to 3.  Let us find the associated eigenvector  
 .  Set 
 
 
Then we must have  
  which translates into
 
 
This reduces to y=x. Hence we may take
 
 
Next we look for the second vector  
 .  The equation giving this
vector is  
  which translates into the
algebraic system
 
 
where
 
 
Clearly the two equations reduce to the equation y - x=1 or y = 1 + x, where x may be chosen to be any number. So if we take x=0 for example, we get
 
 
Therefore the two independent solutions are
 
 
The general solution will then be
 
 
In order to find the solution which satisfies the initial condition
 
 
we must have
 
 
This implies  
  and  
 .  Hence the solution is
 
 
The phase plane with some solutions is given in the picture below:
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Author: Mohamed Amine Khamsi