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(a  ) r a 2 (2 r ) 2
1
2
4
   0  0 (r  5r   0).
b (b  ) r 1 (3 r )
1 2

4
solutions of characteristic equation : r  1 , r  .
1 2
We put in (7.16) r  r  1 . Then :
1
   2  0 ,
1
1

   2  0 .
1
1
1
If to take  1, then    . Consequently, one of
1 1
2
decisions of the system (6.14) is
  1 
Y   1  e t
1
   .
 2 
If in (7.16) to lay down r  r  4
2

  2  2  0 ,
2
2

    0 .
 2 2
We will take  1, then   1. We get the second
2 2
  1 
4t
decision of the system : Y  e .
2  
 1 

The common decision is linear combination of the found
decisions :
 1 
 x    t 1   4t
   C 1 1  e  C 2   e .
1
y
     
 2 
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