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  1 x  1 x  1 x
y  u (x ) e  u (x )   1 e   e (u (x )   1 u  (x ))
2

  1 x  1 x  1 x 2
y   1 e  (u (x )   1 u (x )) e (u (x )   1 u (x )) e (u (x ) 2  u (x )  1 u (x ))
2
1
 
We will put expressions fo y , y , y r in equation (5.6)
2 2 2

(  u  2 (   a )   u ( 2  a   a )u )e  1 x  0 (5.8)
1 1 1 1 1 2

 – solution of equation (5.7), therefore  1 2  a  1  a 2  0. So
1
1
a
multiple of this root, we have      1 , from here
1 2
2
2  a  0 . Then equality (5.8) will acquire a kind:
1 1

u  e  1 x  0 or  u  0.
.
Integrating the got equation twice, obsessed u( x ) Ax  B ,
where А and B - arbitrary constant. As we are interested by
partial decision. Therefore, taking A  , 1 B  0 obsessed
u( x ) x and accordingly partial decision will be written down
y  x e  1  x . It is easily to see that y , y - linearly independent
2 1 2
y
decisions, as 2  x  const .
y 1
General decision will be written down:

y  С e 1  x  С xe 1  x  С (  С x) e 2  x (5.9)
1 2 1 2

Example 5.2 To find the common decision of equation
y  2y  y  0.

 We make characteristic equation
 2  2  1  0

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