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 Lets y (x )  0 – the known partial decision of equation
1
(4.3). We will find second decision of this equation linear
independent with y (x ). We go out from Ostrogradsky -
1
Liouville formula (4.8). We will write down this formula in the
unfolded kind :
  1 p dx
y y   y y  Ce .
2 1 2 1

2
Dividing by y all terms we have:
1
y y   y y C   1 p dx
2 1 2  e ,
y 2 y 2
1 1
that is

 y  C   1 p dx
2
   2 e ; (4.15)
 y 1  y 1

from here, integrating, we find:

1   1 p dx
y  Cy 1 e dx C . (4.16)
2 2 1
y 1

As we are interested by the partial decision, laying down
0
C  1 , C  , have:
1
1   1 p dx
y  y 1 e dx .
2 2
y
1
  y     
As follows from (4.15)   2   0    y 2  const .

  y 1    y 1 
 
That is y and y – linearly independent.
2 1
Then general decision will be written down
 a ( x) dx
e  1
y( x)  C  y ( x)  C  y ( x)  dx .  (4.17)
1 1 2 1  2
y ( x)
1
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