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as
*
y  C 1 ( )x y  C 2 ( )x y , (6.5)
2
1

where C ( )x and C ( )x are some functions which are subject
1 2
of determination. We shall show that these functions can be
found by operations of integration. Basic idea of this method -
not that, how to guess, that the partial decision of heterogeneous
Equation needs to be searched in a kind (6.5) and that in such
kind he can be found.
For finding of two functions C ( )x and C ( )x let we
1 2
differentiate (6.5):
*
y  C y  C y  (C y   C y  )
1 1 2 2 1 1 2 2
On a function C ( )x and C ( )x we will impose an
1 2
additional condition :
C y   C y   0. (6.6)
1 1 2 2
Then:
*
y  C y  C y ,
1 1 2 2
*
y  C y C y  (C y    C y  2 2 )  .
2
2
1 1
1 1
* * *
We put the found values y , y , y in (6.4):
       
C (y  а y  а y ) C (y  а y  а y ) C y  C y  f (x )
1 1 1 1 2 1 2 2 1 2 2 2 1 1 2 2

Expressions in the first and second handles equal a zero
(because y and y are upshots LHDE).
1 2
And we have therefore:

C y    C y    f ( )x . (6.7)
1 1 2 2

For determination C , C we take the system of two
1 2
equations (6.6) and (6.7):

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