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 As well as in a previous example we will lay down
3
3
y y    y 1  , 0 y ) 1 (  , 1  y ) 1 (  3 .
2

Obsessed equation of the first order in relation to р(у) with
dp
3
2
the separated variables: y p  1  . 0 We separate variables
dy
and integrate:


p 2 dp  y 3 dy ,  p 2 dp   y 3 dy ,
p 3 1 3
  c , p  3  3c .
3 2y 2 1 2y 2 1
dy

So if p  y  , we will rewrite the last equation in such
dx
3

type y  3  3c 1 . Preliminary will find the value of
2y 2
arbitrary constant с 1, putting from initial conditions
3
y  , 1  y  3 . Obsessed equation in relation to с 1:
2
3 3
3  3  3c 1 , from where с 1=0. Thus, we come to
2y 2y 2
1
3

equation y  ( ) 3 , which easily gets untied by the
2y 2
separation of variables and subsequent integration:

3y 2 1 dy 3  2
3

dy  ( ) dx ,  dx , 3  y 3 dy  dx ,
2 3y 2 1 2
( ) 3
2
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