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 P  Q

d ln  y  x 
.  . (2.14)
dx Q

Example 2.2 To Find an general integral of equation
2  xyy 3 dx    xx 2 y 2 dy  0 .

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 We have  , yxP   2y  xy 3 ; Q  , yx   x  x 2 y . We
 P  Q
2
will find derivative parts:  2  3xy 2 ;  1 2xy . As
y  x 
 P  Q
 , given equation not in complete differentials. We
y  x 
will consider expression
2
1   P  Q  2  3 xy  1 2 xy 2 1 xy 2 1
      .
Q  y  x    x  x 2 y 2  x 1 xy 2  x


We see that the first case and integrating cofactor which
depends on a variable x only takes place, it is possible to find
d ln  1
from equation of kind (2.12). It will be in this case  ,
dx x
from where consistently:
dx dx
d ln   ;  d ln    ; ln   ln x;   x .
x x

We will increase the given equation on the found
integrating cofactor:

2xy  x 2 y 3 dx  x 2  x 3 y 2 dy  0 .

We will make sure, that the condition of complete
differential is now executed. Indeed

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