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2
 2xy  x 2 y 3   x  x 3 y 2  2 2
  2x  3x y .
y  x 

We find a function  yxu ,  after a formula (2.3),
considering x  ; 0 y  0 :
0 0
x y 1 x x 3 y 3
3
2
u   2xy  x 2 y 3 dx   0  0  y 2 dy  x 2 y  x 3 y 3  x 2 y  .
0 0 3 0 3

Writing down a general integral in a kind (2.9) let as
multiply expression by number 3 for the receipt of more
3
comfortable form: 3x 2 y  x 3 y  . С 

Example 2.3 To find the general integral of equation
xy 2  y dx  xdy  0.

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

We see that the second case and integrating cofactor which
depends on a variable only takes place, it is possible to find
from equation of kind (2.13). It will be in this case,
d ln  2
  from where consistently:
dy y
 2dy dy 1
d ln   ;  d ln   2  ; ln    ln2 ; y  
y y y 2

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