How to solve such fraction differential equation? (x^3 - 2xy^2)dx + 3yx^2dy = xdy - ydx. I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

Juan Lowe

Juan Lowe

Answered question

2022-11-23

How to solve such fraction differential equation?
Here's my first-order differential equation:
( x 3 2 x y 2 ) d x + 3 y x 2 d y = x d y y d x
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

Answer & Explanation

levraijournalk1o

levraijournalk1o

Beginner2022-11-24Added 10 answers

( x 3 2 x y 2 ) d x + 3 y x 2 d y = x d y y d x
( x 3 2 x y 2 ) d x + 2 y x 2 d y + y x 2 d y = x d y y d x
Rearrange terms:
( x 3 d x + x 2 y d y ) 2 x y ( y d x x d y ) = x d y y d x
x 2 ( x d x + y d y ) = ( x d y y d x ) ( 1 2 x y )
x d x + y d y = ( 1 2 x y ) d ( y x )
d ( x 2 + y 2 ) = 2 ( 1 2 x y ) d ( y x )
Divide by x 2 + y 2 and substitute u = x 2 + y 2 and v = y x
1 2 d u u = ( 1 u 2 v v 2 + 1 ) d v
d u d v = 2 4 u v v 2 + 1
u + 4 v v 2 + 1 u = 2
This is a first order linear DE. Try to integrate by integrating factor method:
μ ( v ) = exp 4 v v 2 + 1 d v
μ ( v ) = exp 2 v 2 + 1 d v 2 = ( v 2 + 1 ) 2
The DE becomes:
( u ( v 2 + 1 ) 2 ) = 2 ( v 2 + 1 ) 2
u ( v 2 + 1 ) 2 = 2 ( v 5 5 + 2 v 3 3 + v ) + C
Unsubstitute u and v:
u = x 2 + y 2  and  v = y x

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