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 xI'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.

Question

How to solve such fraction differential equation?Here&#039;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  xI&#039;ve tried to make it fraction, but it isn&#039;t separable differential equation, also it isn&#039;t differential equation in total differentials, so after it I lose any clue for answer.

levraijournalk1o · Accepted Answer

(      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  xRearrange 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  =  2This 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    )    +  CUnsubstitute u and v:  u  =      x    2    +      y    2     and   v  =            y      x