DE with integrable combinations

Question

solve using DE with integrable solution $$ y(y^2-2x)dx + x(y^2+x)dy = 0 $$

Answer 1

Welcome to the site !

You must understand that a lot of people are ready to help you provided that you explain (and show) what you already tried and tell where you are stuck.

As a welcome gift, I shall put you on the track.

Consider the differential equation to be $$y(y^2-2x)x' + x(y^2+x) = 0$$ This can be simplified letting first $$x=\frac 12 y^2+z\implies x'=y+z'$$ Replace in the equation to get $$\frac{3 }{4}y^4-2 y z z'+z^2=0\implies \frac{3 }{4}y^4-y(z^2)'+z^2=0$$ which looks quite nicer.

Just continue with another change of function.

I am sure that you can take it from here.

Answer 2

HINT :

The change of variables $\quad xy=u\quad \text{and}\quad\frac{x^2}{y}=v \quad $ leads to $\quad du-dv=0\quad\to\quad u-v=c$ $$xy-\frac{x^2}{y}=c \quad\to\quad y=\frac{c\pm\sqrt{c^2+4x^3}}{2x}$$

More classical way : the integrating factor method.