<s>[INST] How to solve the following differential equation $(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$ I'd love your help with solving this following differential equation: $$(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$$
I am trying to use the method of finding integrating factor, that is $\frac{1}{Mx+Ny}$. But it is not coming.
Any suggestions?
 [/INST] $$Mx+Ny=3xy(y^2-x^2)$$
$$\frac{y(y^2-2x^2)dx+x(2y^2-x^2)dy}{3xy(y^2-x^2)}=0$$
$$\frac{(y^2-x^2)(ydx+xdy)}{3xy(y^2-x^2)}+\frac{xy(ydy-xdx)}{3xy(y^2-x^2)}=0$$
$$\frac{log{xy}}{3}+\frac{\log(y^2-x^2)}{6}=\frac{logc}{3}$$
$$xy(y^2-x^2)^{1/2}=c$$
</s>