Ecuaciones Diferenciables exactas 2

Esta no es una ecuación diferencial exacta

My = 3y^2

Nx = 3x^2 - y^2

En todo caso es una ecuación homogénea

$$\begin{align}&y^3dx +2(x^3-xy^2)dy=0\\ &\\ &2(x^3-xy^2)dy = -y^3dx\\ &\\ &\frac {dy}{dx}= \frac{y^3}{2(xy^2-x^3)}\\ &\\ &u=\frac yx \implies y=ux\\ &\\ &\frac{dy}{dx}= \frac{du}{dx}·x +u\\ &\\ &\text{y el cambio nos deja esto}\\ &\\ &\frac{du}{dx}·x +u=\frac{(ux)^3}{2[x(ux)^2-x^3]}=\\ &\frac{u^3x^3}{2(x^3u^2-x^3)}=\frac{u^3}{2(u^2-1)}\\ &\\ &resumiendo\\ &\\ &\frac{du}{dx}·x +u= \frac{u^3}{2(u^2-1)}\\ &\\ &\frac{du}{dx}·x = \frac{u^3}{2u^2-2}-u\\ &\\ &\frac{du}{dx}·x = \frac{u^3-2u^3+2u}{2u^2-2}\\ &\\ &\frac{du}{dx}·x = \frac{-u^3+2u}{2u^2-2}\\ &\\ &\frac{2u^2-2}{2u-u^3}du = \frac{dx}{x}\\ &\\ &\frac{2u^2-2}{u^3-2u}du =-\frac {dx}{x}\\ &\\ &\left(\frac{3u^2-2}{u^3-2u}-\frac{u^2}{u^3-2u}\right)du=-\frac{dx}{x}\\ &\\ &ln(u^3-2u)-\int \frac{u}{u^2-2}du=lnC-lnx\\ &\\ &ln(u^3-2u)-\frac 12ln(u^2-2)=ln \frac Cx\\ &\\ &ln \frac{u^3-2u}{\sqrt{u^2-2}}= ln \frac Cx\\ &\\ &\frac{u^3-2u}{\sqrt{u^2-2}}= \frac Cx\\ &\\ &\frac{\frac{y^3}{x^3}-2 \frac yx}{\sqrt{\frac{y^2}{x^2}-2}}= \frac Cx\\ &\\ &\frac{\frac{y^3-2yx^2}{x^3}}{\frac{\sqrt{y^2-2x^2}}{x}}=\frac Cx\\ &\\ &\frac{y^3-2yx^2}{x^2 \sqrt{y^2-2x^2}}= \frac Cx\\ &\\ &\\ &\frac{y(y^2-2x^2)}{x \sqrt{y^2-2x^2}}= C\\ &\\ &y \sqrt{y^2-2x^2}=Cx\end{align}$$

¡Uff!

Y eso es todo.