(x^2-3y^2)dx + 2xydy = 0
I tried everything I just can't separate the variables :X
thank you in advance!
This is a homogeneous equation because both functions in $\displaystyle dx$ and $\displaystyle dy$ are of degree 2. In this case, make the substitution $\displaystyle y=xv$ and $\displaystyle dy = x dv + v dx$. Then your equation becomes
$\displaystyle x^2-x^2 v^2 dx + 2 x^3 v dv = 0$
$\displaystyle x^2(1-v^2) dx = -2x^3 v dv$
$\displaystyle -\frac{x^2}{2 x^3} dx = \frac{v}{1-v^2} dv$
which is separable. Once you solve for $\displaystyle x$ in terms of $\displaystyle v$, you can go back and solve the equation by substituting $\displaystyle v=\frac{y}{x}$ and solve for $\displaystyle y$.