Question

Prove that $x^2-y^2=C(x^2+y^2{)}^2$ is the general solution of differential equation $(x^3-3x{y}^2)dx=(y^3-3x^{2}y)dy,$ where C is a parameter.

Answer 1

Given, differential equation can be rewritten as
$\frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}$ ...(i)
This is a homogeneous equation. So, put y=vx
$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(vx\right)\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes
$v+x\frac{dv}{dx}=\frac{x^3-3x(vx{)}^2}{(vx{)}^3-3x^2(vx)}\Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$