What is the general solution of $(x^{2} - xy)\,dy = (xy + y^{2})\,dx$?

Explanation

1. 01

   $y = vx$: $v + x\frac{dv}{dx} = \frac{v(1 + v)}{1 - v}$, so $x\frac{dv}{dx} = \frac{2v^{2}}{1 - v}$ and $\left(\frac{1}{2v^{2}} - \frac{1}{2v}\right)dv = \frac{dx}{x}$.

2. 02

   $-\frac{1}{2v} - \frac{1}{2}\log\lvert v \rvert = \log\lvert x \rvert + c$; times $-2$: $\frac{1}{v} + \log\lvert v \rvert + 2\log\lvert x \rvert = C$.

3. 03

   $\log\left\lvert \frac{y}{x} \right\rvert + 2\log\lvert x \rvert = \log\lvert xy \rvert$, so $\frac{x}{y} + \log\lvert xy \rvert = C$.