Differential equation with partial fraction

Question

How do you separate this differential equation into a partial fraction?

Solve the following differential equation: $$\frac{dy}{dx}=\frac{2y^2-xy+x^2}{xy-x^2}$$

Answer 1

Let $\frac{y}{x}=v$

$$\frac{{\rm d}y}{{\rm d}x} = v + x\frac{{\rm d}v}{{\rm d}x} =\frac{2v^2-v+1}{v-1} \\ x\frac{{\rm d}v}{{\rm d}x} = \frac{v^2+1}{v-1} \\ \frac{v-1}{v^2+1}dv = \frac {{\rm d}x}{x} \\ \text{Integrating both sides} \\ \int \frac{v-1}{v^2+1}dv =\int \frac {{\rm d}x}{x} \\ \int \frac{v}{v^2+1}dv - \int \frac{1}{v^2+1}dv= \frac {{\rm d}x}{x} \\ \frac{1}{2} \ln(v^2+1) - \tan^{-1}v = lnx + C\\ \frac{1}{2} \ln \left(\frac{y^2+x^2}{x^4}\right) = \tan^{-1}\left( \frac{y}{x}\right) + C $$

Where $C$ is the constant of integration.