The differential equation $\frac{dy}{dx}+\frac{y^2}{x^2}=\frac{y}{x}$ has the solution

We have, $\frac{dy}{dx}+\frac{y^2}{x^2}=\frac{y}{x}$

Putting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we get

$v+x\frac{dv}{dx}+v^2=v\Rightarrow x\frac{dv}{dx}=-v^2\Rightarrow -\frac{1}{v^2}dv=\frac{1}{x}dx$

On integrating, we get 

$\frac{1}{v}=\log x+C\Rightarrow x=y(\log x+C)$