I'm interested in solving the following ODE for $y(x)$, $$\frac{dy}{dx} = \frac{y-x}{x-4y}\:.$$
I tried using the substitution $z = \frac{y}{x}$, which led me to $$\ln|x| = \frac{1}{4} \ln\left|\frac{2z-1}{2z+1}\right| - \frac{1}{2} \ln|4z^2-1| + c$$ but I am now stuck as solving for $x$ gives a nasty looking equation. Any ideas on things to try?
$$\frac{dy}{dx} = \frac{y-x}{x-4y}\:.$$ $$ \frac{dy}{y-x}=\frac{dx}{x-4y}$$ $$ \frac{d(2y+x)}{-2y-x}=\frac{d(2y-x)}{6y-3x}$$ After integration $$\frac 13\ln|2y-x|+\ln|2y+x|=K$$ Implicit form is good too $$(2y+x)^3(2y-x)=C$$