Homogeneous first-order ODE y' = (x-y)/(x+y)

Solve the differential equation $\dfrac{dy}{dx}=\dfrac{x-y}{x+y}$.

$y^2+2xy-x^2=C$. Rearranging, $(x+y)\,dy=(x-y)\,dx$, i.e. $x\,dy+y\,dx+y\,dy-x\,dx=0$, which is $d\!\left(xy+\tfrac{y^2}2-\tfrac{x^2}2\right)=0$. (Substituting $y=vx$ also works.) Explicitly, $y=-x\pm\sqrt{2x^2+C}$.