Keep the solution lost when separating a homogeneous equation

With $v=y/x$, $y^{\prime}=v+xv^{\prime}$. Substitution gives $v+xv^{\prime}=v(1+v)/(v-1)$, hence $xv^{\prime}=2v/(v-1)$. The original equation requires $x\ne0$ and $y\ne x$.

For $v\ne0$, $\int(1-1/v)dv=\int2dx/x$, so $v-\ln|v|=2\ln|x|+C$. Rearranging, $\ln|vx^2|=v-C$, hence $vx^2=Ae^v$. Substituting back gives $xy=Ae^{y/x}$, where the separated derivation has $A\ne0$.

The division by $v$ omitted $v=0$, which means $y=0$. It satisfies the original equation for $x\ne0$ and must be included. Allowing $A=0$ in the final implicit relation restores this branch. All solutions are understood on intervals satisfying the original exclusions $x\ne0$, $y\ne x$; the implicit relation must not be used to add a point where the original right-hand side is undefined.