 Solution

On an interval with $x\ne 0$, this is a homogeneous first-order differential equation. Put $y=xv(x)$, so $y^{\prime }=v+x{v}^{\prime }$. Substitution and division by $x^2$ giveThe constant choices $v=0$ and $v=2$ give $y=0$ and $y=2x$. Otherwise the separable differential equation givesAbsorbing the sign and the exponential integration constant yields $(v-2)/v=C{x}^2$. Relabelling the arbitrary constant then givesThe first family already includes $y=2x$ at $K=0$. Choose any interval where its denominator is nonzero. These give all solutions on intervals avoiding $x=0$. Both families extend through $x=0$ with $y(0)=0$, as required by the original ordinary differential equation. Because it is singular there, differentiable continuations of the nonzero family may use different $K$ on the two sides: both one-sided derivatives at zero equal $2$.