 Solution

Work on an interval with $x\ne 0$. When $n=3$, let $v=y^{-2}$. The resulting linear ordinary differential equation and its integrating factor areIntegration gives $v=2x^3+C{x}^4$. The real nonzero solutions and the separately checked zero solution are thereforeFor $n=1$, separation in the linear ordinary differential equation gives $y^{\prime }/y=x^2-2/x$, henceHere $C=0$ already includes the zero solution. Division by $x$ is why these formulas are stated on intervals excluding zero; the original equation requires $y(0)=0$ if a classical solution is to extend through zero.