Past exam of the mathematical methods I course of the University of Oxford / 2014 / 9 / c

Put $v=y^2$, so $v^{\prime }=2y{y}^{\prime }$. On an interval with $x\ne 0$, the resulting first-order linear differential equation is $v^{\prime }-4v/x=2x$. Its integrating factor $x^{-4}$ gives $(v/x^4{)}^{\prime }=2/x^3$, henceFor real differentiable graphs one needs $C>0$ and $|x|>1/\sqrt{C}$, with either fixed sign on a connected interval. The points $x=0$ are outside the original equation's domain. At $|x|=1/\sqrt{C}$, $y=0$ would make the left side zero while the right side is $x\ne 0$, so those endpoints are also excluded. The identically zero function is not an additional solution.