Past exam of the mathematical methods I course of the University of Oxford / 2022 / 2 / Solution

On an interval with $x\ne 0$, the zero function is one solution. For a nonzero solution, set $u=y^{-2}$. Dividing the equation by two and differentiating $u$ gives the linear ordinary differential equationIts integrating factor is $x^{-2}$, so $(u/x^2{)}^{\prime }=-x^{-5}$ and $u=1/(4x^2)+C{x}^2$. Absorbing the factor four into the integration constant givesFor real solutions, choose a connected interval avoiding $x=0$ on which $1+K{x}^4>0$, with a fixed sign. The special straight solutions $y=\pm 2x$ are already included at $K=0$. The separate zero solution would be lost by the division used in the substitution.