Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 9 / b

The equation requires $x\ne 0$. Put $v=y^2$, giving $v^{\prime }=2y{y}^{\prime }$ andAn integrating factor is $x^{-4}$, so $(v{x}^{-4}{)}^{\prime }=2x^{-3}$. Integrating gives $v{x}^{-4}=C-x^{-2}$, henceReal differentiable branches require $C{x}^4-x^2>0$, with one sign on each connected solution interval. At $y=0$, the original equation would demand $0=x$, contrary to $x\ne 0$; the zeros of the radicand are therefore not regular solution points. In particular, $y\equiv 0$ is not an additional solution.