Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 11 / a / Solution

First retain the equilibrium solution $y=0$. For $y\ne 0$ and $x\ne \pm 1$, separation of variables gives $\int y^{-2}dy=\int x/(1-x^2)dx$, hence $-1/y=-\frac{1}{2}\ln |1-x^2|+C$. Renaming the constant yieldsA nonzero solution is considered on a regular interval avoiding $x=\pm 1$ and zeros of its denominator. The zero solution itself extends through both singular points of the divided equation.