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

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.

The initial value is positive, so set $z=y^{-2}$. The chain rule gives $z^{\prime }=-2y^{-3}{y}^{\prime }=-2(y^{-2}-1)$, making $z^{\prime }+2z=2$. Its solution is $z=1+C{e}^{-2x}$. Since $y(0)=1/2$, $z(0)=4$ and $C=3$. Choosing the positive branch givesIt satisfies the initial condition and approaches the stable positive equilibrium one as $x\to \infty$.

On an interval with $\sin x\ne 0$, divide by $\sin x$ and use integrating factor $1/\sin x$. ThenPut $u=\sin x$ in the integral: $\int {\cos }^{3}xdx/{\sin }^{2}x=\int (u^{-2}-1)du=-u^{-1}-u+C$. Multiplying back givesAlthough the derivation divided by $\sin x$, this final expression is smooth and directly satisfies the original undivided equation at every real $x$, including multiples of $\pi$.