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

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 $x\to \infty$.