Solution
= Solution
The initial value is positive, so set $z=y^{-2}$. The <chain rule> gives $z'=-2y^{-3}y'=-2(y^{-2}-1)$, making $z'+2z=2$. Its solution is $z=1+Ce^{-2x}$. Since $y(0)=1/2$, $z(0)=4$ and $C=3$. Choosing the positive branch gives
$$
\boxed{y(x)=\frac1{\sqrt{1+3e^{-2x}}},\qquad x\in\mathbb R.}
$$
It satisfies the initial condition and approaches the stable positive equilibrium one as $x\to\infty$.