Solution
= Solution
Multiplying the two <Liouville–Green approximation>[Liouville–Green] forms by $e^{x/2}$ gives
$$
\boxed{y_+(x)\sim x^2e^x,\qquad y_-(x)\sim x^{-2}}.
$$
The first expression is in fact exact:
$$
y=x^2e^x,\qquad y'=e^x(x^2+2x),\qquad y''=e^x(x^2+4x+2),
$$
and direct substitution gives
$$
\boxed{y''-y'-\frac{2(x+1)}{x^2}y=0.}
$$