= Solution
Write the normal-form equation as
$$
u''=Q(x)u,
\qquad
Q(x)=\frac1{x^4}-\frac1{4x^2}.
$$
For sufficiently small positive $x$, $Q(x)>0$. Moreover,
$$
\frac{|Q'(x)|}{Q(x)^{3/2}}=O(x)\longrightarrow0,
$$
and the corresponding higher derivative condition also tends to zero, so the amplitude varies slowly relative to the exponential phase and the <WKB approximation>[Liouville-Green approximation] applies.
The phase and amplitude have expansions
$$
\begin{aligned}
\sqrt{Q(x)}
&=x^{-2}\left(1-\frac{x^2}{8}
-\frac{x^4}{128}+O(x^6)\right),\\
Q(x)^{-1/4}
&=x\left(1+\frac{x^2}{16}+O(x^4)\right),\\
\int^x\sqrt{Q(s)}\,ds
&=-\frac1x-\frac x8-\frac{x^3}{384}+O(x^5).
\end{aligned}
$$
Hence two independent Liouville-Green solutions satisfy
$$
\boxed{
u_\pm(x)\sim Q(x)^{-1/4}
\exp\left(\pm\int^x\sqrt{Q(s)}\,ds\right)}.
$$
In particular, after choosing the signs according to growth and decay,
$$
u_{\rm grow}(x)\sim xe^{1/x}\left(1+\frac x8+\frac{9x^2}{128}+\cdots\right),
\qquad
u_{\rm decay}(x)\sim xe^{-1/x}\left(1-\frac x8+\frac{9x^2}{128}+\cdots\right).
$$
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