$$
S^{v}+\left(S^{2}\right)^{2}+v+\frac{1}{2}-\frac{1}{4} x^{2}=0
$$
Making the approximations
$$
S^{*} \propto\left(S^{\prime}\right)^{2}, \quad v+\frac{1}{2} \ll \frac{1}{4} x^{2}, \quad x \rightarrow+\infty,
$$
gives the asymptotic differential equation $\left(S^{\prime}\right)^{2} \sim \frac{1}{4} x^{2}(x \rightarrow+\infty)$ whose solutions are
$$
S(x) \sim \pm \frac{1}{2} x^{2}, \quad x \rightarrow+\infty .
$$
We have now determined that the possible controlling factors of the leading behavior of $y(x)$ are $e^{t^{2} / 4}$,
We leave as an exercise [Prob. 3.38(b)] the verification that the possible leading behaviors of $y(x)$ are
$$
\begin{array}{ll}
y(x) \sim c_{1} x^{-p-1} e^{x^{2} / 4}, & x \rightarrow+\infty \
y(x) \sim c_{2} x^{2} e^{-x^{*} / 4}, & x \rightarrow+\infty
\end{array}
$$
MATH4412 COURSE NOTES ：
we approach a pole we let
$$
\begin{aligned}
&y(x)=v^{-2}(x) \
&y^{\prime}(x)=-\frac{2}{\sqrt{6} v^{3}(x)}-\frac{1}{2} \sqrt{6} x v(x)-3 v^{2}(x)+v^{3}(x) u(x) .
\end{aligned}
$$
Show that this gives the new system of equations
$$
\begin{aligned}
&v^{\prime}=\frac{1}{\sqrt{6}}+\frac{\sqrt{6}}{4} x v^{4}+\frac{3}{2} v^{5}-\frac{1}{2} u v^{6}, \
&u^{\prime}=\frac{3}{4} x^{2} v+\frac{9}{4} \sqrt{6} x v^{2}+(9-\sqrt{6} u x) v^{3}-\frac{15}{2} u v^{4}+\frac{3}{2} u^{2} v^{3}
\end{aligned}
$$
with initial conditions
$$
v=-\frac{1}{\sqrt{y}}, \quad u=\frac{3}{v}+\frac{\sqrt{6 x}}{2 v^{2}}+\frac{y}{v^{3}}+\frac{2}{\sqrt{6} v^{6}},
$$
and that these new equations may be used to integrate past the location of a pole of $y(x)$.