= Solution
{c}
Balance $\epsilon y''$ against $xy$ near the <wave turning point>. With $x=\gamma X$, their equality requires $\epsilon/\gamma^2=\gamma$, hence $\gamma=\epsilon^{1/3}$. The rescaled equation is
$$
Y''+\epsilon^{1/3}(\epsilon^{1/3}X-1)Y'-XY=0.
$$
Its leading equation is the <Airy ordinary differential equation> $Y''=XY$. Thus
$$
Y=\alpha\operatorname{Ai}(X)+\beta\operatorname{Bi}(X).
$$
Let $a_0=\operatorname{Ai}(0)=3^{-2/3}/\Gamma(2/3)$; then $\operatorname{Bi}(0)=\sqrt3\,a_0$. The left boundary condition becomes $a_0(\alpha+\sqrt3\beta)=1$, or
$$
\boxed{\beta=3^{1/6}\Gamma(2/3)-\frac{\alpha}{\sqrt3}.}
$$
This determines one linear relation before the far-boundary matching condition is used.
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