= Solution
{c}
Write $y=e^{S/\delta}$ and divide the equation by $y$:
$$
\frac{\epsilon}{\delta^2}(S')^2+
\frac{\epsilon}{\delta}S''+
\frac1\delta S'+x=0.
$$
Balancing the first and third terms gives $\delta=\epsilon$. With $S=S_0+\epsilon S_1+\cdots$, the terms of order $\epsilon^{-1}$ and one give
$$
S_0'(S_0'+1)=0,\qquad
(2S_0'+1)S_1'+S_0''+x=0.
$$
For the slow branch $S_0'=0$, choose $S_0=0$ and obtain $S_1=-x^2/2$. For the fast branch $S_0'=-1$, choose $S_0=-x$ and obtain $S_1=x^2/2$. Additive integration constants enter the two independent amplitudes. The two-term <WKB approximation> is consequently
$$
y(x)\approx A e^{-x^2/2}+B e^{-x/\epsilon+x^2/2}.
$$
Imposing the <boundary conditions> on these two approximating modes yields
$$
\boxed{
y(x)\approx
\frac{e^{-x^2/2}-e^{-x/\epsilon+x^2/2}}
{e^{-1/2}-e^{-1/\epsilon+1/2}}.}
$$
This uses the first two exponent <functions> $S_0,S_1$ for both branches. The next transport equations, if desired to identify the algebraic error, give $S_2=x-x^3/3$ for the slow mode and $S_2=x+x^3/3$ for the fast mode. Thus each displayed mode has relative amplitude correction $1+O(\epsilon)$ uniformly on the fixed interval; no <turning point> occurs because the two leading exponent roots remain distinct.
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