= Solution
{c}
Setting $\epsilon=0$ gives the reduced <first-order linear differential equation> $y'=y$. Its nonzero solutions cannot satisfy the zero value at $x=2$, so one endpoint needs a <boundary layer>. To determine which endpoint, freeze the coefficient at $x=x_b>0$ in the rapidly varying part: $\epsilon y''-x_by'=0$ has fast mode $e^{x_b(x-x_b)/\epsilon}$. It decays into the interval from the right endpoint but grows into the interval from the left endpoint. Hence
$$
\boxed{\text{the boundary layer is at }x=2,\text{ with thickness }O(\epsilon).}
$$
The <dominant balance> $\epsilon/\delta^2\sim x_b/\delta$ gives $\delta\sim\epsilon/x_b$, so a convenient local thickness is $\epsilon/2$. The left <boundary condition> must select the <outer solution>. Equivalently, a putative left <inner variable> $X=(x-1)/\epsilon$ would give $Y_{XX}-Y_X=0$, whose nonconstant solution grows like $e^X$ and cannot satisfy the <asymptotic matching condition> as $X\to\infty$.
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