= Solution
{c}
The reduced equation and the left <boundary condition> give the leading <outer solution>
$$
y_{\mathrm{out},0}'=y_{\mathrm{out},0},\qquad y_{\mathrm{out},0}(1)=1,
\qquad\boxed{y_{\mathrm{out},0}(x)=e^{x-1}.}
$$
Introduce the right <inner variable> $X=(2-x)/\epsilon\geq0$ and write $Y(X)=y(2-\epsilon X)$. The <chain rule> gives $y'=-Y_X/\epsilon$ and $y''=Y_{XX}/\epsilon^2$. Multiplying the rescaled <differential equation> by $\epsilon$ gives the exact equation
$$
Y_{XX}+(2-\epsilon X)Y_X+\epsilon(2-\epsilon X)Y=0.
$$
At leading order, $Y_{0,XX}+2Y_{0,X}=0$, with <general solution> $Y_0=A+Be^{-2X}$. The right <boundary condition> gives $Y_0(0)=0$. The <asymptotic matching condition> in the <overlap region> $\epsilon\ll2-x\ll1$ gives
$$
\lim_{X\to\infty}Y_0(X)=\lim_{x\to2^-}y_{\mathrm{out},0}(x)=e.
$$
Thus $A=e$ and $B=-e$, yielding the leading <inner expansion>
$$
\boxed{Y_0(X)=e(1-e^{-2X}),\qquad
y_{\mathrm{in},0}(x)=e\left(1-e^{-2(2-x)/\epsilon}\right).}
$$
Equivalently, the <inner variable> $\xi=2(2-x)/\epsilon$ gives $Y_0=e(1-e^{-\xi})$. Both normalizations describe the same <boundary layer>.
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