= Solution
{c}
In the <outer expansion>, omit $\epsilon y''$ to get $y_0'=e^{y_0}$. Integrating and imposing the left boundary condition gives
$$
y_{\rm out}(x)=-\log(C-x),\qquad C=e^{-A}>1.
$$
This is regular throughout $[0,1]$. Its right-end value is $B=-\log(C-1)$, generally different from zero.
At $x=1$ set $X=(x-1)/\epsilon$. The equation is $Y''-Y'+\epsilon e^Y=0$, so the leading <inner expansion> is $Y=C_0+C_1e^X$. Matching as $X\to-\infty$ gives $C_0=B$, while $Y(0)=0$ gives $C_1=-B$. Thus
$$
y_{\rm in}(x)=B\left[1-\exp\left(\frac{x-1}{\epsilon}\right)\right].
$$
The common part is $B$. Subtract it once when forming the <additive composite expansion>:
$$
\boxed{y_c(x)=-\log(e^{-A}-x)+\log(e^{-A}-1)
\exp\left(\frac{x-1}{\epsilon}\right).}
$$
There is one generic boundary layer, at the right endpoint, of width $\epsilon$. A left-end exponential would grow into the interval and cannot supply a decaying matching correction. At $A=-\log2$ the displayed layer amplitude vanishes; the leading outer solution already meets the right condition, though a smaller correction can still be needed at higher order. The composite meets $y(1)=0$ exactly and $y(0)=A$ up to an exponentially small term.
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