= Solution
For the <matched asymptotic expansion>, first determine the <outer expansion>. Setting $\varepsilon=0$ away from $x=0$ gives $y_0'=y_0$, and the right <boundary condition> sets $y_0=e^x$. At first order,
$$
y_1'-y_1=-\frac{e^x}{2x},\qquad y_1(1)=0,
$$
so
$$
\boxed{y_{\mathrm{out}}(x)=e^x\left[1-\frac\varepsilon2\log x\right]+O(\varepsilon^2),\qquad x>0\text{ fixed}.}
$$
There is no outer $\sqrt\varepsilon$ term: its homogeneous coefficient would be forced to zero by the right boundary condition. The logarithmic overlap forces a <switchback term> in the <inner expansion>.
Near zero, balance $\varepsilon y''$ against $2xy'$ to get the <square-root boundary layer at a vanishing drift>:
$$
s=\sqrt\varepsilon,\qquad x=sz,\qquad Y(z)=y(sz),\qquad Y''+2zY'-2szY=0.
$$
For fixed $z$ expand $Y=Y_0+sY_1+s^2Y_2+\cdots$, allowing logarithmic dependence on $s$ in $Y_2$. With $\mathcal L_0=\partial_z^2+2z\partial_z$, the leading equation is $\mathcal L_0Y_0=0$. Its bounded solutions are a constant plus an <error function>. The left boundary condition and the overlap value one fix
$$
\boxed{Y_0(z)=\operatorname{erf}z.}
$$
The following two solution sections complete the half-order and first-order matching; their headings retain the scaffold's organization of the supplied hints.
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