I am given the following ODE $$ \epsilon \frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0 \; , \qquad 1<x<2$$

with the BVPs $$y(1)=0 \qquad , \qquad y(2)=1$$

and where $0<\epsilon \ll1$.

I am to use the method of matching inner and outer expansions to solve this.

By letting $y(x,\epsilon) \sim y\_0(x)+y\_1(x)\epsilon + y\_2(x)\epsilon ^2+\cdots$, I have found the leading-order term of the outer expansion to be $2/x$ (and this satisfies only $y(2)=1$).

In computing the inner expansion, I let $x=\epsilon ^a z$ so that we now have $$\epsilon ^{1-2a}\frac{d^2Y}{dz^2}+z\frac{dY}{dz}+Y=0$$

where $Y(z,\epsilon)=y(x,\epsilon)$.

In order to match the orders, we need to choose $a$ such that $1-2a=0$, thus $a=\frac 12$. Plugging this in, we have $$\frac{d^2Y}{dz^2}+z\frac{dY}{dz}+Y=0$$

Letting $Y(z,\epsilon) \sim Y\_0(z)+\cdots$, the equation at leading order is simply $$\frac{d^2Y\_0}{dz^2}+z\frac{dY\_0}{dz}+Y\_0=0$$ and solving this gives $$Y\_0=e^{-x^2/2}\Bigl(A+B \; \text{erfi} \bigl(\frac{x}{\sqrt 2} \bigr) \Bigr)$$

which doesn't seem to be right. Have I done something wrong? Or, ignoring what I have done, what is the right way to approach this?