Asymptotic Matching for boundary layer problem The question asks to find a global approximation to the leading order of $\epsilon$.
$\epsilon y'' + xy' + \epsilon y =0$, with boundary conditions $y(0)=1,y(1)=-1$.
I assumed it's a boundary layer problem, and tried to work with both $x=0$ and $x=1$.  But did not work out...
(new edit below...)
I start by trying boundary layer at x=0 and x=1 normal.  Outer solution is $y=-1$, for inner solution, do the substitution of $x = \epsilon X$, $Y=y(x/\epsilon)$, then $\frac{d^2 Y}{d X^2} + \epsilon \frac{dY}{dX} + \epsilon ^ 2 Y = 0$,which when take $\epsilon \rightarrow 0$, becomes $\frac{d^2 Y}{d X^2} = 0$, and gives you an inner solution of $Ax/\epsilon +B$.  I can't match this with inner solution...
However if I do boundary layer at $x=1$, the outer solution at $x=0$ would just be $y=1$, for inner solution, I did substitution of $1-x = \epsilon X$, the equation then becomes $\frac{d^2 Y}{d X^2} - \frac{dY}{dX} =0$, which will give the inner solution $y = A + B exp(\frac{1-x}{\epsilon}$.  I can't match this with outer solution either, since when $\frac{1-x}{\epsilon} \rightarrow \infty$, inner solution would blow up.
Somehow I feel I am missing something, even the outer solution might not be something that simple.  since when $\epsilon \rightarrow 0$, the equation becomes $y'=0$, but in that case $\epsilon$ terms aren't negligible any more.
 A: The outer solution $y_{\text{outer}}(x) = -1$ is fine, but it turns out that in this problem the width of the boundary layer isn't $\approx \epsilon$ as you have assumed with your substitution $x = \epsilon X$.  To determine the correct width, instead substitute $x = \epsilon^{\alpha} X$, where $\alpha$ is a constant to be determined.  The equation
$$
\epsilon y'' + xy' + \epsilon y = 0
$$
then becomes
$$
\epsilon^{1-2\alpha} \frac{d^2 Y}{dX^2} + X \frac{dY}{dX} + \epsilon Y = 0.
$$
The third term, $\epsilon Y$, being order $\epsilon$, is certainly smaller than the second term, $X \frac{dY}{dX}$, which is order $1$.  Therefore the third term must not be involved in the dominant balance.  This only leaves the first two terms, which are balanced when $\alpha = 1/2$.  The resulting equation is then, to leading order,
$$
\frac{d^2 Y}{dX^2} + X \frac{dY}{dX} = 0.
$$
After multiplying by the integrating factor $e^{X^2/2}$ we can write the equation as
$$
\frac{d}{dX} \left(e^{X^2/2} \frac{dY}{dX}\right) = 0,
$$
and integrating yields
$$
\frac{dY}{dX} = A e^{-X^2/2}.
$$
Integrating once more we see that
$$
\newcommand{erf}{\operatorname{erf}}
Y(X) = A\erf\left(\frac{X}{\sqrt{2}}\right) + B,
$$
where $\erf$ is the error function.  To satisfy the boundary condition $y(0) = 1$ we must then take $B = 1$.  To match this with the outer solution we need
$$
\lim_{X \to \infty} Y(X) = y_{\text{outer}}(0) = -1,
$$
which is satisfied when $A = -2$.
We now have the inner and outer solutions, namely
$$
y_{\text{outer}}(x) = -1, \\
y_{\text{inner}}(x) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right).
$$
The matched solution is therefore
$$
y(x) \sim y_{\text{inner}}(x) + y_{\text{outer}}(x) - y_{\text{outer}}(0) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right).
$$
Here's a plot which shows the actual solution in blue and this approximate solution in purple for $\epsilon = 1/50$.

