Leading order approximation to differential equation Find a leading order approximation to the solution of 
$\epsilon y'' + 2 y' + e^y = 0$, $y(0)=y(1)=0$
as $\epsilon \to 0$.
I know there is a boundary layer near $x=0$ and not at $x=1$ so I can impose the boundary condition at $x=1$. But the $e^y$ term is confusing me... when I substitute in an series approximation for $y$ and then taylor expand the $e^y$ term, I keep getting $y=0$ as the outer approximation because of the infinite series for $e^y$.
Could anyone help me with a  solution to this problem?
Thanks
 A: To leading order the outer solution satisfies the equation you get when you set $\epsilon = 0$, namely
$$
2y_\text{out}' + e^{y_\text{out}} = 0,
$$
which is separable.  Choose the constant of integration in order to satisfy the boundary condition at the right endpoint, namely $y_\text{out}(1) = 0$.
Now find the inner solution.  Setting $x = \epsilon X$ the equation becomes
$$
\frac{d^2 y}{dX^2} + 2\frac{dy}{dX} + \epsilon e^y = 0
$$
after multiplying through by $\epsilon$, so to leading order the inner solution satisfies
$$
\frac{d^2 y_\text{in}}{dX^2} + 2\frac{dy_\text{in}}{dX} = 0.
$$
Match this inner solution to the outer solution by taking
$$
\lim_{X \to \infty} y_\text{in}(X) = y_\text{out}(0).
$$
We also want the inner solution to satisfy the boundary condition at the left endpoint, namely
$$
y_\text{in}(0) = 0.
$$
These two conditions will determine the two constants of integration.
Finally combine $y_\text{in}$ and $y_\text{out}$ appropriately to get a solution to the original DE.
