Perturbation ODE's outer solution problem The problem is $$\epsilon^2 y''(x)-y(x)=0$$  with boundary conditions $y(0)=1, y(1)=2$
I am asked to use match asymptotes method.
So I started with looking for an outer solution, and there is already problem with this. 
Assuming $y=y_0+\epsilon y_1+ \epsilon^2 y_2+...$ and pluging back to the equation we have $$\epsilon^0: y_0=0 \\ \epsilon^1:y_1=0\\
\epsilon^2:y_0''=y_2$$ etc. but since $y_0=0$ then $y_0''(0)=0$ as well whence $y_2=0$ and it goes the same with higher order terms. So We get $y(x)=0$, but this is not what we want since from explicit solution one can see that it shoud explode as it grows. 
Where did I make mistakes, what is wrong with this example?
 A: The expansion $y=y_0+ϵy_1+ϵ^2y_2+...$ is only valid if under the limit $ϵ\to 0$ the equation not only does have a limit, but also is structurally stable. This one could call a "stable perturbation". This would be the case for $y''-ϵ^2y=0$, as in the limit the structure of a second order ODE is preserved.
However, in the current example the structure is not preserved, the limit of the second order ODEs is a simple functional equation. This is a (structurally) "singular perturbation". So the expansion tried is not sufficient, it does not contain terms that can have infinitely large slopes.
Any solution $y(x;ϵ)$ that has bounded values and derivatives under $ϵ\to 0$ has to be $y=0$, the zero solution. For approximate solutions that bridge the gap between the zero basis solution and the non-zero boundary values, you get a boundary layer.
Doing the analysis for that results in approximate solutions $$y(x;ϵ)=e^{-x/ϵ}+2e^{(x-1)/ϵ},$$
where the first term is essentially zero for $x\gg ϵ$ and the second is zero for $1-x\gg ϵ$. The coefficients of the exact solution differ by terms of size $e^{-1/ϵ}$, which is ridiculously small even for moderately small values of $ϵ$.
