$\epsilon y''=e^{\epsilon y'}+y$ for $0<x<1$
where $y(0)=1$ and $y(1)=-1$. Compute the first term matched asymptotic expansion for the equation.
The outer expansion expansion gives me $y(x)=-1$?
How do you approach this question? Confused cause of the exponential terms power.
The size of $y$ is finite, so you can replace $e^{ϵy'}$ with its series and apply the balancing of scales to the series.
For the inner solution, set $x=x_0+\delta X$, $Y(X)=y(x)$ and find $$ ϵY''(X)=δ^2\sum_{k=0}^\infty\frac{(ϵY'(X))^k}{δ^kk!}+δ^2Y(X)=δ^2Y(X)+δ^2+δϵY'(X)+\frac12ϵ^2Y'(X)^2+... $$ Obviously, $δ\ggϵ$ and terms with $δϵ,ϵ^2$ etc. get dominated by the left side. The only way to have balancing terms, that is, more than one dominating term, is when $ϵ=δ^2$ with the leading equation for the inner solution $$ Y''(X)=1+Y(X) $$
The solutions do not look very exciting, neither the numerical nor the outer+inner approximations