Outer and Inner expansions of odes I have been asked to determine the first two terms of the outer and inner expansions  of the ODE
 $$\epsilon \frac{d^2y}{dx^2}+\frac{dy}{dx}+\frac{1}{1+x}y=0$$ on $0<x<1$ with $0 < \epsilon \ll 1$ with boundary conditions $y(1)=1, y(0)=0$. 
I've worked out the boundary layer is at $x=0$ but I cant get the right answer even for the outer perturbation. I've been told to maybe Taylor expand it but I still can't seem to get it using the expansion $y=y_0+\epsilon y_1$.
The solution is meant to be $y_0=\frac{2}{1 + x}$ which I can get and then I get lost getting $$y_1=\frac{-2}{1+x}+\frac{4}{(1+x)^2}$$ Can anyone see how you can get to this?
Using the expansion $y=y_0+\epsilon y_1$ I have got the equation to be $\epsilon 
 y_0'' + y_0' + \epsilon y_1' + \frac{1}{1+x} y_0 + \frac{1}{1+x} \epsilon y_1=0$.
The peeling off order one I have the equation $y_0'+ \frac{1}{1+x} y_0=0$, which I solved using the integrating factor to get $y_0=\frac{A}{1 + x}$ then used the BC at $x=1$ to get $A=2$.
Next is when I got problems getting the answer as I got order $\epsilon$ to be $y_0''+y_1'+\frac{1}{1+x} y_1=0$.
I got $y_0''=\frac{4}{(1+x)^3}$, so the equation becomes $y_1'+\frac{1}{1+x} y_1=\frac{-4}{(1+x)^3}$ but this is where I am struggling?
 A: The integrating factor is $(1+x)$, so that
$$
\frac{d}{dx}((1+x)y_1(x))=(1+x)y_1'(x)+y_1(x)=-\frac4{(1+x)^2}
$$
which you can integrate.

Another fast way that sometimes works is the WKB approximation $y=\exp(S/\delta)$ that balances here to $δ=ϵ$ to give as equation for $S$
$$
ϵS''+S'^2+S'+\fracϵ{1+x}=0
$$
Now insert the perturbation expansion $S=S_0+ϵS_1+ϵS_2+\dots$ into the equation and compare powers of $ϵ$. The first is
$$
S_0'^2+S_0'=0
$$
with the two solutions $S_0'=0$ corresponding to the outer solution as it has no singular term in $ϵ$ and $S_0'=-1$ which gives the inner solution. Then for the next term $S_1$ we get
$$
2S_0'S_1'+S_1'+S_0''+\frac1{1+x}=0
\implies\begin{cases}
S_0=0:&S_1'=-\frac1{1+x},& S_1=-\ln(1+x)\\
S_0=-x:&S_1'=\frac1{1+x},& S_1=\ln(1+x)
\end{cases}
$$
Up to this point the approximation combining both basis solutions is 
$$
y=A\frac1{1+x}+B(1+x)e^{-x/ϵ}+O(ϵ).
$$ 
In the next step to determine $S_2$ the equation is
$$
S_1'^2+(2S_0'+1)S_2'+S_1''=0
\implies\begin{cases}
S_0=0:&S_2'=-\frac2{(1+x)^2},& S_2=\frac2{1+x}\\
S_0=-x:&S_2'=0,& S_2=0
\end{cases}
$$
and the approximation improves to
$$
y=A\frac1{1+x}\exp\left(\frac{2ϵ}{1+x}\right)+B(1+x)\exp\left(-\frac xϵ\right)+O(ϵ^2).
$$
The initial conditions in accuracy $O(ϵ^2)$ lead to
$$
0=y(0)=Ae^{2ϵ}+B,~ 1=y(1)=\frac A2e^{ϵ}\implies A=2e^{-ϵ},~ B=-2e^{ϵ}
\\
y(x)=\frac2{1+x}\exp\left(ϵ\frac{1-x}{1+x}\right)-2(1+x)\exp\left(ϵ-\frac xϵ\right)+O(ϵ^2)
$$

A: The solution $$y_1'+\frac{1}{1+x} y_1=\frac{-4}{(1+x)^3}$$ you can use a variation of parameters: $y_1(x)=\frac{A(x)}{1+x}$ so:
$$\frac{A'(x)}{1+x}=\frac{-4}{(1+x)^3}$$
so:
$$A(x)=\frac{4}{(1+x)}+c$$
$$y_1(x)=\frac{4}{(1+x)^2}+\frac{c}{1+x}$$
The next step is to find $c$ using boundaries conditions.
As you have shown the boundary layer is at $x=0$, there is no boundary layer at $x=1$ so you must have:
$$(y_0+\epsilon y_1)(1)=1+0 \times \epsilon$$
which leads to:
$$y_1(1)=0$$
so $4/4+c/2=0$ thus $c=-2$ i.e:$$y_1(x)=\frac{4}{(1+x)^2}+\frac{-2}{1+x}$$
