In this problem
$\epsilon y''+(x+\frac{1}{2})y'+y=0$ for $0 < x <1 $
$y(0)=2, y(1)=3$
In the outer expansion,
$y\approx y_0(x)+\epsilon y_1(x)+\cdots$
I found the $O(1)$ problem to be: $(x+\frac{1}{2})y'+y=0$ which upon solving gives us $y_0(x)=\frac{c_1}{2x+1}$. I'm getting stuck with find while solving the $O(\epsilon)$ problem for the outer expansion. Would really appreciate the help! Thank you!
For the local expansion, consider first $x=x_0+\delta X$. Balancing finds $δ=ϵ$. The first order approximation is a solution of $Y''+(x_0+\frac12)Y'=0$.
As $x_0+\frac12$ is positive on the given interval, the inner solution is a simple exponential $Y(X)=c_2+c_3\exp(-(x_0+\frac12)X)$. To be able to match this to the outer solution, you need boundedness in the interval. As $Y(-\infty)=\infty$, this is satisfied only at the left boundary $x_0=0$.
The right BC gives $c_1=9$ in the outer solution, the jump at the left has to fill the gap from left BC $2$ to the value $9=\frac{c_1}{2\cdot 0+1}$ of the outer solution, so $c_2=0$, $c_3=-7$, the solution approximation is $y(x)\approx\frac{9}{2x+1}-7\exp(-\frac{x}{2ϵ})$.
You can always test the computed approximations against the exact solution. The given ODE can be integrated once without further preparations, $$ ϵy'+(x+\tfrac12)y=C\implies (ϵe^{(x^2+x)/(2ϵ)}y)'=Ce^{(x^2+x)/(2ϵ)} $$ so that after one further integration $$ e^{(x^2+x)/(2ϵ)}y(x)-y(0)=\frac{C}ϵ\int_0^{x}e^{(s^2+s)/(2ϵ)}\,ds =\frac{C}ϵ\int_0^{x}e^{(x^2-2xs+s^2+x-s)/(2ϵ)}\,ds \\~\\ y(x)=y(0)e^{-(x^2+x)/(2ϵ)}+\frac{C}ϵ\int_0^{x}e^{(-(2x+1-s)s)/(2ϵ)}\,ds $$ Then at the second boundary, with $y(0)=2$, $$ 3=y(1)=2e^{-1/ϵ}+C\int_0^{\frac1ϵ}e^{(-3s+ϵs^2)/2}\,ds\approx \frac{C}3 $$ so that at the end $$ y(x)\approx 2e^{-x(1+x)/(2ϵ)}+9\int_0^{x/ϵ}e^{(-(2x+1)s+ϵs^2)/2}\,ds $$