Use singular perturbation methods to obtain a uniform approximate solution to the following problems. In each case assume $0 < \epsilon \ll 1$ and $0<x<1$. $$ \epsilon y''-(2-x^2)y=-1.$$ with $y'(0)=0$ and $y(1)=1$.
I find a problem in the Book of Applied Mathematics of J. David Logan in the page 121, exercise 1.h. I solved with finite differences in order to understand how it looks the real the solution but my method of perturbations says that the solution is so different. I solved it like this
There are two layers near $x = −1$ and $x = 1$. The outer solution, which is valid in the interval (−1, 1), away from the layers is $$y_0=\frac{1}{2-x^2}$$ In the layer near $x=1$ set $\tau=(1−x)/\delta(\epsilon)$. We find $\delta= \sqrt{\epsilon}$ with inner equation, $Y_i''-Y_i=1$ to leading order, $$Y_i(\tau)=1+ae^{\tau}-(1+a)e^{\tau}$$ In the layer near $x=−1$ set $\tau=(1−x)/\delta(\epsilon)$ .We find $\delta= \sqrt{\epsilon}$ with inner equation, to leading order, $(Y_i^*)'' − Y_{i}^* = −1$. The inner solution is $$Y_i(\tau)=1+be^{\tau}-(1+b)e^{\tau}$$ Matching gives $a = b = 1$ and the uniform approximation is $$y(t)=\frac{1}{2-x^2}-e^{(x-1)/\sqrt{\epsilon}}-e^{(x+1)/\sqrt{\epsilon}}$$ This solution is so different from the one I found with the numerical method.
Since the DE involves only $y'$ and $y''$, not $y$, it's just a first-order equation in $v = y'$: $\epsilon v' - (2-x^2) v = -1$, with initial condition $v(0)=0$. That has the solution $v(x) = - \epsilon^{-1} e^{-x(x^2-6)/(3\epsilon)} F(x)$ where $$ F(x) = \int_0^x e^{z(z^2-6)/(3\epsilon)}\; dz $$ Now I think you can use steepest descent to approximate the integral.