Singular Perturbation Approx. for $\epsilon y'' + \frac{2 \epsilon}{t} y'-y=0$ Use singular perturbation techniques to find the leading order uniform approximation to the solution to the boundary value problem
$$\epsilon y'' + \frac{2 \epsilon}{t} y'-y=0$$
$0<t<1$ and $y(0)=0  ,  y'(1)=1$
This has a boundary layer near $t=1$
I am having trouble figuring out how to compute the inner approximation. I am used to doing these problems with $t=0$ as the boundary layer.
Using these methods, I get the outer approximation to be $y_0=0$, which I know is correct, but I'm at a loss on how to proceed with the inner approximation. I would appreciate if someone could work out the inner layer so that I could see how it is done.
 A: First, something about the boundary conditions. The general solution to the ODE is 
\begin{equation}
y(t) = c_1 \frac{1}{t}e^{\frac{t}{\sqrt{\epsilon}}}+ c_2 \frac{1}{t}e^{-\frac{t}{\sqrt{\epsilon}}}, \qquad (1)
\end{equation}
which has a singularity at $t=0$ for general $c_{1,2}$. Furthermore, you can show that, regardless of approximation techniques, the only solution obeying $y(0) = 0$ is the trivial solution. However, the solution for which $y'(1)=1$ and which doesn't blow up at $t=0$ is exponentially small in $\sqrt{\epsilon}$ near the origin. To summarise: you should take $y(0) = 0$ with a grain of salt, it should be $y(0) = \mathcal{O}(e^{-\frac{1}{\sqrt{\epsilon}}})$. The value of $y$ at $0$ is smaller than any (positive) power of $\epsilon$ though, so from the singular perturbation point of view, it's indistinguishable from $0$.
That being said, your approach is correct. The leading order equation gives $y = 0$, so you're correct to take that as an outer approximation. Then, following @David's advice, you can look at values of $t$ close to $1$, i.e. write
\begin{equation}
t = 1 + \epsilon^{\alpha} T,
\end{equation}
for some as yet undetermined $\alpha > 0$, where $T$ is now the free variable. You can invert this equation to get an expression for $T$ in terms of $t$, yielding
\begin{equation}
T(t) = \frac{1-t}{\epsilon^{\alpha}}.
\end{equation}
Now, again following @David, you would like to study how the solution depends on this 'zoomed in' variable $T$. So, you write
\begin{equation}
y(t) = Y(T(t)) = Y\left(\epsilon^{-\alpha}(1-t)\right),
\end{equation}
and substitute that in the ODE. The result is
\begin{equation}
\epsilon^{1-2\alpha} \frac{\text{d}^2 Y}{\text{d} T^2} + \frac{2 \epsilon^{1-\alpha}}{1+\epsilon^{\alpha} T} \frac{\text{d} Y}{\text{d} T} - Y = 0. \qquad (2)
\end{equation}
Since $\alpha > 0$ (and $0 < \epsilon \ll 1$), the term $\epsilon^{1-2\alpha}$ is larger than the term $\epsilon^{1-\alpha}$, so dominant balance implies that the term $\epsilon^{1-2\alpha} \frac{\text{d}^2 Y}{\text{d} T^2}$ should be of the same order as the third term, $-Y$. This implies that $\alpha = \frac{1}{2}$; note that this choice of time scaling is also present in the general solution to the ODE (1).
Substituting $\alpha = \frac{1}{2}$ in the rescaled ODE (2), we obtain
\begin{equation}
\frac{\text{d}^2 Y}{\text{d} T^2} - Y + \mathcal{O}(\sqrt{\epsilon}) = 0.
\end{equation}
This leading order equation is easy to solve; we obtain
\begin{equation}
Y(T) = c_1 e^{T} + c_2 e^{-T}.
\end{equation}
Now the only thing left is to incorporate the boundary conditions. That I leave up to you, but keep in mind that the boundary conditions are given in terms of $t$, instead of $T$. If you can reformulate the boundary conditions in terms of $T$, you should be able to solve the layer problem for $Y$ and then give a leading order expression for the solution $y$.
