Obtain the leading order uniform approximation of the solution to: $\epsilon y'' +(1+x)^2y'+y=0, y(0)=0 y(1)=1$ as $\epsilon \rightarrow 0$. I am completely lost.
Am i right in doing this? Since $(1+x)^2>0$ the boundary layer is at x=0? Then $$y(x;\epsilon)=y_0(x)+\epsilon y_1(x)+...$$ so the leading order solutions satisfies: $$(1+x)^2y_0'+y_0=0 \mbox{ which has solutions } y_0=e^{\frac{1}{(1+x)}+c}$$ Then using a stretch variable to obtain the inner solution we use $X=x/\delta$ and the principle of dominant balance to obtain:
$$Y''(X)+(1+\epsilon x)^2Y'(X)+\epsilon Y(X)=0$$ but what do I do now? How do i complete this if I have done it all correctly so far?
You're off to a good start. The first thing you want to do is find the right constant $c$ for your outer solution,
$$ y_\text{outer}(x) = e^{1/(1+x)+c}. $$
This should resemble the true solution away from the point $x=0$, so the only boundary condition it can hope to satisfy is $y(1) = 1$. So
$$ y_\text{outer}(1) = 1 \quad \Longrightarrow \quad c = -\frac{1}{2}, $$
and thus
$$ y_\text{outer}(X) = e^{1/(1+x)-1/2}. $$
Now for the inner solution you let $X = x/\delta$ (you wrote $x = X/\delta$ but I assume this was a typo), deduced that $\delta = \epsilon$, and the equation you ended up with is
$$ Y''(X)+(1+\epsilon X)^2Y'(X)+\epsilon Y(X)=0 $$
(note the $\epsilon X$ inside the parentheses; you wrote $\epsilon x$ but I assume this is also a typo). Now, to leading order (i.e. ignoring the terms involving $\epsilon$) this equation is
$$ Y''(X) + Y'(X) = 0. $$
The solution to this equation is called the inner solution.
One of your boundary conditions is $y(0) = 0$, and the above equation is relevant only for $x \approx 0$, so one of the boundary conditions for this equation has to be
$$ Y(0) = 0. \tag{inner bc1} $$
To find the other boundary condition, we want the inner solution to match with the outer solution as you leave the boundary layer, and this means that you want
$$ \lim_{X \to \infty} Y(X) = y_\text{outer}(0). \tag{inner bc2} $$
These two boundary conditions uniquely determine $Y(X)$. We'll call
$$ y_\text{inner}(x) = Y(x/\epsilon). $$
Finally you match these two solutions by
$$ y(x) \sim y_\text{inner}(x) + y_\text{outer}(x) - y_\text{outer}(0). $$