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I want to find a two term asymptotic expansion, for small $epsilon$, of the solution of the following problem:
$$
y'' - epsilon y' - y = 1tag1, , y(0) = 0, ,y(1) = 1
$$
My approach: I assume that the solution has an asymptotic expansion that looks like this:
$$
ysim y_0 + epsilon^alpha y_1 + epsilon^beta y_2 + ldots tag2
$$
with $0<alpha <beta<ldots$.
If I substitute $(2)$ into $(1)$ I get the following equation:
$$
(y_0 + epsilon y_1 + ldots)'' - epsilon(y_0 + epsilon y_1 + ldots)' - (y_0 + epsilon y_1 + ldots) = 1
$$
I will now try to find $y_0, y_1$ by inspecting different order terms:

$mathcalO(1):$ In order to have balance in the equation, $y_0$ must satisfy
$$
y_0'' - y_0 = 1
$$
Solving this equation for $y_0$ gives
$$y_0 = c_1exp((frac12 + sqrt5/2)t) + c_2exp((frac12
) - sqrt5/2)t)$$
This solution needs to satisfy $y(0) = 0$. Substituting $t = 0$ and setting $y_0(0) = 0$ leads to the solution:
$$
y_0 = -dfrac1e^1/2 - sqrt5/2exp((1/2 + sqrt5/2)t) + dfrac1e^1/2 - sqrt5/2exp((1/2 - sqrt5/2)t)
$$
Furthermore, to have balance we need $alpha = 1$.

$mathcalO(epsilon):$ In order to have balance in the equation $y_1$ must satisfy
$$
y_1'' - y_1 = -y_0'
$$
Or
$$
y_1'' - y_1 = dfracddt(-dfrac1e^1/2 - sqrt5/2exp((1/2 + sqrt5/2)t) + dfrac1e^1/2 - sqrt5/2exp((1/2 - sqrt5/2)t))
$$
While I could plug this into wolfram to find a solution for $y_1$, I'm not sure whether this is the correct way to find a two term asymptotic expansion for the initial problem.

Question: Am I on the right track? It feels like there must be an easier way to do this. Could someone point me in the right direction?

My calculation is different from yours. Let
$$
y=y_0 + epsilon y_1 +O(epsilon^2)
$$
in the equation to get
$$ (y''_0 + epsilon y''_1 +O(epsilon^2))-epsilon(y'_0 + epsilon y'_1 +O(epsilon^2))-(y_0 + epsilon y_1 +O(epsilon^2))=1. tag1 $$
Then the boundary conditions $y(0)=0,y(1)=1$ become
$$ y_0(0)=y_1(0)=0, y_0(1)=1,y_1(0)=0.$$
$O(1)$:
$$ y''_0-y_0=1, y_0(0)=0,y_0(1)=1 $$
which has the solution
$$ y_0= frace^-x left(e^x-e^2 x-e^x+2+2 e^2 x+1-2 e+e^2right)e^2-1. $$
$O(epsilon)$:
$$ y''_1-y_1=y_0', y_1(0)=0,y_1(1)=0 $$
which has the solution
begineqnarray* y_1&=& frac12 left(e^2-1right)^2bigg[e^-x (-e^2 x+2 (x-4)+2 e^x-4 e^x+2+2 e^x+4+e^2 x (x-2)+2 e^2 x+3 (x-2)\
&&-2 e^2 x+1 x+e^2 x-2 e x-e^4 (x+2)+2 e^3 (x+2))bigg].
endeqnarray*
Thus
begineqnarray*
y&=&y_0+epsilon y_1+O(epsilon^2)\
&=&frace^-x left(e^x-e^2 x-e^x+2+2 e^2 x+1-2 e+e^2right)e^2-1\
&& +fracepsilon2 left(e^2-1right)^2bigg[e^-x (-e^2 x+2 (x-4)+2 e^x-4 e^x+2+2 e^x+4+e^2 x (x-2)+2 e^2 x+3 (x-2)\
&&-2 e^2 x+1 x+e^2 x-2 e x-e^4 (x+2)+2 e^3 (x+2))bigg].
endeqnarray*