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I want to find a two term asymptotic expansion, for small $\epsilon$, of the solution of the following problem: $$ y'' - \epsilon y' - y = 1\tag{1}, \, y(0) = 0, \,y(1) = 1 $$ My approach: I assume that the solution has an asymptotic expansion that looks like this: $$ y\sim y_0 + \epsilon^\alpha y_1 + \epsilon^\beta y_2 + \ldots \tag{2} $$ 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:

$\mathcal{O}(1):$ In order to have balance in the equation, $y_0$ must satisfy $$ y_0'' - y_0 = 1 $$ Solving this equation for $y_0$ gives $$y_0 = c_1\exp((\frac{1}{2} + \sqrt{5}/2)t) + c_2\exp((\frac{1}{2} ) - \sqrt{5}/2)t)$$ This solution needs to satisfy $y(0) = 0$. Substituting $t = 0$ and setting $y_0(0) = 0$ leads to the solution: $$ y_0 = -\dfrac{1}{e^{1/2 - \sqrt{5}/2}}\exp((1/2 + \sqrt{5}/2)t) + \dfrac{1}{e^{1/2 - \sqrt{5}/2}}\exp((1/2 - \sqrt{5}/2)t) $$ Furthermore, to have balance we need $\alpha = 1$.

$\mathcal{O}(\epsilon):$ In order to have balance in the equation $y_1$ must satisfy $$ y_1'' - y_1 = -y_0' $$ Or $$ y_1'' - y_1 = \dfrac{d}{dt}(-\dfrac{1}{e^{1/2 - \sqrt{5}/2}}\exp((1/2 + \sqrt{5}/2)t) + \dfrac{1}{e^{1/2 - \sqrt{5}/2}}\exp((1/2 - \sqrt{5}/2)t)) $$ 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?

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My calculation is different from yours. Let $$ y=y_0 + \epsilon y_1 +O(\epsilon^2) $$ in the equation to get $$ (y''_0 + \epsilon y''_1 +O(\epsilon^2))-\epsilon(y'_0 + \epsilon y'_1 +O(\epsilon^2))-(y_0 + \epsilon y_1 +O(\epsilon^2))=1. \tag{1} $$ 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 $$ y_0= \frac{e^{-x} \left(e^x-e^{2 x}-e^{x+2}+2 e^{2 x+1}-2 e+e^2\right)}{e^2-1}. $$ $O(\epsilon)$: $$ y''_1-y_1=y_0', y_1(0)=0,y_1(1)=0 $$ which has the solution \begin{eqnarray*} y_1&=& \frac1{2 \left(e^2-1\right)^2}\bigg[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]. \end{eqnarray*} Thus \begin{eqnarray*} y&=&y_0+\epsilon y_1+O(\epsilon^2)\\ &=&\frac{e^{-x} \left(e^x-e^{2 x}-e^{x+2}+2 e^{2 x+1}-2 e+e^2\right)}{e^2-1}\\ && +\frac{\epsilon}{2 \left(e^2-1\right)^2}\bigg[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]. \end{eqnarray*}

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  • $\begingroup$ The solution for the ODE for $y_1$is $$y_1(x) = \frac {e^{-x} (2 (x - 1) e^{2 x + 1} - x e^{2 x} + e (e - 2) x + 2 e)} {2 (e^2 - 1)}.$$ $\endgroup$
    – Maxim
    Apr 1, 2019 at 14:03

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