I have the following BVP that I am trying to solve, $$ay^{\prime}-\epsilon y^{\prime\prime}=1$$ with $y(0)=0$ and $y(1)=0$. First I solve the homogenous equation $ay^{\prime}-\epsilon y^{\prime\prime}=0$, using the characteristic equation, $$-\epsilon r^{2}+ar=0\Rightarrow x_{1}=0, x_{2}=\frac{a}{\epsilon},$$ which gives the homogenous solution of $y_{h}(x)=c_{1}+c_{2}e^{\frac{a}{\epsilon}x}$. Now the particular solution will simply be a constant $A$. Combining the constants $c_{1}+A=c_{3}$, I get, $$y(x)=c_{3}+c_{2}e^{\frac{a}{\epsilon}x}.$$ Now I want to apply the boundary values to determine the constants, which gives me the system of equations, $$\begin{split}c_{3}+c_{2}&=0,\\ c_{3}+c_{2}e^{\frac{a}{\epsilon}}&=0.\end{split}$$ Solving this gives me $c_{2}=c_{3}=0$, which is of course not the solution that I want. Is there a method I can apply to not get this solution? Or did I make a mistake in my computations?
Edit: Following the comments I now consider that $y_{p}(x)=Ax$ for some constant $A$. Plugging this into the differential equation gives, the particular solution $y_{p}(x)=\frac{1}{a}x$. This gives me the solution, $$y(x)=\frac{1}{a}x+c_{1}+c_{2}e^{\frac{a}{\epsilon}x}.$$ Now I apply the boundary conditions to get the system, $$ \begin{split} c_{1}+c_{2}&=0, \\ \frac{1}{a}+c_{1}+c_{2}e^{\frac{a}{\epsilon}}&=0, \end{split} $$ which results in $c_{1}=\frac{1}{a(e^{\frac{a}{\epsilon}}-1)}$ and $c_{2}=-\frac{1}{a(e^{\frac{a}{\epsilon}}-1)}$. So finally I get, $$y(x)=\frac{1}{a}\left(\frac{1-e^{\frac{a}{\epsilon}x}}{e^{\frac{a}{\epsilon}}-1}+x\right).$$
Solve this as a first order diff eq.
Make the substitution $u=y', u' = y''$ if you must.
$y' = \frac {1}{\alpha} + Ce^{\frac {\alpha}{\epsilon}x}$
And integrate $y'$
$y = \frac {1}{\alpha}x + C_1e^{\frac {\alpha}{\epsilon}x}+C_2$