$y''+\epsilon y'=\epsilon$, where $y(0)=1$, $y'(0)=0$ I am trying to solve $y''+\epsilon y'=\epsilon$, where $y(0)=1$, $y'(0)=0$ using perturbation theory.
Using the substitution $y=y_{0}+y_{1}\epsilon$ I got the series $y=1+\epsilon(1+\frac{x^{2}}{2})+O(\epsilon ^{2})$.
However wolframalpha tells me the exact solution involves an exponential, (see http://m.wolframalpha.com/input/?i=y%22%2B0.1y%27%3D0.1 where I set $\epsilon=0.1$).
Am I on the right track with the solution? I'd then like to determine the validity of the solution as $x\rightarrow\infty$ but I'm not confident with this.

The exaxt solution has exponental 
$$y''+\epsilon y'=\epsilon$$
Just integrate 
$$y'+\epsilon y=\epsilon x+ K_1$$
$$e^{\epsilon x}y'+e^{\epsilon x}\epsilon y=e^{\epsilon x}(\epsilon x+ K_1)$$
$$(e^{\epsilon x}y)=\int e^{\epsilon x}(\epsilon x+ K_1)dx$$
$$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+e^{-\epsilon x}\int e^{\epsilon x}\epsilon xdx)$$
$$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+ x  - \frac 1 {\epsilon}$$
$$\boxed {y=K_1+e^{-\epsilon x}K_2 + x }$$
Use conditions to get the constants
$$y'+\epsilon y=\epsilon x+ K_1 \text { and } y'0)=0 \to K_1=\epsilon$$
$$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+ x  - \frac 1 {\epsilon} \text { and } y(0)=1 \to K_2=\frac 1 {\epsilon}$$
$$\boxed {y(x)=\frac {e^{-\epsilon x}-1}{\epsilon} + x +1}$$
