$y''+\epsilon y'=\epsilon$, 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.

I'm new to perturbation theory, but since the other answers have addressed the exact solving of the DE, I thought i'd chip in.
Taking $y=y(x,\epsilon)$, then $y\sim y(x,0)+\epsilon\frac{\partial y}{\partial \epsilon}|_{\epsilon=0}$
Solving $y(x,0)$ gives $y(x,0)=c_1x+c_2$.
Now for $\frac{\partial y}{\partial \epsilon}$, one has to solve: $\frac{\partial}{\partial \epsilon}\left( \frac{\partial^2}{\partial x^2}y+\epsilon\frac{\partial y}{\partial x}\right)\bigg|_{\epsilon=0}=\frac{\partial}{\partial \epsilon}\epsilon\bigg|_{\epsilon=0}=1\bigg|_{\epsilon=0}=1$
This gives 
$$\begin{align}
\frac{\partial}{\partial \epsilon}\left( \frac{\partial^2}{\partial x^2}y+\epsilon\frac{\partial y}{\partial x}\right)\bigg|_{\epsilon=0}&=1\\
\frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\frac{\partial}{\partial \epsilon}\epsilon\frac{\partial y}{\partial x}\bigg|_{\epsilon=0}&=1\\
\frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\left(\frac{\partial y}{\partial x}+\epsilon\frac{\partial}{\partial x}\frac{\partial y}{\partial \epsilon}\right)\bigg|_{\epsilon=0}&=1\\
\frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\frac{\partial y(x,0)}{\partial x}&=1\\
\frac{\partial^2}{\partial x^2}\upsilon(x)+c_1&=1 &&\text{where 
 }\upsilon(x)=\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}
\end{align}$$
Then $\upsilon(x)=\frac{1-c_1}{2}x^2+c_3x+c_4$
Imposing the initial conditions and solving for the coefficients gives: $$
y(x,\epsilon)\sim 1-c_3x+\epsilon\left(c_3x+x^2\frac{1+c_3}{2}\right)
$$
Knowing the exact solution i believe $c_3=0$, so actually $y\sim 1+\frac{1}{2}\epsilon x^2$
