$y''+\epsilon y'=\epsilon$, 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.
 A: Zeroth order: $y_0''=0$, with these boundary conditions you get $y_0=1$.
First order: $y_1''+y_0'=1,y_1(0)=0$. (This boundary condition comes because you have already fully taken into account the boundary condition in computing $y_0$, so you want all corrections to leave the boundary conditions alone.) Thus $y_1''=1$ so $y_1=x^2/2$. This matches the first order series expansion of the exact solution, which is$x+\exp(-\varepsilon x)/\varepsilon-(1/\varepsilon-1)$.
A: If
$$
y''+\varepsilon y'=\varepsilon, \quad y(0)=0,\,y'(0)=1
$$
then integrating the ODE over $[0,x]$ we obtain
$$
y'+\varepsilon y=1+\varepsilon x
$$
and hence
$$
\mathrm{e}^{\varepsilon x}\big(y'+\varepsilon y\big)=\mathrm{e}^{\varepsilon x}(1+\varepsilon x)
$$
or
$$
(\mathrm{e}^{\varepsilon x}y)'=\frac{1}{\varepsilon^2}((\varepsilon x-(\varepsilon-1))e^{ex})'
$$
and finally
$$
y=\frac{1}{\varepsilon^2}\big(\varepsilon x-(\varepsilon-1)\big)+c\mathrm{e}^{-\varepsilon x}
$$
and since $y(0)=0$, then $c=(\varepsilon-1)/\varepsilon^2$, i.e.,
$$
y=\frac{1}{\varepsilon^2}\big(\varepsilon x-(\varepsilon-1)\big)+\frac{(\varepsilon-1)\mathrm{e}^{-\varepsilon x}}{\varepsilon^2}
$$
A: 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
$$y'+\epsilon y=\epsilon x+ K_1 \text { and } y'0)=0 \to K_1=\epsilon$$
$$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+ x  - \frac 1 {\epsilon} \text { and } y(0)=1 \to K_2=\frac 1 {\epsilon}$$
$$\boxed {y(x)=\frac {e^{-\epsilon x}-1}{\epsilon} + x +1}$$
A: 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 $y\sim 1+\frac{1}{2}\epsilon x^2$
