Nonlinear Differential Equation question I have a nonlinear Diffeq:
$$\frac{d^2x}{dt^2}+\beta \frac{dx}{dt}+\epsilon \times e^{- \lambda x} = f(t) $$
where $f(t)$ is a function that is known, and $\beta$ and $\lambda$ are constants that are known. Also, we know that $\epsilon$ is a constant parameter that is small.
I first need to obtain the zero order solution $x_0$, before finding the first order solution $x_1$
The first thing that I need to do is to use asymptotic expansions to obtain solutions of order $\epsilon=0$ and (TYPO)
Note that general solution for f(t) that will have two unknown constants. 
UPDATE: After the first order term is solve, it needs to be plugged back in. The exponential needs to be linearized and things should start cancelling out. I am not sure how to do this, I just know this is what needs to be done.
UPDATE2: Correction, $\epsilon = 1$ was a typo. It should be $\epsilon^1$
I need to find a solution in the form:
$$x(t)=x_0(t)+\epsilon^1x_1(t)+\epsilon^2x_2 (t) + ... $$
So initially, $\epsilon$ needs to be set to 0 in order to obtain $x_0$. To find $x_1$, I need $\epsilon^1$
UPDATE3: I know now that I need to plug:
$$x=x_0+\epsilon_1x_1 $$ back into the original equation
Thus:
$$\frac{d^2}{dt^2}(x_0+\epsilon_1x_1) + \beta\frac{d}{dt}(x_0+\epsilon_1x_1)+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1))  $$
Then
$$\frac{d^2}{dt^2}x_0+\frac{d^2}{dt^2}\epsilon_1x_1+\beta \frac{d}{dt}x_0 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda x_0))+\epsilon \times exp(-\lambda \epsilon_1 x_1)$$
I think then the $x_0$ terms may cancel with f(t) or something like that? It may be some sort of approximation.
I still need to linearize the exponential. Any help is appreciated. 
Update4: Taking the solution a little but further...
We know that:
$$\frac{d^2x_0}{dt^2}+\beta \frac{dx_0}{dt} = f(t) $$
So, those terms all cancel. And now we have:
$$\frac{d^2}{dt^2}\epsilon_1x_1 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1))=0$$
But we dont want $\epsilon^2$ terms, to part of the exponential goes away as well.
We are left with:
$$\frac{d^2}{dt^2}\epsilon_1x_1 +\beta \frac{d}{dt}\epsilon_1x_1+\epsilon \times exp(-\lambda x_0)=0$$
Where we know $x_0$. This now means that the exponential is no longer a function of arbitrary x. I feel like the solution should be trivial now, but I am having a hard time finding it. Any ideas?
Can this be solved with the method of undetermined coefficients?
UPDATE5: Well I have updated this problem several times with very little response. As a latch ditch effort, is there anyone who can offer any advice on how to solve:
$$\frac{d^2x_1}{dt^2}  +\beta \frac{dx_1}{dt} +  exp{-\lambda x_0}=0$$
where $x_0$ is known
 A: A possible solution is as follows:
Write the equation as
${\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +\beta\,{\frac {d}{dt}}x
 \left( t \right) +\epsilon\,{{\rm e}^{-\lambda\,x \left( t \right) }}
=f_{{0}} \left( t \right) +f_{{1}} \left( t \right) \epsilon
$
Now look for a solution of the form
$x \left( t \right) =x_{{0}} \left( t \right) +x_{{1}} \left( t
 \right) \epsilon
$
Replacing this solution in the equation we have
${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) + \left( {\frac {d^{
2}}{d{t}^{2}}}x_{{1}} \left( t \right)  \right) \epsilon+\beta\,
 \left( {\frac {d}{dt}}x_{{0}} \left( t \right) + \left( {\frac {d}{dt
}}x_{{1}} \left( t \right)  \right) \epsilon \right) +\epsilon\,{
{\rm e}^{-\lambda\, \left( x_{{0}} \left( t \right) +x_{{1}} \left( t
 \right) \epsilon \right) }}=f_{{0}} \left( t \right) +f_{{1}} \left( 
t \right) \epsilon
$
expanding the exponential we obtain
${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) +\beta\,{\frac {d}{d
t}}x_{{0}} \left( t \right) + \left( {\frac {d^{2}}{d{t}^{2}}}x_{{1}}
 \left( t \right) +\beta\,{\frac {d}{dt}}x_{{1}} \left( t \right) +{
{\rm e}^{-\lambda\,x_{{0}} \left( t \right) }} \right) \epsilon=f_{{0}
} \left( t \right) +f_{{1}} \left( t \right) \epsilon
$
Then we deduce that
${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) +\beta\,{\frac {d}{d
t}}x_{{0}} \left( t \right) =f_{{0}} \left( t \right) 
$
and
${\frac {d^{2}}{d{t}^{2}}}x_{{1}} \left( t \right) +\beta\,{\frac {d}{d
t}}x_{{1}} \left( t \right) +{{\rm e}^{-\lambda\,x_{{0}} \left( t
 \right) }}=f_{{1}} \left( t \right) 
$
The solution at zero order is
$x_{{0}} \left( t \right) =\int _{0}^{t}\! \left( \int _{0}^{\tau}\!f_{
{0}} \left( \sigma \right) {{\rm e}^{\beta\,\sigma}}{d\sigma}+C_{{1}}
 \right) {{\rm e}^{-\beta\,\tau}}{d\tau}+C_{{2}}
$
Replacing this solution in the equation at first order we obtain a linear equation which can be solved formally as
$x_{{1}} \left( t \right) =\int _{0}^{t}\! \left( \int _{0}^{\tau}\!(f_{
{1}} \left( \sigma \right) - {{\rm e}^{-\lambda\,x_{{0}} \left( \sigma 
 \right) }} ){{\rm e}^{\beta\,\sigma}}{d\sigma}+C_{{3}}
 \right) {{\rm e}^{-\beta\,\tau}}{d\tau}+C_{{4}}$
