ordinary differential equation solving I have a diffeq:

I have a nonlinear Diffeq:

$$\frac{d^2x}{dt^2}+\beta \frac{dx}{dt}+\varepsilon e^{- \lambda x} = f(t) $$

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:
$$\frac{d^2}{dt^2}\epsilon x_1 +\beta \frac{d}{dt}\epsilon x_1+\epsilon \times exp(-\lambda x_0)=0$$

Where we know $x_0$. This now means that the exponential is no longer a function of arbitrary x.

$$\frac{d^2x_1}{dt^2} +\beta \frac{dx_1}{dt}+ e^{-\lambda x_0}=0$$

Where we know $x_0$. Should I use method of undetermined coeffs?
solving this 2nd order ODE

you solved this kind of 2nd order diff eqn before for x0

consider e^-ramda xo as a f(t) 

the solution will be similar
solving this 2nd order ODE

you solved this kind of 2nd order diff eqn before for x0

consider e^-ramda xo as a f(t) 

the solution will be similar