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


$$\frac{d^2}{dt^2}(x_0+\epsilon_1x_1) + \beta\frac{d}{dt}(x_0+\epsilon_1x_1)+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1)) $$


$$\frac{d^2}{dt^2}x_0+\frac{d^2}{dt^2}\epsilon_1x_1+\beta \frac{d}{dt}x_0 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda x_0))+\epsilon \times exp(-\lambda \epsilon_1 x_1)$$

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?

  • $\begingroup$ Is this exactly how the problem is stated? I would expect $$\frac{d^2x}{dt^2} +\beta \frac{dx}{dt}+ e^{-\lambda t}=0$$ where $x(0) = x_0$ $\endgroup$
    – graydad
    Dec 11, 2014 at 17:35
  • $\begingroup$ The original function had $e^ {\lambda x}$ However, I have solved the solution to this point. More details can be found:math.stackexchange.com/questions/1055041/… $\endgroup$ Dec 11, 2014 at 17:37
  • $\begingroup$ In that case, please edit your question to state the original problem and then show the work you did to get to where you are now. As your question is currently stated, I have a hunch of how you can solve this but am unsure without more information. $\endgroup$
    – graydad
    Dec 11, 2014 at 17:38
  • $\begingroup$ @graydad What hunch did you have on how to solve this? $\endgroup$ Dec 11, 2014 at 17:55
  • $\begingroup$ @graydad Note that I left out the solution for solving $x_0$ Note that this has been solved and is thus, known $\endgroup$ Dec 11, 2014 at 18:04

1 Answer 1


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

  • 1
    $\begingroup$ So method of integrating factor? Could you expand on your solution? $\endgroup$ Dec 11, 2014 at 20:44
  • $\begingroup$ yes integrating factor, just substitute e^ramda*xo to f(t) in your solution of x0 $\endgroup$
    – eric
    Dec 11, 2014 at 20:47
  • $\begingroup$ could you show this in your answer with LATEX? $\endgroup$ Dec 11, 2014 at 21:10
  • $\begingroup$ x1=∫▒(∫▒〖〖-e〗^(-ramdax0) e^(betat) 〗 dt)/e^(beta*t) dt $\endgroup$
    – eric
    Dec 11, 2014 at 21:16
  • $\begingroup$ this is a same form of your x0 solution and this x1 is in terms of x0 $\endgroup$
    – eric
    Dec 11, 2014 at 21:16

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