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Consider the nonlinear equation
$$\frac{d^2x}{dt^2}+\epsilon\sin(x)=0,~~\epsilon \ll 1\\ x(0)=0,~~\dot{x}(0)=1$$

and find...
A. The value of $x_0$ as $\epsilon$ goes to $0$
B. The first order term $x_1$ and the second order term $x_2$
C. Plot $x(t)$ for $x_0$, $x_1$, and $x_2$
D. Assemble the solution $x=x_0(t)+\epsilon x_1(t)+\epsilon^2 x_2(t)$. Plot $x(t)=x_0(t)$, $x=x_0(t)+\epsilon x_1(t)$ and $x+x_0(t)+\epsilon x_1(t)+\epsilon^2 x_2(t)$ on the same graph.

So the thought is that you can expand the sine function around $x=x_0$ to obtain the equations you need. Then, using Laplacian transforms, $x_1 and x_2$ are solvable.

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closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, Namaste, KReiser Dec 8 '18 at 0:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer. $\endgroup$ – the_candyman Dec 7 '18 at 18:14
  • $\begingroup$ Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola. $\endgroup$ – AmbretteOrrisey Dec 7 '18 at 18:38
  • $\begingroup$ Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed. $\endgroup$ – Pascal Dec 7 '18 at 20:12
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    $\begingroup$ @Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!! $\endgroup$ – AmbretteOrrisey Dec 7 '18 at 20:48
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You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get $$ \ddot x_0=0,~~x_0(0)=0,~\dot x_0(0)=1\\ \ddot x_1=-\sin x_0,~~x_1(0)=0,~\dot x_1(0)=0\\ \ddot x_2=-\cos(x_0)\,x_1,~~x_2(0)=0,~\dot x_2(0)=0\\ $$


In another approach, multiply with $2\dot x$ and integrate to find $$ 1=\dot x^2+2ϵ(1-\cos x)=\dot x^2+4ϵ\sin^2\frac x2 $$ This is a circle equation that can be parametrized via $\dot x=\cos\phi(t)$, $2\sqrtϵ\sin\frac x2=\sin\phi(t)$ or $x=2\arcsin\frac{\sin\phi(t)}{2\sqrtϵ}$. Compare the expressions for the first derivative $$ \cosϕ(t)=\dot x=\frac{\cosϕ(t)\,\dot ϕ(t)}{\sqrt{ϵ-\frac14\sin^2ϕ(t)}} $$ Now you can expand $1=\dot ϕ(t)\left(ϵ-\frac14\sin^2ϕ(t)\right)^{-1/2}$ in powers of $ϵ$ and integrate.

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