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I didn't really know how to put this in the title as I am not too familiar with solving differential equations. I am trying to solve:

$x''(t) = -Asin(x(t))$

I'm not sure how to even begin solving this equation and have been searching online and can't find anything (although I may be searching the wrong thing).

Many thanks.

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  • $\begingroup$ How about multiplying both sides by $x'$ and then integrating twice? $\endgroup$ – Moo Jun 22 '17 at 19:45
  • $\begingroup$ How would one integrate sin(x(t)) with respect to time? $\endgroup$ – Benjamin Rogers-Newsome Jun 22 '17 at 19:46
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    $\begingroup$ $\dfrac{1}{2} (x')^2 = -a \cos(x(t)) + c$. Take square roots, but next integral is ugly result! $\endgroup$ – Moo Jun 22 '17 at 19:47
  • $\begingroup$ I think this is the pendulum equation, no closed form solution, but a good deal known. $\endgroup$ – Will Jagy Jun 22 '17 at 19:48
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    $\begingroup$ This is the pendulum equation and the best solution that is possible is leaving it in terms of an integral. It is easy to study its behavior using Moo's formula. Otherwise, I recommend small angle approximation $(x(t)<<1)$ thus $\sin(x(t))\approx x(t)$ like most physicists would use. $\endgroup$ – MasterYoda Jun 22 '17 at 20:11
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It's easy enough to integrate once: multiply both sides by $2x'(t)$ and integrating gives $$ x'(t)^2 = 2A(\cos{x(t)}-\cos{x_0}). $$ Taking the square root and dividing gives $$ 1 = \frac{x'(t)}{\sqrt{2A(\cos{x}-\cos{x_0})}}. $$ This doesn't have an elementary integral: one can express $t$ in terms of the elliptic integral of the first kind, $$ \pm \sqrt{A}(t-t_0) = (\csc{(x_0/2)})F( x/2 , \csc{(x_0/2)} ) $$ This has an inverse, the Jacobi amplitude $\operatorname{am}$. Specifically, $$ x = \pm 2\operatorname{am}(\sqrt{A}(t-t_0)\sin{(x_0/2)},\csc{(x_0/2)}). $$

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