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.

  • $\begingroup$ How about multiplying both sides by $x'$ and then integrating twice? $\endgroup$
    – Moo
    Jun 22, 2017 at 19:45
  • $\begingroup$ How would one integrate sin(x(t)) with respect to time? $\endgroup$ Jun 22, 2017 at 19:46
  • 2
    $\begingroup$ $\dfrac{1}{2} (x')^2 = -a \cos(x(t)) + c$. Take square roots, but next integral is ugly result! $\endgroup$
    – Moo
    Jun 22, 2017 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, 2017 at 19:48
  • 1
    $\begingroup$ 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, 2017 at 20:11

1 Answer 1


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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