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.
Best Answer
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, $$ x = \pm 2\operatorname{am}(\sqrt{A}(t-t_0)\sin{(x_0/2)},\csc{(x_0/2)}). $$