I am considering the ODE $x''=\sin(x)$, where $'$ means $\frac{d}{dt}$. This equation is equivalent to the system $\dot{x}=y$, $\dot{y}=\sin(x)$. I need to find the eigenvalues and eigenvectors of the monodromy matrix of the first variational equation of this system along the solution which satisfies $(\frac{\pi}{2},0)$ at $t=0$.
This is a Hamiltonian system, so I can draw the phase diagram and I know that the periodic solution $p(t)$ which satisfies the condition is given by $y=\pm \sqrt{-2 \cos(x)}$. Also, I know that the first variational equation is $$\dot{X} = D_{p(t)}V \, X \, , \, \, \mbox{where} \, \, V(x,y) = (y,\sin(x)) \, .$$
I have been trying to find an explicit expression for $p(t)$, but I have not found it yet. My questions are: How can I find the solution $p(t)$? Are there another ways to compute the first variational equation along $p(t)$?
Finally, I am asked to solve the first variational equation for the solution which satisfies $(0,2)$ at $t=0$. I know that this curve is given by $y=\sqrt{6-2 \cos(x)}$, but again I don't understand how to find an explicit expression for this solution.
I deeply appreciate your help.