Since for any periodic equation $A(t)$, the differential equation $x'(t)=A(t)x(t)$ has the principal matrix solution.

It's trivial for one-dim case, we have $x=\exp(\int_{t_{0}}^{t}a(s)ds)$ where $t_{0}$ is the inital state.

for the equation $x''(t)+\omega^{2}(t)x(t)=0$, suppose $y(t)=x'(t)$, we have:

$\left(\begin{array}{c} y'\\ x' \end{array}\right)+\left(\begin{array}{cc} 0 & \omega^{2}\\ -1 & 0 \end{array}\right)\left(\begin{array}{c} y\\ x \end{array}\right)=0$. How to solve this for $\omega=\sin(t)+c$ where $c\in\mathbb{R}$ by Floquet's Theorem?


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