Q: Finding the characteristic exponents (Floquet/Lyapunov exponents) of ODEs with periodic coefficients I am interested in finding the characteristic exponents $\lambda_i$ of the ODE $\pmb{x}'\left(t\right) = \pmb{A}\left(t\right)\pmb{x}\left(t\right)$, where $\pmb{A}\left(t\right)$ is periodic with period $T$. In particular, consider the following $2$D case
$\left[ {\begin{array}{cc}
   x \\
   y \\
  \end{array} } \right]' = \left[ {\begin{array}{cc}
   a_1 + b_1 \sin(\omega t) & a_2 + b_2 \sin(\omega t) \\
   a_3 + b_3 \sin(\omega t) & a_4 + b_4 \sin(\omega t) \\
  \end{array} } \right]\left[ {\begin{array}{cc}
   x \\
   y \\
  \end{array} } \right]$,
where $a_i$, $b_i$, and $\omega$ are all constants.
Although we have useful relations such as $$\lambda_1+\lambda_2= \frac{1}{T}\int_0^T {\mathrm tr}\boldsymbol{A}(\tau)\,\mathrm{d}\tau,$$
in order to find the values of individual $\lambda_1$ and $\lambda_2$, we still need to calculate the principal fundamental matrix, which as far as I know requires solving the original ODE.
I know for the $1$D case $x' = \left(a_1 + b_1 \sin(\omega t)\right) x$, the solution is $x(t) = c_1\mathrm{e}^{-\frac{b_1}{\omega}\cos(\omega t)+a_1 t}$. But what is the solution to the $2$D case above?
I tried something like
$$x(t) = c_1\mathrm{e}^{p_1\cos(\omega t)+q_1 t} + c_2\mathrm{e}^{p_2\cos(\omega t)+q_2 t},$$
$$y(t) = c_3\mathrm{e}^{p_1\cos(\omega t)+q_1 t} + c_4\mathrm{e}^{p_2\cos(\omega t)+q_2 t},$$
but it doesn't seem to work.
