I have problems finding the fundamental matrix for this exercise of Floquet theory
\begin{align*}
\dot{x} &= − \sin(2t)x + (\cos(2t) − 1)y\\
\dot{y} &= (\cos(2t) + 1)x + \sin(2t)y
\end{align*}
I noticed that if I rearrange this system into the form of $\dot{x}=A(t)x$ then $trace(A)=0$ and using the formula of the Wronskian I have
$$W(t)=1,$$
I think the matrix is something like
$$\Phi(t)=\begin{bmatrix} e^{t}\cos(2t) & -e^{-t}\sin(2t)\\
e^{t}\sin(2t) & e^{-t}\cos(2t) \end{bmatrix},$$
but that matrix does not satisfy that
$$\dot{\Phi}(t)=A(t) \Phi(t).$$
Any help?
Best Answer
You can extract common factors in both equations of the system to get \begin{align} \dot x &=-2\sin(t)(\cos(t)x+\sin(t)y)\\ \dot y &= 2\cos(t)(\cos(t)x+\sin(t)y) \end{align} The common factor $z=\cos(t)x+\sin(t)y$ has the derivative \begin{align} \dot z&=-\sin(t)x+\cos(t)y+\cos(t)\dot x+\sin(t)\dot y\\ &=-\sin(t)x+\cos(t)y=w\\ \ddot z=\dot w&=-z+2z=z \end{align} The solution can now be traced backwards.