Given $$ \dot{y} = A(t)y,\quad A(t)= \begin{pmatrix} 1 + \frac{\cos t}{2 + \sin t} & 0 \\ 1 & -1 \end{pmatrix}, \quad y\in\mathbb{R}^2, $$
I have calculated the $(0,t)$-flow ($\Phi^{t,0}$) and the monodromy operator $M=\Phi^{2\pi,0}$.
We have $$\Phi^{t,0}=\Psi(t)= \begin{pmatrix} \frac{1}{2}e^t(\sin(t)+2) & 0 \\ \frac{1}{2}(e^{t} - \frac{1}{5} e^{t} (\cos t - 2 \sin t) - \frac{4}{5}e^{-t}) & e^{-t} \end{pmatrix},$$ and therefore $$M=\Phi^{2\pi,0}= \begin{pmatrix} e^{2\pi} & 0 \\ \frac{2}{5}(e^{2\pi}-e^{-2\pi}) & e^{-2\pi} \end{pmatrix}, $$ where $\Psi(t)$ is the fundamental matrix of $A$.
Now the exercise wants me to find a matrix $\Omega$ such that $M=e^{2\pi\Omega}$.
I already know that $\Omega=\begin{pmatrix}1 & 0\\\frac{4}{5} & -1\end{pmatrix}$ (it was given so we'd know when we'd have found the correct answer), but I have to know how to get it, and also how to check that it is true (which I know by using Matlab, but cannot replicate myself).
This is where I think I have to look:
By Floquet's theorem I know that there exists a substitution of variables $$y=\Lambda(t)z(t), t\in\mathbb{R},$$ with $\Lambda:\mathbb{R}\to M_n(\mathbb{C}), \Lambda(t+\tau)=\Lambda(t)$ (with $\tau=2\pi$ in our case?), such that the system at the start of this question becomes the system with constant coefficients: $$ \frac{dz}{dt}=\Omega z, z\in\mathbb{C}^n, \Omega\in M_n(\mathbb{C}). $$
But I don't know $\Omega$, $z$, $\Lambda$, and for $y$ I'm also not sure (one of two columns of fundamental matrix?)
I have already found that the eigenvalues of $M$ are $\mu_1=e^{2\pi}$ and $\mu_2=e^{-2\pi}$.
Only in the next step, where I already had to know the answer to this question, do they tell me that $\Lambda(t)=\Phi^{t,0}e^{-t\Omega}$, which they ask me to calculate in this next step. (But that's a question for another post)
After reading on the Jordan canonical form, I discovered an answer myself
Since $M\in M_2(\mathbb{C})$ has two distinct eigenvalues, we have that $M$ is diagonalizable. We write $$M=PJP^{-1},\quad J=\begin{pmatrix}e^{-2\pi} & 0 \\ 0 & e^{2\pi}\end{pmatrix},\quad P\in M_2(\mathbb{C}),\quad\det(P)\neq0.$$ So $J$ is a diagonal matrix with the eigenvalues on its diagonal. To find $P$ we solve the system of equations $$MP=PJ.$$ Since this system has two equations in four unknowns, we have end up with two degrees of freedom, and therefore we have $$P_{a,b}=\begin{pmatrix}0 & 5a \\ b & 2a \end{pmatrix},$$ for any $a,b\in\mathbb{R}$. We choose $P=P_{1,1}$, and see $\det(P)\neq0$, so $P$ has an inverse given by $$ P^{-1}=\begin{pmatrix}-\frac{2}{5} & 1 \\ \frac{1}{5} & 0\end{pmatrix}. $$ Because $J$ is a diagonal matrix we have $$ J = \begin{pmatrix} e^{-2\pi} & 0 \\ 0 & e^{2\pi} \end{pmatrix} = e^{ \begin{pmatrix} -2\pi & 0 \\ 0 & 2\pi \end{pmatrix} } = e^{ 2\pi\hat{J}},\quad \hat{J}=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}. $$
So we now have an alternate expression for $M$: $$ M=PJP^{-1}=Pe^{ 2\pi\hat{J}} P^{-1} = e^{2\pi P\hat{J}P^{-1}}, $$
where
$$ P\hat{J}P^{-1} = \begin{pmatrix} 0 & 5 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} -\frac{2}{5} & 1\\ \frac{1}{5} & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ \frac{4}{5} & -1 \end{pmatrix} = \Omega, $$ and that is exactly as was to be shown.