Fundamental Matrix (Floquet theory) Let $\begin{pmatrix} \dot{x}_1 \\\dot{x}_2\end{pmatrix}=A(t)\begin{pmatrix}x_1\\x_2 \end{pmatrix}$ where $$A(t)=\begin{pmatrix}\alpha(t)+\cos(t)&\sin(t)\\ -\sin(t)& \alpha(t)+\cos(t)\end{pmatrix}$$ for a $\alpha:\mathbb{R} \rightarrow \mathbb{R}$ continuous and $2\pi$ periodic.
I want to show that $\Phi(t):=\exp(\int_0^\infty A(s)\;\mathrm{d}s$) defines a fundamental matrix. In order to do so, I have calculated that 
$$A(t)=\begin{pmatrix}i &-i\\1 &1\end{pmatrix}\begin{pmatrix} \cos(t)+ \alpha(t)+i\sin(t) &0\\0 &\cos(t)+ \alpha(t)-i\sin(t)\end{pmatrix}\begin{pmatrix}i &-i\\1 &1\end{pmatrix}^{-1}$$
So $A(t_1)A(t_2)=A(t_2)A(t_1)$ for all $t_1, t_2 \in \mathbb{R}$. No I am stuck proving that for all $v \in \mathbb{R}^2$ $$x(t):=\Phi(t)v$$ is a solution to the differential equation.
Thank you.
 A: If matrices $B$ and $C$ commute, then $\exp(B+C) = \exp(B) \exp(C)$.  One way to see this is that $\exp(t(B+C))$ and $\exp(tB) \exp(tC)$ are both 
solutions of the differential equation $\dfrac{d}{dt} \Psi(t) = (B + C) \Psi(t)$ with $\Psi(0)=I$.  
Now if all $A(s)$ commute, we have
$$\eqalign{\Phi(t+h) \Phi(t)^{-1} &= \exp\left(\int_0^{t+h} A(s)\; ds\right) \exp\left(-\int_0^{t} A(s)\; ds\right)\cr
&=\exp\left(\int_0^{t+h} A(s)\; ds - \int_0^t A(s)\; ds\right)\cr &= \exp\left(\int_t^{t+h} A(s)\; ds\right)\cr
  &= h A(t) + O(h^2)}$$
so that
$$ \Phi'(t) = \lim_{h \to 0} \dfrac{\Phi(t+h) - \Phi(t)}{h} = A(t) \Phi(t) $$
A: Having shown that $A(t_1) A(t_2) = A(t_2) A(t_1)$ has for consequence that $A(t) \int \limits _0 ^t A(s) \Bbb d s = \int \limits _0 ^t A(s) \Bbb d s A(t)$. From this follows that $\Big( \big( \int \limits _0 ^t A(s) \Bbb d s \big) ^n \Big) ' = n A(t) \big( \int \limits _0 ^t A(s) \Bbb d s \big) ^{n-1}$ (just apply Leibniz's rule for the derivation of a product; this will give you a sum of $n$ terms; in each such term, thanks to the commutation property shown above, bring an $A(t)$ to the front). Now, using this in $\exp \big( \int \limits _0 ^t A(s) \Bbb d s \big) = \sum \limits _{n=0} ^\infty \frac 1 {n!} \big( \int \limits _0 ^t A(s) \Bbb d s \big) ^n$ to derive term by term (I shall ignore matters of convergence here), you get $\Phi ' (t) = A(t) \exp \big( \int \limits _0 ^t A(s) \Bbb d s \big) = A(t) \Phi (t)$. Therefore, $x'(t) = A(t) x(t)$.
