I am trying to compute the state transition matrix $\Phi(t,0)$ for the following linear time-varying system of the form
$$\dot{x}(t) = A(t)x(t) + Bu(t)$$ with a periodicity in $A(t)$.
The state transition matrix $\Phi(t,0)$ is a matrix that describes the state transition of $x(t)$ to the system of equations above such that
$$ x(t) = \Phi(t,0)x(0) + \int_0^{t}\Phi(t,\tau)B(\tau)u(\tau)d\tau. $$
My system has $x(t) \in \mathbb{R}^2$, $u(t) \in \mathbb{R}$, $B = [0, 1]^{\top}$, and the matrix $A(t)$ is periodic as below
$$ A(t) =\begin{bmatrix} 0 & 1 \\ -sin(\pi t) & -1 \end{bmatrix}. $$
Since it's not time-invariant, $\Phi(t,0)$ is not a simple matrix exponential of $A(t)$.
My futile attempt was as follows:
Using the fact that $\frac{d\Phi(t,0)}{dt} = A(t)\Phi(t,0), \quad \Phi(0,0) = I,\quad$ where $I$ is an identity matrix,
I can set up four equations
$$ \dot{\Phi}_{11}(t,0) = \Phi_{21}(t,0), \quad \Phi_{11}(0,0) = 1, $$ $$ \dot{\Phi}_{12}(t,0) = \Phi_{22}(t,0), \quad \Phi_{12}(0,0) = 0, $$ $$ \dot{\Phi}_{21}(t,0) = -\Phi_{21}(t,0) - \Phi_{11}(t,0)\sin(\pi t), \quad \Phi_{21}(0,0) = 0, $$ $$ \dot{\Phi}_{22}(t,0) = -\Phi_{22}(t,0) - \Phi_{12}(t,0)\sin(\pi t), \quad \Phi_{22}(0,0) = 1. $$
However, it doesn't make them easier to solve.
Also, $A(t)$ is not integral commutative, $A(t)\int_{\tau}^tA(\sigma)d\sigma \neq \big(\int_{\tau}^tA(\sigma)d\sigma\big) A(t)$, and so I cannot compute it using $\Phi(t,0) = e^{\int_{\tau}^t A(\sigma)d\sigma}$, either.
I also tried to compute a few terms of the Peano-Baker series and don't think this is helpful.
Seems that the periodicity of $A(t)$ should help but I am not sure how in obtaining the transition matrix.
I am looking for an analytic solution but if there is none, I want to understand why there is none (if only can be solved numerically).