Calculating the state transition matrix of an LTV system using the Fundamental Matrix. I'm trying to determine the state transition matrix,$\Phi(t,t_0)$, of the following system:
$$
\begin{bmatrix}x'_1\\x'_2\end{bmatrix}=
\begin{bmatrix}-\sin(t)&t\\0&-\cos(t)\end{bmatrix}
\times \begin{bmatrix}x_1\\x_2\end{bmatrix}
$$
with initial condition $$x(t_0)=\begin{bmatrix}x_1(t_0)\\x'_2(t_0)\end{bmatrix}$$
The fundamental matrix I got from this system is:
$$
\begin{bmatrix}(t^2/2)+\cos(t)&1-\sin(t)\\t^2&2-\sin(t)\end{bmatrix}$$
and from that, I got $$\Phi(t,t_0)=
\begin{bmatrix}1&0\\0&1\end{bmatrix}
$$
which I don't think is correct. Can someone tell me where I'm going wrong? Is it that my fundamental matrix is incorrect?
The general Fundamental Matrix that I got is:
$$
\begin{bmatrix}(t^2-t^2_0/2)+\cos(t-t_0)&1-\sin(t-t_0)\\t^2-t_0^2&2-\sin(t-t_0)\end{bmatrix}$$
 A: From $x_2'(t)=- x_2(t)\cos t$, you obtain (say, by separation of variables), $x_2(t)=x_2(t_0)e^{\sin t_0}e^{-\sin t}$.
Substitute in the first equation and you obtain $$x_1'(t)+x_1(t)\sin t=x_2(t_0)e^{\sin t_0}te^{-\sin t}$$
The fundamental function of (the homogenous part of) this last ODE is $e^{\cos t}$, and therefore \begin{align}&x_1(t)=x_1(t_0)e^{-\cos t_0}e^{\cos t}+e^{\cos t}\int_{t_0}^t x_2(t_0)e^{\sin t_0}ue^{-\cos u-\sin u}du=\\=&\left(x_1(t_0)e^{-\cos t_0}+x_2(t_0)e^{\sin t_0}\int_{t_0}^0 ue^{-\cos u-\sin u}du\right)e^{\cos t}+x_2(t_0)e^{\sin t_0}e^{\cos t}\int_{0}^t ue^{-\cos u-\sin u}du\end{align}
Which means that $$\begin{pmatrix}x_1(t)\\ x_2(t)\end{pmatrix}=\begin{pmatrix}A e^{\cos t}+Be^{\cos t}\int_0^t ue^{-\cos u-\sin u}\,du\\ Be^{-\sin t}\end{pmatrix}$$
With $A=x_1(t_0)e^{-\cos t_0}+x_2(t_0)e^{\sin t_0}\int_{t_0}^0 ue^{-\cos u-\sin u}\,du$ and $B=x_2(t_0)e^{\sin t_0}$. This means that a possible choice for $\Psi$ is $$\Psi(t)=\begin{pmatrix}e^{\cos t}&e^{\cos t}\int_0^tue^{-\cos u-\sin u}\,du\\ 0&e^{-\sin t}\end{pmatrix}$$
And now you can calculate easily $\Psi(\tau)(\Psi(t))^{-1}$.
