Given the following equations
\begin{cases} \dot x = - x - y + \cos t,\\ \dot y = x - y + \sin t. \end{cases}
I am looking for Poincaré map for the orbit through $(1,1)$ and the section $\Sigma = \{(s,s)|s>0\}$. I managed to find the general solution for some initial conditions $x_0,y_0$ \begin{equation} \begin{pmatrix} x \\ y \end{pmatrix} = e^{-t} \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix} \begin{pmatrix} x_0-1 \\ y_0 \end{pmatrix} + \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} \end{equation}
However, I think that for finding the Poincaré map I would have to solve the autonomous equation, while the one I have is non-autonomous.
But autonomous form of that equation is non-homogeneous... right? I am stuck here, Thank you.
As the system has many parts corresponding to a rotation in the plane, it can be interesting to explore it as a dynamic in the complex plane. Set $z=x+iy$, then $$ \dot z=(-1+i)z+e^{it}\\ z=Ae^{(-1+i)t}+Be^{it}\\ B(i+1-i)=1\implies B=1\\ z(0)=1+i=A+1\implies A=i $$ The task is now to find the points where $\arg(z(t))=\frac\pi4$, $$ \frac\pi4+2k\pi=\arg(e^{it}(ie^{-t}+1))=t+\arctan(e^{-t}) $$ for some $k\in\Bbb Z$. This is not symbolically solvable, so some kind of approximation is necessary.
For large $t>0$, the inverse tangent will have a value close to zero, so $t=t_k\approx \frac\pi4+2k\pi$. For large $t<0$, the inverse tangent will be close to $\frac\pi2$, so $t=t_k\approx-\frac{\pi}4+2k\pi$. This approximation should also be good for $k=\pm1$, but the correction to the exact root will be more noticeable.