Exercise :
Find the matrix $A$, such that the following function is a solution to a $2\times 2$ differential equation system :
$$x(t) = \begin{bmatrix} e^{-t}(\cos t + 2\sin t) \\ e^{-t}\cos t \end{bmatrix}$$
Attempt :
I understand we need the following :
$$\begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$
so we get the system :
$$\begin{cases} x_1' = ax_1 + bx_2 \\ x_2' = cx_1 + dx_2\end{cases}$$
which translates to :
$$\begin{cases} x_1' = a[e^{-t}(\cos t + 2\sin t)] + b e^{-t}\cos t \\ x_2' = c[e^{-t}(\cos t + 2\sin t)] + d e^{-t}\cos t\end{cases}$$
but this leads me in a $2\times 2$ system whilst having $4$ variables to find.
I seem to be a little stuck on this problem, so I would really appreciate some hints/tips or a solution. This is an exercise that is part of a chapter involving stability and stationary points, so maybe there's a way to work through them to calculate the matrix $A$ ? I tried checking it by eye but didn't see anything.
I would really appreciate any help !