For $\alpha \geq 0$ and $\beta \in \mathbb{R}$, we consider the system
$$x'(t)+\alpha x(t)=-\beta y(t), t \geq 0, \\ y'(t)+\alpha y(t)=\beta x(t), t \geq 0, \\ x(0)=1 \\ y(0)=2.\\$$
I want to state the trapezoid method for this problem. Also I want to show that the computation of the approximation vector requires the solution of a linear system. And I want to find the matrix of this system.
The trapezoid method is the following, right?
$$y_{n+1}=y_n+\frac{h}{2} [f(x_n, y_n)+f(x_{n+1}, y_{n+1})]$$
But how can we state the trapezoid method for the given problem?
Also, the rpoblem can be also written in the followng form, right?
$$\binom{x'(t)}{y'(t)}=\begin{pmatrix} -\alpha & -\beta\\ \beta & -\alpha \end{pmatrix} \binom{x(t)}{y(t)}$$
So this is the wanted linear system that we have to solve, to compute the approximation vector and the matrix of the system is the following, right?
$$\begin{pmatrix} -\alpha & -\beta\\ \beta & -\alpha \end{pmatrix} $$
with $z_k = (x_k, \ y_k)'$ and $A=\left( \begin{array}{cc} -\alpha & -\beta \\ \beta & -\alpha \\ \end{array} \right)$ and following the Winter indications we have
$$ z_{k+1} = z_k +\frac h2(A z_k+A z_{k+1}) $$
or
$$ z_{k+1} = \left(I_2-\frac h2\right)^{-1} \left(I_2+\frac h2 A\right)z_k $$
or
$$ z_{k+1} = \left( \begin{array}{cc} \frac{4 \alpha h+8}{\left(\alpha ^2+\beta ^2\right) h^2+4 \alpha h+4}-1 & -\frac{4 \beta h}{\left(\alpha ^2+\beta ^2\right) h^2+4 \alpha h+4} \\ \frac{4 \beta h}{\left(\alpha ^2+\beta ^2\right) h^2+4 \alpha h+4} & \frac{4 \alpha h+8}{\left(\alpha ^2+\beta ^2\right) h^2+4 \alpha h+4}-1 \\ \end{array} \right)z_k $$