How can the following coupled system of ODE be solved?

Question

The following system is a coupled ODEs that are dependant on each other.

$$ Y''(x)+A_1 (Y(x) - Z(x)) = A_2 Y(x) - H(x -A_3)$$ $$ Z''(x)+B_1 (Z(x) - Y(x)) = B_2 Z(x) - H(x -B_3)$$

Where:

H(x) is the Unit Step Function

Answer 1

Let $\begin{cases} y_1(x) = Y(x)\\ y_2(x) = y_1'(x)\\ z_1(x) = Z(x)\\ z_2(x) = z_1'(x) \end{cases}$, you may rewrite as,

$\begin{cases} y_1(x)' = y_2(x)\\ y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\\ z_1(x)' = z_2(x)\\ z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\\ h_1(x) = H(x - A_3)\\ h_2(x) = H(x - B_3) \end{cases}$

and

$ \left( \begin{array}{c} y_1(x)'\\ y_2(x)'\\ z_1(x)'\\ z_2(x)'\\ h_1(x)\\ h_2(x) \end{array} \right) = \left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0\\ (A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0\\ B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} Y(x)\\ Y'(x)\\ Z(x)\\ Z'(x)\\ H(x - A_3)\\ H(x - B_3) \end{array} \right) $