Consider the second-order system of differential equations \ddot{x}=-2 x+6 y, \quad \ddot{y}=-x+5 y \text { The coefficient matrix } \left.\left[\begin{array}{ll} -2 & 6 \\ -1 & 5 \end{array}\right] \text {

has eigenvectors }\left[\begin{array}{l} 6 \\ 1 \end{array}\right] \text { and } \mid \begin{array}{l} 1 \\ 1 \end{array}\right], \text { and the } corresponding eigenvalues are –1 and 4. \mathbf{B} \quad\left[\begin{array}{l} x \\ y \end{array}\right]=C_{1}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] e^{t}+C_{2}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] e^{-t}+C_{3}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] \cos (2 t)+C_{4}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] \sin (2 t) \mathbf{C}\left[\begin{array}{l} x \\ y \end{array}\right]=C_{1}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] e^{t}+C_{2}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] e^{-t}+C_{3}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \cos (2 t)+C_{4}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \sin (2 t) \mathbf{D}\left[\begin{array}{l} x \\ y \end{array}\right]=C_{1}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \cos t+C_{2}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \sin t+C_{3}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] e^{2 t}+C_{4}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] e^{-2 t} \mathbf{A} \quad\left[\begin{array}{l} x \\ y \end{array}\right]=C_{1}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] \cos t+C_{2}\left[\begin{array}{l} 6 \\ 1 \end{array}\right] \sin t+C_{3}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] e^{2 t}+C_{4}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] e^{-2 t}