Question
\left[\begin{array}{ll} 9 & 0 \\ 0 & 1 \end{array}\right] \ddot{\mathbf{x}}(t)+\left[\begin{array}{cc} 27 & -3 \\ -3 & 3 \end{array}\right] \mathbf{x}(t)=\mathbf{0} \mathbf{x}(0)=\left[\begin{array}{c} -\frac{1}{3} \\ 1 \end{array}\right] \dot{\mathbf{x}}(0)=\mathbf{0}
Question image 1Question image 2Question image 3