Find the general solution of the given system. (dx)/(dt)=6x-y (dy)/(dt)=36x-6y langle x(t),y(t)rangle = square

\langle x(t), y(t) \rangle = \langle c_1 e^{6\sqrt{2}t} + c_2 e^{-6\sqrt{2}t}, (6\sqrt{2} - 6)c_1 e^{6\sqrt{2}t} + (-6\sqrt{2} - 6)c_2 e^{-6\sqrt{2}t} \rangle Explanation 1. Write the system in matrix form The system can be written as \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 & -1 \\ 36 & -6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. 2. Find eigenvalues of the coefficient matrix Solve \det\left(\begin{bmatrix} 6 & -1 \\ 36 & -6 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 0. This gives (6-\lambda)(-6-\lambda) + 36 = 0. 3. Simplify and solve the characteristic equation The characteristic equation is \lambda^2 - 0\lambda - 72 = 0, which simplifies to \lambda^2 = 72. Thus, \lambda = \pm 6\sqrt{2}. 4. Find eigenvectors for each eigenvalue For \lambda_1 = 6\sqrt{2}, solve \begin{bmatrix} 6 - 6\sqrt{2} & -1 \\ 36 & -6 - 6\sqrt{2} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. A possible eigenvector is \begin{bmatrix} 1 \\ 6\sqrt{2} - 6 \end{bmatrix}. For \lambda_2 = -6\sqrt{2}, solve \begin{bmatrix} 6 + 6\sqrt{2} & -1 \\ 36 & -6 + 6\sqrt{2} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. A possible eigenvector is \begin{bmatrix} 1 \\ -6\sqrt{2} - 6 \end{bmatrix}. 5. Construct the general solution The general solution is x(t) = c_1 e^{6\sqrt{2}t} + c_2 e^{-6\sqrt{2}t} and y(t) = (6\sqrt{2} - 6)c_1 e^{6\sqrt{2}t} + (-6\sqrt{2} - 6)c_2 e^{-6\sqrt{2}t}.