Let's solve the system step by step: ### 1. **Write the system in matrix form** Let \( \mathbf{X} = \begin{pmatrix} x \\ y \end{pmatrix} \), then: \[ \frac{d\mathbf{X}}{dt} = A\mathbf{X} \] where \[ A = \begin{pmatrix} 3 & 4 \\ 8 & 7 \end{pmatrix} \] --- ### 2. **Find the eigenvalues of \(A\)** The eigenvalues \( \lambda \) satisfy: \[ \det(A - \lambda I) = 0 \] \[ \det\begin{pmatrix} 3-\lambda & 4 \\ 8 & 7-\lambda \end{pmatrix} = 0 \] \[ (3-\lambda)(7-\lambda) - 32 = 0 \] \[ 21 - 3\lambda - 7\lambda + \lambda^2 - 32 = 0 \] \[ \lambda^2 - 10\lambda - 11 = 0 \] Solve the quadratic equation: \[ \lambda^2 - 10\lambda - 11 = 0 \] \[ \lambda = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-11)}}{2} \] \[ = \frac{10 \pm \sqrt{100 + 44}}{2} \] \[ = \frac{10 \pm \sqrt{144}}{2} \] \[ = \frac{10 \pm 12}{2} \] So, \[ \lambda_1 = \frac{10 - 12}{2} = -1 \] \[ \lambda_2 = \frac{10 + 12}{2} = 11 \] **Eigenvalues:** \[ \boxed{-1, 11} \] --- ### 3. **Find eigenvectors** #### For \( \lambda_1 = -1 \): Solve \( (A - (-1)I)\mathbf{K}_1 = 0 \): \[ A + I = \begin{pmatrix} 4 & 4 \\ 8 & 8 \end{pmatrix} \] Let \( \mathbf{K}_1 = \begin{pmatrix} a \\ b \end{pmatrix} \): \[ 4a + 4b = 0 \implies a = -b \] So one eigenvector is: \[ \mathbf{K}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] --- #### For \( \lambda_2 = 11 \): Solve \( (A - 11I)\mathbf{K}_2 = 0 \): \[ A - 11I = \begin{pmatrix} 3-11 & 4 \\ 8 & 7-11 \end{pmatrix} = \begin{pmatrix} -8 & 4 \\ 8 & -4 \end{pmatrix} \] \[ -8a + 4b = 0 \implies b = 2a \] So one eigenvector is: \[ \mathbf{K}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] --- ### 4. **General Solution** The general solution is: \[ \mathbf{X}(t) = c_1 \mathbf{K}_1 e^{\lambda_1 t} + c_2 \mathbf{K}_2 e^{\lambda_2 t} \] \[ = c_1 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{11t} \] --- ## **Final Answers** **Eigenvalues:** \[ \boxed{-1, 11} \] **Eigenvectors:** \[ \mathbf{K}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] \[ \mathbf{K}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] **General solution:** \[ \boxed{ \mathbf{X}(t) = c_1 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{11t} } \]