Let's solve **Problem 8** step by step using the eigenvalue method: Given: \[ \begin{cases} x_1' = -5x_1 + 2x_2 \\ x_2' = 6x_1 - 5x_2 \end{cases} \] --- ### **Step 1: Write the System in Matrix Form and Find Eigenvalues** Write as: \[ \mathbf{x}' = A\mathbf{x} \] where \[ A = \begin{pmatrix} -5 & 2 \\ 6 & -5 \end{pmatrix} \] Find eigenvalues \(\lambda\): \[ \text{det}(A - \lambda I) = 0 \] \[ \left| \begin{matrix} -5 - \lambda & 2 \\ 6 & -5 - \lambda \end{matrix} \right| = 0 \] \[ (-5-\lambda)^2 - 12 = 0 \implies (-5-\lambda)^2 = 12 \] \[ -5-\lambda = \pm 2\sqrt{3} \] \[ \lambda_1 = -5 + 2\sqrt{3}, \quad \lambda_2 = -5 - 2\sqrt{3} \] **Explanation:** We rewrote the system as a matrix equation and found the eigenvalues by solving the characteristic equation. --- ### **Step 2: Find Eigenvectors** For each eigenvalue, solve \((A - \lambda I) \mathbf{v} = 0\). #### For \(\lambda_1 = -5 + 2\sqrt{3}\): \[ A - \lambda_1 I = \begin{pmatrix} -5 - (-5 + 2\sqrt{3}) & 2 \\ 6 & -5 - (-5 + 2\sqrt{3}) \end{pmatrix} = \begin{pmatrix} -2\sqrt{3} & 2 \\ 6 & -2\sqrt{3} \end{pmatrix} \] Set up: \[ -2\sqrt{3} v_1 + 2 v_2 = 0 \implies v_2 = \sqrt{3} v_1 \] Let \(v_1 = 1\): eigenvector is \(\begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}\). #### For \(\lambda_2 = -5 - 2\sqrt{3}\): \[ A - \lambda_2 I = \begin{pmatrix} 2\sqrt{3} & 2 \\ 6 & 2\sqrt{3} \end{pmatrix} \] Set up: \[ 2\sqrt{3} v_1 + 2 v_2 = 0 \implies v_2 = -\sqrt{3} v_1 \] Let \(v_1 = 1\): eigenvector is \(\begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix}\). **Explanation:** We found the eigenvectors corresponding to each eigenvalue by solving the system \((A - \lambda I)\mathbf{v} = 0\). --- ### **Step 3: Write the General Solution** The general solution is: \[ \mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} \] Substitute values: \[ \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} e^{(-5 + 2\sqrt{3}) t} + c_2 \begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix} e^{(-5 - 2\sqrt{3}) t} \] Or, written explicitly: \[ \boxed{ \begin{aligned} x_1(t) &= c_1 e^{(-5 + 2\sqrt{3}) t} + c_2 e^{(-5 - 2\sqrt{3}) t} \\ x_2(t) &= c_1 \sqrt{3} e^{(-5 + 2\sqrt{3}) t} - c_2 \sqrt{3} e^{(-5 - 2\sqrt{3}) t} \end{aligned} } \] **Explanation:** We constructed the general solution by combining the eigenvalues and eigenvectors found in the previous steps. --- ## **Final Answer** \[ \boxed{ x_1(t) = c_1 e^{(-5 + 2\sqrt{3}) t} + c_2 e^{(-5 - 2\sqrt{3}) t} } \] \[ \boxed{ x_2(t) = c_1 \sqrt{3} e^{(-5 + 2\sqrt{3}) t} - c_2 \sqrt{3} e^{(-5 - 2\sqrt{3}) t} } \] --- **Direction Field:** To sketch the direction field, plot the vectors \((x_1', x_2')\) at various points \((x_1, x_2)\). The solution curves will approach the origin, since both eigenvalues are negative, but at different rates depending on the initial condition. The solution will generally decay to zero as \(t\) increases, with the direction determined by the eigenvectors.