Let's solve the given system step-by-step: Given system: \[ \vec{y}' = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \vec{y} \] We're asked for the **complementary solution** (homogeneous part): --- ### **Step 1: Find the eigenvalues of the matrix** Matrix \( A = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \) The characteristic equation is: \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} -1 - \lambda & 4 \\ -2 & 5 - \lambda \end{vmatrix} = 0 \] \[ (-1 - \lambda)(5 - \lambda) - (4)(-2) = 0 \] \[ (-1 - \lambda)(5 - \lambda) + 8 = 0 \] \[ ( -1 \cdot 5 -1 \cdot (-\lambda) - \lambda \cdot 5 + \lambda^2 ) + 8 = 0 \] \[ (-5 + \lambda -5\lambda + \lambda^2) + 8 = 0 \] \[ \lambda^2 - 4\lambda + 3 = 0 \] Solving the quadratic equation: \[ (\lambda - 3)(\lambda - 1) = 0 \] So, **eigenvalues** are: \(\lambda_1 = 3, \lambda_2 = 1\) **Explanation:** The eigenvalues represent the exponential rates in the solution to the homogeneous system. --- ### **Step 2: Find the eigenvectors for each eigenvalue** #### For \(\lambda_1 = 3\): Substitute \(\lambda = 3\) into \(A - \lambda I\): \[ A - 3I = \begin{bmatrix} -1-3 & 4 \\ -2 & 5-3 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 2 \end{bmatrix} \] Set up the equation for the eigenvector \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\): \[ \begin{bmatrix} -4 & 4 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 \] This gives: \(-4x_1 + 4x_2 = 0\) So, \(x_2 = x_1\) So, the eigenvector is: \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) #### For \(\lambda_2 = 1\): Substitute \(\lambda = 1\) into \(A - \lambda I\): \[ A - 1I = \begin{bmatrix} -1-1 & 4 \\ -2 & 5-1 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ -2 & 4 \end{bmatrix} \] Set up the equation: \[ -2x_1 + 4x_2 = 0 \implies x_2 = \frac{1}{2}x_1 \] So, the eigenvector is: \(\begin{bmatrix} 1 \\ 0.5 \end{bmatrix}\) **Explanation:** The eigenvectors give the directions of the solution components for each exponential. --- ### **Final Answer:** The complementary solution is: \[ \vec{y}_c(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 1 \\ 0.5 \end{bmatrix} e^{t} \] So, fill the boxes as follows: \[ \vec{y}_c(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 1 \\ 0.5 \end{bmatrix} e^{t} \]

Let's solve the problem step-by-step: --- **Step 1: Find the eigenvalues of the matrix** Given matrix: \[ A = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \] Characteristic equation: \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} -1 - \lambda & 4 \\ -2 & 5 - \lambda \end{vmatrix} = 0 \] Calculating the determinant: \[ (-1 - \lambda)(5 - \lambda) - (4)(-2) = 0 \] \[ (-1 - \lambda)(5 - \lambda) + 8 = 0 \] Expanding: \[ (-1)(5) + (-1)(-\lambda) + (-\lambda)(5) + \lambda^2 + 8 = 0 \] \[ -5 + \lambda - 5\lambda + \lambda^2 + 8 = 0 \] \[ \lambda^2 - 4\lambda + 3 = 0 \] Solving the quadratic: \[ \lambda^2 - 4\lambda + 3 = 0 \] \[ (\lambda - 3)(\lambda - 1) = 0 \] **Eigenvalues:** \[ \boxed{\lambda_1 = 3,\quad \lambda_2 = 1} \] --- **Step 2: Find the eigenvectors corresponding to each eigenvalue** **For \(\lambda_1 = 3\):** Compute \(A - 3I\): \[ A - 3I = \begin{bmatrix} -1-3 & 4 \\ -2 & 5-3 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 2 \end{bmatrix} \] Set up \(\left(A - 3I\right)\mathbf{v} = 0\): From the first row: \[ -4x + 4y = 0 \implies y = x \] Choose \(x=1\), then \(y=1\). Eigenvector: \[ \boxed{\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}} \] **For \(\lambda_2 = 1\):** Compute \(A - I\): \[ A - I = \begin{bmatrix} -1-1 & 4 \\ -2 & 5-1 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ -2 & 4 \end{bmatrix} \] Set up: \[ -2x + 4y = 0 \implies y = \frac{1}{2} x \] Choose \(x=1\), then \(y=0.5\). Eigenvector: \[ \boxed{\mathbf{v}_2 = \begin{bmatrix} 1 \\ 0.5 \end{bmatrix}} \] --- ### **Final answer:** The **complementary solution** (general solution to the homogeneous system) is: \[ \boxed{ \vec{y}_c(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 1 \\ 0.5 \end{bmatrix} e^{t} } \] This represents the general solution to the homogeneous differential system.