Let's solve the given system step by step: Given: \[ \vec{y}' = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \vec{y} + \begin{bmatrix} e^{-4t} \\ 0 \end{bmatrix} \] with initial condition \(\vec{y}(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\). --- ### **a. Complementary Solution \((\vec{y}_C)\) to the Homogeneous Equation** Set the nonhomogeneous term to zero: \[ \vec{y}' = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \vec{y} \] Let \(\vec{y} = \vec{v}e^{\lambda t}\). Find eigenvalues \(\lambda\): \[ \text{det} \begin{bmatrix} -1-\lambda & 4 \\ -2 & 5-\lambda \end{bmatrix} = 0 \] Compute determinant: \[ (-1-\lambda)(5-\lambda) + 8 = 0 \] \[ (-1-\lambda)(5-\lambda) = (-1)(5-\lambda) - \lambda(5-\lambda) = -5 + \lambda -5\lambda + \lambda^2 = \lambda^2 - 4\lambda - 5 \] So: \[ \lambda^2 - 4\lambda - 5 + 8 = \lambda^2 - 4\lambda + 3 = 0 \] \[ (\lambda-1)(\lambda-3) = 0 \] So \(\lambda_1 = 1\), \(\lambda_2 = 3\). #### **Eigenvectors** **For \(\lambda=1\):** \[ \begin{bmatrix} -1-1 & 4 \\ -2 & 5-1 \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ -2 & 4 \end{bmatrix} \] So: \[ -2v_1 + 4v_2 = 0 \implies v_1 = 2v_2 \] Take \(v_2 = 1\): \(\vec{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\) **For \(\lambda=3\):** \[ \begin{bmatrix} -1-3 & 4 \\ -2 & 5-3 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 2 \end{bmatrix} \] So: \[ -4v_1 + 4v_2 = 0 \implies v_1 = v_2 \] Take \(v_2 = 1\): \(\vec{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) #### **Complementary Solution** \[ \boxed{ \vec{y}_C(t) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} } \] --- ### **b. Particular Solution** Assume \(\vec{y}_P(t) = e^{-4t} \vec{a}\) where \(\vec{a}\) is a constant vector. Compute derivative: \[ \vec{y}_P'(t) = -4e^{-4t}\vec{a} \] Plug into ODE: \[ -4e^{-4t}\vec{a} = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} e^{-4t}\vec{a} + \begin{bmatrix} e^{-4t} \\ 0 \end{bmatrix} \] Divide both sides by \(e^{-4t}\): \[ -4\vec{a} = \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \vec{a} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] Move all terms to one side: \[ \left(-4 I - \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} \right) \vec{a} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] Calculate \(-4I - A\): \[ -4I - A = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} - \begin{bmatrix} -1 & 4 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} -3 & -4 \\ 2 & -9 \end{bmatrix} \] So: \[ \begin{bmatrix} -3 & -4 \\ 2 & -9 \end{bmatrix} \vec{a} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \] Let \(\vec{a} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}\): \[ -3a_1 - 4a_2 = 1 \tag{1} \] \[ 2a_1 - 9a_2 = 0 \implies a_1 = \frac{9}{2} a_2 \tag{2} \] Substitute (2) into (1): \[ -3\left(\frac{9}{2}a_2\right) - 4a_2 = 1 \implies -\frac{27}{2}a_2 - 4a_2 = 1 \] \[ -\frac{27}{2}a_2 - \frac{8}{2}a_2 = 1 \implies -\frac{35}{2} a_2 = 1 \implies a_2 = -\frac{2}{35} \] \[ a_1 = \frac{9}{2} \left(-\frac{2}{35}\right) = -\frac{9}{35} \] So: \[ \boxed{ \vec{y}_P(t) = e^{-4t} \begin{bmatrix} -\frac{9}{35} \\ -\frac{2}{35} \end{bmatrix} } \] --- ### **c. General Solution and Apply Initial Condition** General solution: \[ \vec{y}(t) = \vec{y}_C(t) + \vec{y}_P(t) \] \[ = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + e^{-4t} \begin{bmatrix} -\frac{9}{35} \\ -\frac{2}{35} \end{bmatrix} \] Apply initial condition (\(t=0\)): \[ \vec{y}(0) = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \begin{bmatrix} -\frac{9}{35} \\ -\frac{2}{35} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] Set up two equations: \[ 2c_1 + c_2 - \frac{9}{35} = 0 \\ c_1 + c_2 - \frac{2}{35} = 0 \] Subtract second equation from the first: \[ (2c_1 + c_2 - \frac{9}{35}) - (c_1 + c_2 - \frac{2}{35}) = 0 \\ c_1 - \frac{7}{35} = 0 \implies c_1 = \frac{7}{35} \] Now use in the second equation: \[ \frac{7}{35} + c_2 - \frac{2}{35} = 0 \implies c_2 = \frac{2}{35} - \frac{7}{35} = -\frac{5}{35} = -\frac{1}{7} \] --- ### **Final Solution** \[ \boxed{ \vec{y}(t) = \frac{7}{35} \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{t} - \frac{1}{7} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + e^{-4t} \begin{bmatrix} -\frac{9}{35} \\ -\frac{2}{35} \end{bmatrix} } \] Or, componentwise: \[ \boxed{ \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} = \begin{bmatrix} \frac{14}{35} e^{t} - \frac{1}{7} e^{3t} - \frac{9}{35} e^{-4t} \\ \frac{7}{35} e^{t} - \frac{1}{7} e^{3t} - \frac{2}{35} e^{-4t} \end{bmatrix} } \] --- **Summary:** - **Complementary solution:** \( c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix}e^t + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix}e^{3t} \) - **Particular solution:** \( e^{-4t} \begin{bmatrix} -\frac{9}{35} \\ -\frac{2}{35} \end{bmatrix} \) - **General solution:** As boxed above.