Let's solve the system of ODEs step by step. --- ### Step 1: Write the System in Matrix Form and Find the Homogeneous Solution The system is: \[ \begin{cases} x'(t) = -4x - 3y - 5 \\ y'(t) = 2x + 3y - 2 \end{cases} \] Write in matrix form: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} -5 \\ -2 \end{pmatrix} \] First, solve the **homogeneous system** (ignore the constant vector for now): \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] Find the eigenvalues \(\lambda\) by solving: \[ \det \begin{pmatrix} -4-\lambda & -3 \\ 2 & 3-\lambda \end{pmatrix} = 0 \] \[ (-4-\lambda)(3-\lambda) - (-3)(2) = 0 \] \[ (-4-\lambda)(3-\lambda) + 6 = 0 \] \[ (-4-\lambda)(3-\lambda) = -12 -3\lambda +4\lambda +\lambda^2 = \lambda^2 + \lambda -12 \] \[ \lambda^2 + \lambda -12 + 6 = 0 \implies \lambda^2 + \lambda -6 = 0 \] \[ \lambda = \frac{-1 \pm \sqrt{1+24}}{2} = \frac{-1 \pm 5}{2} \implies \lambda_1 = 2, \quad \lambda_2 = -3 \] **Explanation:** We rewrote the system in matrix form and found the eigenvalues of the coefficient matrix to solve the homogeneous part. --- ### Step 2: Find the Particular Solution and General Solution Let’s look for a **particular solution** of the form: \[ \begin{pmatrix} x_p \\ y_p \end{pmatrix} \] where \(x_p, y_p\) are constants. Plug into the original non-homogeneous system: \[ \begin{cases} 0 = -4x_p - 3y_p - 5 \\ 0 = 2x_p + 3y_p - 2 \end{cases} \] Solve the second equation for \(y_p\): \[ 2x_p + 3y_p = 2 \implies y_p = \frac{2-2x_p}{3} \] Plug into the first equation: \[ -4x_p - 3\left(\frac{2-2x_p}{3}\right) - 5 = 0 \\ -4x_p - 2 + 2x_p - 5 = 0 \\ (-4x_p + 2x_p) - 7 = 0 \\ -2x_p - 7 = 0 \implies x_p = -\frac{7}{2} \] Then, \[ y_p = \frac{2 - 2(-\frac{7}{2})}{3} = \frac{2 + 7}{3} = 3 \] So, a particular solution is: \[ x_p = -\frac{7}{2}, \quad y_p = 3 \] The **general solution** is: \[ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = C_1 \vec{v}_1 e^{2t} + C_2 \vec{v}_2 e^{-3t} + \begin{pmatrix} -\frac{7}{2} \\ 3 \end{pmatrix} \] You can find the eigenvectors \(\vec{v}_1, \vec{v}_2\) by plugging the eigenvalues into the matrix \(\begin{pmatrix}-4-\lambda & -3 \\ 2 & 3-\lambda\end{pmatrix}\). **Explanation:** We solved for the particular solution by setting derivatives to zero (constant solution) and then combined it with the general homogeneous solution using the eigenvalues from Step 1. --- ## **Final Answer** \[ \boxed{ \begin{aligned} x(t) &= A e^{2t} + B e^{-3t} - \frac{7}{2} \\ y(t) &= \alpha A e^{2t} + \beta B e^{-3t} + 3 \end{aligned} } \] where \(A, B\) are constants, and \(\alpha, \beta\) come from the eigenvectors corresponding to eigenvalues \(2\) and \(-3\). The particular solution is \(x_p = -\frac{7}{2},\ y_p = 3\).