Let's solve the given system of ODEs step by step: Given: \[ \begin{cases} x'(t) = -4x - 3y - 5 \\ y'(t) = 2x + 3y - 2 \end{cases} \] --- ### **Step 1: Find the Steady-State (Particular Solution)** Set derivatives to zero (for steady-state): \[ \begin{cases} 0 = -4x - 3y - 5 \\ 0 = 2x + 3y - 2 \end{cases} \] Solve for \(x\) and \(y\): Add the two equations to eliminate \(y\): \[ (-4x - 3y - 5) + (2x + 3y - 2) = 0 \\ -2x - 7 = 0 \implies x = -\frac{7}{2} \] Substitute \(x = -\frac{7}{2}\) into the second equation: \[ 2\left(-\frac{7}{2}\right) + 3y - 2 = 0 \\ -7 + 3y - 2 = 0 \\ 3y = 9 \implies y = 3 \] **Explanation:** We set the derivatives to zero to find the equilibrium (constant) solutions, also called the particular solution. **Steady-state solution:** \[ (x_p, y_p) = \left(-\frac{7}{2}, 3\right) \] --- ### **Step 2: Solve for the Homogeneous System** Let \(x = X + x_p\), \(y = Y + y_p\) where \(X, Y\) are new variables representing the deviation from the steady state. So, substitute \(x = X - \frac{7}{2}\), \(y = Y + 3\): Rewrite the system: \[ \begin{cases} X' = -4X - 3Y \\ Y' = 2X + 3Y \end{cases} \] (This is because the constant terms cancel out using the steady-state values.) This is a homogeneous linear system. Write in matrix form: \[ \frac{d}{dt} \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} \] Find eigenvalues \(\lambda\) by solving: \[ \begin{vmatrix} -4 - \lambda & -3 \\ 2 & 3 - \lambda \end{vmatrix} = 0 \] \[ (-4-\lambda)(3-\lambda) - (-3)(2) = 0 \\ (-4-\lambda)(3-\lambda) + 6 = 0 \\ (-4-\lambda)(3-\lambda) + 6 = 0 \] Expand: \[ (-4-\lambda)(3-\lambda) = -12 + 4\lambda -3\lambda + \lambda^2 = \lambda^2 + \lambda -12 \] So: \[ \lambda^2 + \lambda -12 + 6 = 0 \implies \lambda^2 + \lambda -6 = 0 \] Solve quadratic: \[ \lambda = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2} \] Thus, \[ \lambda_1 = 2,\quad \lambda_2 = -3 \] Find corresponding eigenvectors for each \(\lambda\): **For \(\lambda_1 = 2\):** \[ \begin{pmatrix} -6 & -3 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 0 \] \[ -6a - 3b = 0 \implies 2a + b = 0 \implies b = -2a \] **For \(\lambda_2 = -3\):** \[ \begin{pmatrix} -1 & -3 \\ 2 & 6 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 0 \] \[ -a - 3b = 0 \implies a = -3b \] So, the general homogeneous solution is: \[ \begin{pmatrix} X \\ Y \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} -3 \\ 1 \end{pmatrix} e^{-3t} \] **Explanation:** We solved the shifted homogeneous system by finding eigenvalues and eigenvectors, which yields the general solution for deviations from the steady state. --- ## **Final Answer** Combine the particular (steady-state) solution and the homogeneous solution, and substitute back for \(x\) and \(y\): \[ \boxed{ \begin{aligned} x(t) &= -\frac{7}{2} + c_1 e^{2t} - 3c_2 e^{-3t} \\ y(t) &= 3 - 2c_1 e^{2t} + c_2 e^{-3t} \end{aligned} } \] where \(c_1\) and \(c_2\) are constants determined by initial conditions.

Let's approach the problem in two clear steps, explaining each carefully. --- ### **Step 1: Find the steady-state (particular solution)** **Given system:** \[ \begin{cases} x'(t) = -4x - 3y - 5 \\ y'(t) = 2x + 3y - 2 \end{cases} \] **Explanation:** To find the particular solution (steady-state), set derivatives to zero (since at steady state, variables are constant): \[ \begin{cases} 0 = -4x - 3y - 5 \\ 0 = 2x + 3y - 2 \end{cases} \] Solve these algebraic equations: 1. From the second equation: \[ 2x + 3y = 2 \implies 2x = 2 - 3y \implies x = \frac{2 - 3y}{2} \] 2. Substitute \(x\) into the first equation: \[ 0 = -4 \left( \frac{2 - 3y}{2} \right) - 3y - 5 \] Simplify: \[ 0 = -2(2 - 3y) - 3y - 5 \] \[ 0 = -4 + 6y - 3y - 5 \] \[ 0 = -4 + 3y - 5 \] \[ 0 = -9 + 3y \] \[ 3y = 9 \implies y = 3 \] 3. Find \(x\): \[ x = \frac{2 - 3(3)}{2} = \frac{2 - 9}{2} = \frac{-7}{2} \] **Result:** \[ \boxed{ x_{ss} = -\frac{7}{2}, \quad y_{ss} = 3 } \] **This is the particular (steady-state) solution.** --- ### **Step 2: Find the general solution of the homogeneous system** **Change variables:** \[ X = x - x_{ss} = x + \frac{7}{2}, \quad Y = y - y_{ss} = y - 3 \] Rewriting the original system in terms of \(X, Y\): \[ \begin{cases} X' = x' = -4x - 3y - 5 \\ Y' = y' = 2x + 3y - 2 \end{cases} \] But since \(x = X - \frac{7}{2}\), \(y = Y + 3\), substitute: \[ X' = -4 \left( X - \frac{7}{2} \right) - 3 (Y + 3) - 5 \] \[ Y' = 2 \left( X - \frac{7}{2} \right) + 3 (Y + 3) - 2 \] Simplify: \[ X' = -4X + 14 - 3Y - 9 - 5 = -4X - 3Y \] \[ Y' = 2X - 7 + 3Y + 9 - 2 = 2X + 3Y \] **Homogeneous system:** \[ \begin{cases} X' = -4X - 3Y \\ Y' = 2X + 3Y \end{cases} \] Expressed in matrix form: \[ \frac{d}{dt} \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} -4 & -3 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} \] Find eigenvalues \(\lambda\): \[ \det \begin{bmatrix} -4 - \lambda & -3 \\ 2 & 3 - \lambda \end{bmatrix} = 0 \] Calculate: \[ (-4 - \lambda)(3 - \lambda) - (-3)(2) = 0 \] \[ (-4 - \lambda)(3 - \lambda) + 6 = 0 \] Expand: \[ (-4)(3) + 4\lambda + 3\lambda + \lambda^2 + 6 = 0 \] \[ -12 + 7\lambda + \lambda^2 + 6 = 0 \] \[ \lambda^2 + 7\lambda - 6 = 0 \] Solve quadratic: \[ \lambda = \frac{-7 \pm \sqrt{49 - 4 \times 1 \times (-6)}}{2} = \frac{-7 \pm \sqrt{49 + 24}}{2} = \frac{-7 \pm \sqrt{73}}{2} \] Eigenvalues: \[ \boxed{ \lambda_1 = \frac{-7 + \sqrt{73}}{2}, \quad \lambda_2 = \frac{-7 - \sqrt{73}}{2} } \] Corresponding eigenvectors can be found for each eigenvalue, but for simplicity, the general solution involves exponentials with these eigenvalues. --- ### **Final solution:** \[ \boxed{ \begin{aligned} x(t) &= -\frac{7}{2} + C_1 e^{\lambda_1 t} \mathbf{v}_1^{(x)} + C_2 e^{\lambda_2 t} \mathbf{v}_2^{(x)} \\ y(t) &= 3 + C_1 e^{\lambda_1 t} \mathbf{v}_1^{(y)} + C_2 e^{\lambda_2 t} \mathbf{v}_2^{(y)} \end{aligned} } \] where \(\mathbf{v}_1, \mathbf{v}_2\) are eigenvectors corresponding to \(\lambda_1, \lambda_2\). **In summary:** - The particular (steady-state) points are \(\left(-\frac{7}{2}, 3\right)\). - The homogeneous solution involves exponentials with eigenvalues \(\frac{-7 \pm \sqrt{73}}{2}\). --- **Note:** The constants \(C_1, C_2\) are determined by initial conditions. --- **This completes the solution in two steps with explanations.**