Let's solve the system of differential equations step by step: Given: \[ \frac{dx}{dt} = x - 2y \quad (1) \] \[ \frac{dy}{dt} = 5x + 3y \quad (2) \] --- ### Step 1: Write the system in matrix form We can express the system as: \[ \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \] Let: \[ \mathbf{X} = \begin{bmatrix} x \\ y \end{bmatrix} \] and the coefficient matrix: \[ A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} \] --- ### Step 2: Find eigenvalues of matrix \(A\) Eigenvalues \(\lambda\) satisfy: \[ \det(A - \lambda I) = 0 \] Compute the determinant: \[ \det \begin{bmatrix} 1 - \lambda & -2 \\ 5 & 3 - \lambda \end{bmatrix} = (1 - \lambda)(3 - \lambda) - (-2)(5) \] Simplify: \[ (1 - \lambda)(3 - \lambda) + 10 = 0 \] Expand: \[ (1 \times 3) - 1 \lambda - 3 \lambda + \lambda^2 + 10 = 0 \] \[ 3 - 4 \lambda + \lambda^2 + 10 = 0 \] \[ \lambda^2 - 4 \lambda + 13 = 0 \] Solve quadratic: \[ \lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 13}}{2} \] \[ = \frac{4 \pm \sqrt{16 - 52}}{2} \] \[ = \frac{4 \pm \sqrt{-36}}{2} \] \[ = \frac{4 \pm 6i}{2} \] \[ = 2 \pm 3i \] **Eigenvalues:** \[ \boxed{\lambda_1 = 2 + 3i, \quad \lambda_2 = 2 - 3i} \] --- ### Step 3: Find eigenvectors Choose \(\lambda = 2 + 3i\). Find eigenvector \(\mathbf{v}\) such that: \[ (A - \lambda I) \mathbf{v} = 0 \] Compute \(A - \lambda I\): \[ A - (2 + 3i) I = \begin{bmatrix} 1 - (2 + 3i) & -2 \\ 5 & 3 - (2 + 3i) \end{bmatrix} = \begin{bmatrix} -1 - 3i & -2 \\ 5 & 1 - 3i \end{bmatrix} \] Set up the equations: \[ (-1 - 3i) v_1 - 2 v_2 = 0 \] \[ 5 v_1 + (1 - 3i) v_2 = 0 \] From the first: \[ (-1 - 3i) v_1 = 2 v_2 \Rightarrow v_2 = \frac{(-1 - 3i)}{2} v_1 \] Choose \(v_1 = 1\): \[ v_2 = \frac{-1 - 3i}{2} \] Eigenvector: \[ \mathbf{v} = \begin{bmatrix} 1 \\ \frac{-1 - 3i}{2} \end{bmatrix} \] Similarly, for \(\lambda = 2 - 3i\), eigenvector will be the complex conjugate. --- ### Step 4: Write the general solution The solution to a system with complex eigenvalues is: \[ \mathbf{X}(t) = e^{\alpha t} \left[ C_1 \mathbf{Re}(\mathbf{v} e^{i \beta t}) + C_2 \mathbf{Im}(\mathbf{v} e^{i \beta t}) \right] \] where: - \(\lambda = \alpha + i \beta\), - \(\mathbf{v}\) is the eigenvector for \(\lambda\), - \(C_1, C_2\) are constants. Here: \[ \alpha = 2, \quad \beta = 3 \] and the eigenvector (for \(\lambda=2+3i\)) is: \[ \mathbf{v} = \begin{bmatrix} 1 \\ \frac{-1 - 3i}{2} \end{bmatrix} \] --- ### **Step 5: Express the solution in real form** Let’s write the solution explicitly: \[ \mathbf{X}(t) = e^{2 t} \left[ \mathbf{A} \cos 3 t + \mathbf{B} \sin 3 t \right] \] where \(\mathbf{A}\) and \(\mathbf{B}\) are real vectors derived from the real and imaginary parts of the eigenvector. --- ### **Step 6: Find the real eigenvector components** Eigenvector: \[ \mathbf{v} = \begin{bmatrix} 1 \\ \frac{-1 - 3i}{2} \end{bmatrix} = \begin{bmatrix} 1 \\ -\frac{1}{2} - \frac{3}{2} i \end{bmatrix} \] Real part: \[ \mathbf{v}_R = \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix} \] Imaginary part: \[ \mathbf{v}_I = \begin{bmatrix} 0 \\ -\frac{3}{2} \end{bmatrix} \] Thus, the general real solution: \[ \boxed{ \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = e^{2 t} \left[ C_1 \begin{bmatrix} 1 \\ -\frac{1}{2} \end{bmatrix} \cos 3 t + C_2 \begin{bmatrix} 0 \\ -\frac{3}{2} \end{bmatrix} \sin 3 t \right] } \] Simplify: \[ x(t) = e^{2 t} \left[ C_1 \cos 3 t \right] \] \[ y(t) = e^{2 t} \left[ -\frac{1}{2} C_1 \cos 3 t - \frac{3}{2} C_2 \sin 3 t \right] \] Constants \(C_1, C_2\) are determined by initial conditions. --- ### **### **Conclusion:** The solutions to the system are of the form: \[ \boxed{ \begin{cases} x(t) = e^{2 t} \left[ A \cos 3 t + B \sin 3 t \right] \\ y(t) = e^{2 t} \left[ -\frac{1}{2} A \cos 3 t - \frac{3}{2} B \sin 3 t \right] \end{cases} } \] where \(A, B\) are arbitrary constants based on initial conditions. **This indicates the solutions are exponentially growing with oscillatory behavior, due to the complex eigenvalues.**