Let's solve the given linear system: \[ x' = x - 2y \] \[ y' = 2x - 3y \] We are to find the general solution in 3 steps, explaining each step. --- ### **Step 1: Write the system in matrix form and find eigenvalues** Write the system as: \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] Let \(A = \begin{pmatrix} 1 & -2 \\ 2 & -3 \end{pmatrix}\). To solve, we need the eigenvalues \(\lambda\) of \(A\): \[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} 1 - \lambda & -2 \\ 2 & -3 - \lambda \end{vmatrix} = (1-\lambda)(-3-\lambda) - (2)(-2) = 0 \] \[ = (1-\lambda)(-3-\lambda) + 4 = 0 \] \[ = [-(3+\lambda) + \lambda(3+\lambda)] + 4 = (1-\lambda)(-3-\lambda) + 4 \] Multiply out: \[ (1-\lambda)(-3-\lambda) = 1\cdot(-3-\lambda) - \lambda(-3-\lambda) = -3 - \lambda + 3\lambda + \lambda^2 = -3 + 2\lambda + \lambda^2 \] So the characteristic equation is: \[ -3 + 2\lambda + \lambda^2 + 4 = 0 \] \[ \lambda^2 + 2\lambda + 1 = 0 \] \[ (\lambda+1)^2 = 0 \] So, \(\lambda = -1\) is a repeated eigenvalue. **Explanation:** We rewrote the system in matrix form and found that it has a repeated eigenvalue \(\lambda = -1\). --- ### **Step 2: Find eigenvector and generalized eigenvector** For \(\lambda = -1\): \(A - (-1)I = A + I = \begin{pmatrix} 2 & -2 \\ 2 & -2 \end{pmatrix}\) Solve \((A+I)\vec{v} = 0\): \[ \begin{pmatrix} 2 & -2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] So \(2v_1 - 2v_2 = 0\) or \(v_1 = v_2\). Let \(v_1 = 1\), \(v_2 = 1\), so the eigenvector is \(\vec{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\). Next, find generalized eigenvector \(\vec{w}\): \[ (A + I)\vec{w} = \vec{v} \] \[ \begin{pmatrix} 2 & -2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \] This gives: \[ 2w_1 - 2w_2 = 1 \implies w_1 - w_2 = \frac{1}{2} \] Let \(w_2 = 0\), then \(w_1 = \frac{1}{2}\). So, a generalized eigenvector is \(\vec{w} = \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\). **Explanation:** We found the (regular) eigenvector \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and a generalized eigenvector \(\begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\), needed because the eigenvalue is repeated. --- ### **Step 3: Write the general solution** For a repeated eigenvalue with one linearly independent eigenvector, the general solution is: \[ \vec{x}(t) = c_1 e^{-t} \vec{v} + c_2 e^{-t}(t\vec{v} + \vec{w}) \] Plug in the found vectors: - \(\vec{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) - \(\vec{w} = \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\) So: \[ \vec{x}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-t} \left[ t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix} \right] \] \[ = e^{-t} \left[ c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \left( t\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix} \right) \right] \] Component form: \[ x(t) = e^{-t} \left[ c_1 + c_2 t + \frac{1}{2}c_2 \right] = e^{-t} \left[ c_1 + c_2 t + \frac{c_2}{2} \right] \] \[ y(t) = e^{-t} \left[ c_1 + c_2 t \right] \] Or, more simply: \[ \boxed{ \begin{aligned} x(t) &= e^{-t} \left( c_1 + c_2 t + \frac{c_2}{2} \right) \\ y(t) &= e^{-t} \left( c_1 + c_2 t \right) \end{aligned} } \] **Explanation:** We used the structure for repeated eigenvalues to write the general solution for the system in terms of exponentials and polynomials, using the eigenvector and generalized eigenvector. --- ## **Final Answer** \[ \boxed{ \begin{aligned} x(t) &= e^{-t} \left( c_1 + c_2 t + \frac{c_2}{2} \right) \\ y(t) &= e^{-t} \left( c_1 + c_2 t \right) \end{aligned} } \] where \(c_1\) and \(c_2\) are arbitrary constants.