1. **State the problem:** We need to find the general solution of the system of differential equations: $$\frac{dx}{dt} = 3x - y$$ $$\frac{dy}{dt} = 9x - 3y$$ 2. **Write the system in matrix form:** $$\frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ 3. **Find the eigenvalues:** The eigenvalues $\lambda$ satisfy: $$\det\left( \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = 0$$ which is $$\det\begin{pmatrix} 3 - \lambda & -1 \\ 9 & -3 - \lambda \end{pmatrix} = 0$$ Calculate the determinant: $$ (3 - \lambda)(-3 - \lambda) - (-1)(9) = 0 $$ $$ = -(3 - \lambda)(3 + \lambda) + 9 = 0 $$ $$ = - (9 - \lambda^2) + 9 = 0 $$ $$ = -9 + \lambda^2 + 9 = 0 $$ $$ \lambda^2 = 0 $$ So the repeated eigenvalue is: $$ \lambda = 0 $$ 4. **Find eigenvectors:** Solve: $$ \left( A - 0I \right) \mathbf{v} = 0 $$ which is $$ \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ From the first row: $$ 3v_1 - v_2 = 0 \implies v_2 = 3v_1 $$ From the second row: $$ 9v_1 - 3v_2 = 0 \implies 9v_1 - 3(3v_1) = 0 \implies 9v_1 - 9v_1 = 0 $$ which is consistent. Choose $v_1 = 1$, then eigenvector: $$ \mathbf{v} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ 5. **Find generalized eigenvector:** Since eigenvalue is repeated and matrix is not diagonalizable, find $\mathbf{w}$ such that: $$ (A - 0I) \mathbf{w} = \mathbf{v} $$ Solve: $$ \begin{pmatrix} 3 & -1 \\ 9 & -3 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ From first row: $$ 3w_1 - w_2 = 1 $$ From second row: $$ 9w_1 - 3w_2 = 3 $$ Multiply first row by 3: $$ 9w_1 - 3w_2 = 3 $$ which matches second row, so infinite solutions. Express $w_2$ from first row: $$ w_2 = 3w_1 - 1 $$ Choose $w_1 = 0$ for simplicity: $$ w_2 = -1 $$ So generalized eigenvector: $$ \mathbf{w} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} $$ 6. **Write the general solution:** For repeated eigenvalue $\lambda=0$, the general solution is: $$ \mathbf{x}(t) = c_1 e^{0t} \mathbf{v} + c_2 t e^{0t} \mathbf{v} + c_2 e^{0t} \mathbf{w} = c_1 \mathbf{v} + c_2 (t \mathbf{v} + \mathbf{w}) $$ Substitute $\mathbf{v}$ and $\mathbf{w}$: $$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 \left( t \begin{pmatrix} 1 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -1 \end{pmatrix} \right) = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 \begin{pmatrix} t \\ 3t - 1 \end{pmatrix} $$ **Final answer:** $$ x(t) = c_1 + c_2 t $$ $$ y(t) = 3 c_1 + c_2 (3t - 1) $$ where $c_1$ and $c_2$ are arbitrary constants.