1. **State the problem:** Solve the system of differential equations with initial conditions: $$\frac{dy}{dt} = 2x + y$$ $$\frac{dx}{dt} = 2x + 3y$$ with $$x(0) = -4$$ and $$y(0) = -1$$. 2. **Write the system in matrix form:** $$\frac{d}{dt}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}2 & 3 \\ 2 & 1\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix}$$ 3. **Find eigenvalues $$\lambda$$ of the coefficient matrix:** $$A = \begin{bmatrix}2 & 3 \\ 2 & 1\end{bmatrix}$$ Solve characteristic equation: $$\det(A - \lambda I) = 0$$ $$\det\begin{bmatrix}2 - \lambda & 3 \\ 2 & 1 - \lambda\end{bmatrix} = (2 - \lambda)(1 - \lambda) - 6 = 0$$ $$ (2 - \lambda)(1 - \lambda) - 6 = 0 $$ $$ 2 - 2\lambda - \lambda + \lambda^2 - 6 = 0 $$ $$ \lambda^2 - 3\lambda - 4 = 0 $$ 4. **Solve quadratic:** $$\lambda = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}$$ 5. **Eigenvalues:** $$\lambda_1 = 4, \quad \lambda_2 = -1$$ 6. **Find eigenvectors:** For $$\lambda_1 = 4$$: $$(A - 4I)\mathbf{v} = 0$$ $$\begin{bmatrix}2-4 & 3 \\ 2 & 1-4\end{bmatrix} \begin{bmatrix}v_1 \\ v_2\end{bmatrix} = \begin{bmatrix}-2 & 3 \\ 2 & -3\end{bmatrix} \begin{bmatrix}v_1 \\ v_2\end{bmatrix} = 0$$ From first row: $$-2v_1 + 3v_2 = 0 \Rightarrow 3v_2 = 2v_1 \Rightarrow v_2 = \frac{2}{3}v_1$$ Choose $$v_1 = 3$$, then $$v_2 = 2$$. Eigenvector: $$\mathbf{v}_1 = \begin{bmatrix}3 \\ 2\end{bmatrix}$$ For $$\lambda_2 = -1$$: $$(A + I)\mathbf{v} = 0$$ $$\begin{bmatrix}3 & 3 \\ 2 & 2\end{bmatrix} \begin{bmatrix}v_1 \\ v_2\end{bmatrix} = 0$$ From first row: $$3v_1 + 3v_2 = 0 \Rightarrow v_1 = -v_2$$ Choose $$v_2 = 1$$, then $$v_1 = -1$$. Eigenvector: $$\mathbf{v}_2 = \begin{bmatrix}-1 \\ 1\end{bmatrix}$$ 7. **General solution:** $$\begin{bmatrix}x(t) \\ y(t)\end{bmatrix} = c_1 e^{4t} \begin{bmatrix}3 \\ 2\end{bmatrix} + c_2 e^{-t} \begin{bmatrix}-1 \\ 1\end{bmatrix}$$ 8. **Apply initial conditions:** At $$t=0$$: $$\begin{bmatrix}x(0) \\ y(0)\end{bmatrix} = c_1 \begin{bmatrix}3 \\ 2\end{bmatrix} + c_2 \begin{bmatrix}-1 \\ 1\end{bmatrix} = \begin{bmatrix}-4 \\ -1\end{bmatrix}$$ This gives system: $$3c_1 - c_2 = -4$$ $$2c_1 + c_2 = -1$$ 9. **Solve for $$c_1$$ and $$c_2$$:** Add equations: $$(3c_1 - c_2) + (2c_1 + c_2) = -4 + (-1)$$ $$5c_1 = -5 \Rightarrow c_1 = -1$$ Substitute $$c_1$$ into second equation: $$2(-1) + c_2 = -1 \Rightarrow -2 + c_2 = -1 \Rightarrow c_2 = 1$$ 10. **Final solution:** $$x(t) = -1 \cdot 3 e^{4t} + 1 \cdot (-1) e^{-t} = -3 e^{4t} - e^{-t}$$ $$y(t) = -1 \cdot 2 e^{4t} + 1 \cdot 1 e^{-t} = -2 e^{4t} + e^{-t}$$