1. **State the problem:** Solve the system of differential equations given by $$X' = AX + B$$ where $$A = \begin{bmatrix} 2 & 1 \\ -4 & 2 \end{bmatrix}$$ and $$B$$ is a constant vector (not fully provided). 2. **Recall the formula:** The general solution to the nonhomogeneous system $$X' = AX + B$$ is $$X(t) = X_h(t) + X_p$$ where $$X_h(t)$$ is the solution to the homogeneous system $$X' = AX$$ and $$X_p$$ is a particular solution to the nonhomogeneous system. 3. **Solve the homogeneous system:** - Find eigenvalues $$\lambda$$ by solving $$\det(A - \lambda I) = 0$$. Calculate: $$\det\left(\begin{bmatrix} 2 - \lambda & 1 \\ -4 & 2 - \lambda \end{bmatrix}\right) = (2 - \lambda)^2 - (-4)(1) = (2 - \lambda)^2 + 4 = 0$$ Expand: $$(2 - \lambda)^2 = \lambda^2 - 4\lambda + 4$$ So: $$\lambda^2 - 4\lambda + 4 + 4 = \lambda^2 - 4\lambda + 8 = 0$$ 4. **Find eigenvalues:** Use quadratic formula: $$\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 8}}{2} = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} = \frac{4 \pm 4i}{2} = 2 \pm 2i$$ 5. **Form homogeneous solution:** Since eigenvalues are complex $$2 \pm 2i$$, the homogeneous solution is: $$X_h(t) = e^{2t} \left( C_1 \begin{bmatrix} \cos 2t \\ ? \end{bmatrix} + C_2 \begin{bmatrix} \sin 2t \\ ? \end{bmatrix} \right)$$ To find eigenvectors, solve $$ (A - \lambda I)v = 0 $$ for $$\lambda = 2 + 2i$$: $$\begin{bmatrix} 2 - (2 + 2i) & 1 \\ -4 & 2 - (2 + 2i) \end{bmatrix} = \begin{bmatrix} -2i & 1 \\ -4 & -2i \end{bmatrix}$$ From first row: $$-2i v_1 + v_2 = 0 \Rightarrow v_2 = 2i v_1$$ Choose $$v_1 = 1$$, then $$v = \begin{bmatrix} 1 \\ 2i \end{bmatrix}$$. 6. **Write homogeneous solution explicitly:** Using Euler's formula: $$X_h(t) = e^{2t} \left[ C_1 \begin{bmatrix} \cos 2t \\ -2 \sin 2t \end{bmatrix} + C_2 \begin{bmatrix} \sin 2t \\ 2 \cos 2t \end{bmatrix} \right]$$ 7. **Find particular solution:** Since $$B$$ is not fully given, assume $$B = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$$ constant. Try constant solution $$X_p$$ such that: $$0 = A X_p + B \Rightarrow A X_p = -B$$ Solve: $$\begin{bmatrix} 2 & 1 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} x_p \\ y_p \end{bmatrix} = \begin{bmatrix} -b_1 \\ -b_2 \end{bmatrix}$$ This linear system can be solved for $$x_p, y_p$$ if $$A$$ is invertible. 8. **Summary:** The general solution is: $$X(t) = e^{2t} \left[ C_1 \begin{bmatrix} \cos 2t \\ -2 \sin 2t \end{bmatrix} + C_2 \begin{bmatrix} \sin 2t \\ 2 \cos 2t \end{bmatrix} \right] + X_p$$ where $$X_p$$ solves $$A X_p = -B$$. **Note:** To complete the particular solution, the vector $$B$$ must be specified.