1. **State the problem:** Solve the system of differential equations given by $$\mathbf{X}' = A\mathbf{X} + \mathbf{B}$$ where $$A = \begin{pmatrix} 2 & 1 \\ -4 & 2 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} te^{2t} \\ -e^{2t} \end{pmatrix}.$$ 2. **General approach:** The solution to the nonhomogeneous system is the sum of the homogeneous solution $$\mathbf{X}_h$$ and a particular solution $$\mathbf{X}_p$$: $$\mathbf{X} = \mathbf{X}_h + \mathbf{X}_p.$$ 3. **Solve the homogeneous system:** $$\mathbf{X}' = A\mathbf{X}.$$ Find eigenvalues $$\lambda$$ from $$\det(A - \lambda I) = 0$$: $$\det\begin{pmatrix} 2-\lambda & 1 \\ -4 & 2-\lambda \end{pmatrix} = (2-\lambda)^2 + 4 = 0.$$ 4. Simplify characteristic equation: $$(2-\lambda)^2 = -4 \implies 2-\lambda = \pm 2i \implies \lambda = 2 \mp 2i.$$ 5. **Eigenvalues:** $$\lambda_1 = 2 + 2i, \quad \lambda_2 = 2 - 2i.$$ 6. **Find eigenvector for $$\lambda_1$$:** Solve $$(A - \lambda_1 I)\mathbf{v} = 0,$$ $$\begin{pmatrix} 2-(2+2i) & 1 \\ -4 & 2-(2+2i) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -2i & 1 \\ -4 & -2i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0.$$ 7. From first row: $$-2i v_1 + v_2 = 0 \implies v_2 = 2i v_1.$$ Choose $$v_1 = 1$$, so $$\mathbf{v} = \begin{pmatrix} 1 \\ 2i \end{pmatrix}.$$ 8. **Write homogeneous solution:** $$\mathbf{X}_h = e^{2t} \left[ C_1 \begin{pmatrix} \cos 2t \\ -2 \sin 2t \end{pmatrix} + C_2 \begin{pmatrix} \sin 2t \\ 2 \cos 2t \end{pmatrix} \right].$$ 9. **Find particular solution $$\mathbf{X}_p$$:** Use variation of parameters or undetermined coefficients. Since $$\mathbf{B}$$ involves $$te^{2t}$$ and $$e^{2t}$$, try $$\mathbf{X}_p = e^{2t} \begin{pmatrix} a t + b \\ c t + d \end{pmatrix}$$ where $$a,b,c,d$$ are constants to find. 10. Compute $$\mathbf{X}_p'$$: $$\mathbf{X}_p' = 2 e^{2t} \begin{pmatrix} a t + b \\ c t + d \end{pmatrix} + e^{2t} \begin{pmatrix} a \\ c \end{pmatrix} = e^{2t} \begin{pmatrix} 2(a t + b) + a \\ 2(c t + d) + c \end{pmatrix}.$$ 11. Substitute into the system: $$\mathbf{X}_p' = A \mathbf{X}_p + \mathbf{B}$$ $$e^{2t} \begin{pmatrix} 2(a t + b) + a \\ 2(c t + d) + c \end{pmatrix} = A e^{2t} \begin{pmatrix} a t + b \\ c t + d \end{pmatrix} + \begin{pmatrix} t e^{2t} \\ -e^{2t} \end{pmatrix}.$$ 12. Divide both sides by $$e^{2t}$$: $$\begin{pmatrix} 2(a t + b) + a \\ 2(c t + d) + c \end{pmatrix} = A \begin{pmatrix} a t + b \\ c t + d \end{pmatrix} + \begin{pmatrix} t \\ -1 \end{pmatrix}.$$ 13. Compute $$A \begin{pmatrix} a t + b \\ c t + d \end{pmatrix}$$: $$= \begin{pmatrix} 2 & 1 \\ -4 & 2 \end{pmatrix} \begin{pmatrix} a t + b \\ c t + d \end{pmatrix} = \begin{pmatrix} 2(a t + b) + (c t + d) \\ -4(a t + b) + 2(c t + d) \end{pmatrix} = \begin{pmatrix} (2a + c) t + (2b + d) \\ (-4a + 2c) t + (-4b + 2d) \end{pmatrix}.$$ 14. Equate components: First component: $$2(a t + b) + a = (2a + c) t + (2b + d) + t,$$ which simplifies to $$2a t + 2b + a = (2a + c + 1) t + (2b + d).$$ 15. Equate coefficients of $$t$$ and constants: $$2a = 2a + c + 1 \implies c = -1,$$ $$2b + a = 2b + d \implies a = d.$$ 16. Second component: $$2(c t + d) + c = (-4a + 2c) t + (-4b + 2d) - 1,$$ which is $$2 c t + 2 d + c = (-4a + 2 c) t + (-4 b + 2 d - 1).$$ 17. Equate coefficients: $$2 c = -4 a + 2 c \implies -4 a = 0 \implies a = 0,$$ $$2 d + c = -4 b + 2 d - 1 \implies c = -4 b - 1.$$ 18. From step 15, $$a = d = 0$$ and from step 17, $$c = -1$$, so $$-1 = -4 b - 1 \implies -4 b = 0 \implies b = 0.$$ 19. **Summary of constants:** $$a = 0, b = 0, c = -1, d = 0.$$ 20. **Particular solution:** $$\mathbf{X}_p = e^{2t} \begin{pmatrix} 0 \cdot t + 0 \\ -1 \cdot t + 0 \end{pmatrix} = e^{2t} \begin{pmatrix} 0 \\ -t \end{pmatrix} = \begin{pmatrix} 0 \\ -t e^{2t} \end{pmatrix}.$$ 21. **Final solution:** $$\mathbf{X} = e^{2t} \left[ C_1 \begin{pmatrix} \cos 2t \\ -2 \sin 2t \end{pmatrix} + C_2 \begin{pmatrix} \sin 2t \\ 2 \cos 2t \end{pmatrix} \right] + \begin{pmatrix} 0 \\ -t e^{2t} \end{pmatrix}.$$