1. **Problem 1: Compute the Wronskian for** $$\vec{y}_1(t) = \begin{bmatrix} \cos(2t) \\ -\sin(2t) \end{bmatrix}, \quad \vec{y}_2(t) = \begin{bmatrix} -2 \sin(2t) \\ -2 \cos(2t) \end{bmatrix}$$ We want to determine if these two vector functions are linearly independent by computing the Wronskian. 2. **Formula for the Wronskian** For two vector functions in $\mathbb{R}^2$, the Wronskian is the determinant of the matrix formed by placing the vectors as columns: $$W(t) = \det \begin{bmatrix} y_{11}(t) & y_{21}(t) \\ y_{12}(t) & y_{22}(t) \end{bmatrix}$$ where $y_{ij}(t)$ is the $j$th component of $\vec{y}_i(t)$. 3. **Calculate the Wronskian matrix:** $$\begin{bmatrix} \cos(2t) & -2 \sin(2t) \\ -\sin(2t) & -2 \cos(2t) \end{bmatrix}$$ 4. **Compute the determinant:** $$W(t) = (\cos(2t))(-2 \cos(2t)) - (-\sin(2t))(-2 \sin(2t))$$ $$= -2 \cos^2(2t) - 2 \sin^2(2t)$$ 5. **Simplify using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:** $$W(t) = -2 (\cos^2(2t) + \sin^2(2t)) = -2 (1) = -2$$ 6. **Interpretation:** Since $W(t) = -2 \neq 0$ for all $t$, the functions $\vec{y}_1(t)$ and $\vec{y}_2(t)$ are linearly independent. Therefore, the solutions to the system $$\vec{y}' = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \vec{y}$$ form a fundamental set of solutions. --- 7. **Problem 2: Verify solution $\vec{y}(t) = \begin{bmatrix} 2e^{t} \\ -e^{t} \\ -e^{t} \end{bmatrix}$ for the system:** $$y_1' = 2y_1 + y_2 + y_3,$$ $$y_2' = y_1 + y_2 + 2y_3,$$ $$y_3' = y_1 + 2y_2 + y_3.$$ 8. **Calculate derivatives:** $$y_1' = \frac{d}{dt} (2e^{t}) = 2e^{t}$$ $$y_2' = \frac{d}{dt} (-e^{t}) = -e^{t}$$ $$y_3' = \frac{d}{dt} (-e^{t}) = -e^{t}$$ 9. **Evaluate right sides:** a. For $y_1'$: $$2y_1 + y_2 + y_3 = 2(2e^{t}) + (-e^{t}) + (-e^{t}) = 4e^{t} - e^{t} - e^{t} = 2e^{t}$$ b. For $y_2'$: $$y_1 + y_2 + 2y_3 = 2e^{t} + (-e^{t}) + 2(-e^{t}) = 2e^{t} - e^{t} - 2e^{t} = -e^{t}$$ c. For $y_3'$: $$y_1 + 2y_2 + y_3 = 2e^{t} + 2(-e^{t}) + (-e^{t}) = 2e^{t} - 2e^{t} - e^{t} = -e^{t}$$ 10. **Conclusion:** The derivatives match the right sides exactly, so $\vec{y}(t)$ is a solution. --- 11. **Problem 3: Rewrite system as $\vec{y}' = P \vec{y}$** Given: $$y_1' = 2y_1 + y_2 + y_3,$$ $$y_2' = y_1 + y_2 + 2y_3,$$ $$y_3' = y_1 + 2y_2 + y_3.$$ Matrix form: $$\begin{bmatrix} y_1' \\ y_2' \\ y_3' \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$$ So, $$P = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 1 \end{bmatrix}$$ --- 12. **Problem 4: Convert second order system to first order system** Given: $$y'' = t^{-1} y' + 4y - t z + (\sin t) z' + e^{5t},$$ $$z'' = y - 3 z'.$$ Define variables: $$y_1 = y, \quad y_2 = y', \quad y_3 = z, \quad y_4 = z'$$ Then, $$y_1' = y_2,$$ $$y_2' = t^{-1} y_2 + 4 y_1 - t y_3 + (\sin t) y_4 + e^{5t},$$ $$y_3' = y_4,$$ $$y_4' = y_1 - 3 y_4.$$ Matrix form: $$\begin{bmatrix} y_1' \\ y_2' \\ y_3' \\ y_4' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 4 & t^{-1} & -t & \sin t \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix} + \begin{bmatrix} 0 \\ e^{5t} \\ 0 \\ 0 \end{bmatrix}$$ --- 13. **Problem 5: Compute Wronskian for** $$\vec{y}_1(t) = \begin{bmatrix} 2 e^{3t} - 4 e^{-t} \\ 3 e^{3t} - 10 e^{-t} \end{bmatrix}, \quad \vec{y}_2(t) = \begin{bmatrix} -6 e^{3t} + 2 e^{-t} \\ -9 e^{3t} + 5 e^{-t} \end{bmatrix}$$ Wronskian matrix: $$\begin{bmatrix} 2 e^{3t} - 4 e^{-t} & -6 e^{3t} + 2 e^{-t} \\ 3 e^{3t} - 10 e^{-t} & -9 e^{3t} + 5 e^{-t} \end{bmatrix}$$ Compute determinant: $$W(t) = (2 e^{3t} - 4 e^{-t})(-9 e^{3t} + 5 e^{-t}) - (3 e^{3t} - 10 e^{-t})(-6 e^{3t} + 2 e^{-t})$$ Expand terms: First product: $$2 e^{3t} \cdot (-9 e^{3t}) = -18 e^{6t}$$ $$2 e^{3t} \cdot 5 e^{-t} = 10 e^{2t}$$ $$-4 e^{-t} \cdot (-9 e^{3t}) = 36 e^{2t}$$ $$-4 e^{-t} \cdot 5 e^{-t} = -20 e^{-2t}$$ Sum first product: $$-18 e^{6t} + 10 e^{2t} + 36 e^{2t} - 20 e^{-2t} = -18 e^{6t} + 46 e^{2t} - 20 e^{-2t}$$ Second product: $$3 e^{3t} \cdot (-6 e^{3t}) = -18 e^{6t}$$ $$3 e^{3t} \cdot 2 e^{-t} = 6 e^{2t}$$ $$-10 e^{-t} \cdot (-6 e^{3t}) = 60 e^{2t}$$ $$-10 e^{-t} \cdot 2 e^{-t} = -20 e^{-2t}$$ Sum second product: $$-18 e^{6t} + 6 e^{2t} + 60 e^{2t} - 20 e^{-2t} = -18 e^{6t} + 66 e^{2t} - 20 e^{-2t}$$ Subtract second from first: $$W(t) = (-18 e^{6t} + 46 e^{2t} - 20 e^{-2t}) - (-18 e^{6t} + 66 e^{2t} - 20 e^{-2t}) = 46 e^{2t} - 20 e^{-2t} - 66 e^{2t} + 20 e^{-2t} = -20 e^{2t}$$ Since $W(t) = -20 e^{2t} \neq 0$ for all $t$, the functions are linearly independent. Therefore, the solutions to $$\vec{y}' = \begin{bmatrix} 9 & -4 \\ 15 & -7 \end{bmatrix} \vec{y}$$ form a fundamental set. --- 14. **Problem 6: Compute Wronskian for** $$\vec{y}_1(t) = \begin{bmatrix} t^2 - 2t \\ 2t \end{bmatrix}, \quad \vec{y}_2(t) = \begin{bmatrix} t - 1 \\ 1 \end{bmatrix}$$ Wronskian matrix: $$\begin{bmatrix} t^2 - 2t & t - 1 \\ 2t & 1 \end{bmatrix}$$ Compute determinant: $$W(t) = (t^2 - 2t)(1) - (2t)(t - 1) = t^2 - 2t - 2t^2 + 2t = -t^2$$ Since $W(t) = -t^2$, which is zero only at $t=0$, the functions are linearly independent for $t \neq 0$. Therefore, the solutions to $$\vec{y}' = \begin{bmatrix} 2 t^{-2} & 1 - 2 t^{-1} + 2 t^{-2} \\ -2 t^{-2} & 2 t^{-1} - 2 t^{-2} \end{bmatrix} \vec{y}$$ form a fundamental set for $t \neq 0$. --- **Summary:** - Problems 1, 5, and 6 involve computing Wronskians to check linear independence. - Problem 2 verifies a solution to a system. - Problem 3 rewrites a system in matrix form. - Problem 4 converts a second order system to first order.