91影视

Variation-of-Parameters Formula# The formula for the variation-of-parameters is: \[ \begin{bmatrix} U'(t) \\ V'(t) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ -1 & -2 \end{bmatrix}^{-1} \begin{bmatrix} -3e^t \\ 6te^t \end{bmatrix} \] To solve this, we first need to find the inverse of the matrix:

Calculate U'(t) and V'(t)# \[ \begin{bmatrix} U'(t) \\ V'(t) \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -3e^t \\ 6te^t \end{bmatrix} = \begin{bmatrix} -6e^t + 6te^t \\ 3e^t-6te^t \end{bmatrix} \] To find U(t) and V(t), we integrate: \[ \begin{bmatrix} U(t) \\ V(t) \end{bmatrix} = \begin{bmatrix} -6\int e^tdt + 6 \int te^t dt \\ 3\int e^t dt - 6 \int te^t dt \end{bmatrix}\] Using integration by parts for the terms with \(te^t\): \[\int te^tdt = te^t - \int e^tdt = te^t - e^t + C\] Thus, we get: \[ \begin{bmatrix} U(t) \\ V(t) \end{bmatrix} = \begin{bmatrix} -6(e^t - 1) + 6 (te^t - e^t + 1) \\ 3 (e^t - 1) - 6 (te^t - e^t + 1) \end{bmatrix} = \begin{bmatrix} 6t e^t - 6 \\ -9 e^t + 6 t e^t +3 \end{bmatrix} \] Now, we can find the particular solution, \(\mathbf{x}_{p}\), using: \[ \mathbf{x}_{p} = U(t) \mathbf{v}_1 + V(t) \mathbf{v}_2 = \begin{bmatrix} 6t e^t - 6 \\ -9 e^t + 6 t e^t +3 \end{bmatrix} \] #Step 3: Combine 饾懃鈧� and 饾懃鈧� to get the general solution# Finally, we can combine the complementary solution, \(\mathbf{x}_h\), and the particular solution, \(\mathbf{x}_p\), to obtain the general solution: \(\mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t)\) \(\mathbf{x}(t) = C_1 e^{-1t}\begin{bmatrix} 1 \\ -1 \end{bmatrix} + C_2 e^{3t}\begin{bmatrix} 1 \\ -2 \end{bmatrix} + \begin{bmatrix} 6t e^t - 6 \\ -9 e^t + 6 t e^t +3 \end{bmatrix}\) This is the general solution to the given system of differential equations.