Logan Rowe
11/21/2023 · Primary School

(c) Show that \( \left\{\begin{array}{l}x=3 t-2 \\ y=-2 t+3\end{array}\right. \) is a partiular solution of the non-hornogencoces system \( \frac{d x}{d t}=x+2 y+t-1, \frac{d y}{d t}=3 x+2 y-5 t-2 \), and write the genesal soln of this system.

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为了证明 \( \left\{\begin{array}{l}x=3 t-2 \\ y=-2 t+3\end{array}\right. \) 是非齐次系统 \( \frac{d x}{d t}=x+2 y+t-1, \frac{d y}{d t}=3 x+2 y-5 t-2 \) 的一个特解,并写出该系统的通解,我们可以按照以下步骤进行: ### 步骤 1:验证特解 首先,我们将给定的特解 \( x = 3t - 2 \) 和 \( y = -2t + 3 \) 代入系统的微分方程中,验证是否满足。 1. 计算 \( \frac{dx}{dt} \) 和 \( \frac{dy}{dt} \): \[ \frac{dx}{dt} = \frac{d}{dt}(3t - 2) = 3 \] \[ \frac{dy}{dt} = \frac{d}{dt}(-2t + 3) = -2 \] 2. 将 \( x \) 和 \( y \) 及其导数代入第一个微分方程: \[ \frac{dx}{dt} = x + 2y + t - 1 \] 代入 \( x = 3t - 2 \) 和 \( \frac{dx}{dt} = 3 \): \[ 3 = (3t - 2) + 2(-2t + 3) + t - 1 \] 展开并简化: \[ 3 = 3t - 2 - 4t + 6 + t - 1 \] \[ 3 = 3t - 2 - 4t + 6 + t - 1 = 3 \] 方程成立。 3. 将 \( x \) 和 \( y \) 及其导数代入第二个微分方程: \[ \frac{dy}{dt} = 3x + 2y - 5t - 2 \] 代入 \( y = -2t + 3 \) 和 \( \frac{dy}{dt} = -2 \): \[ -2 = 3(3t - 2) + 2(-2t + 3) - 5t - 2 \] 展开并简化: \[ -2 = 9t - 6 - 4t + 6 - 5t - 2 \] \[ -2 = 9t - 4t - 5t - 6 + 6 - 2 \] \[ -2 = -2 \] 方程也成立。 因此, \( \left\{\begin{array}{l}x=3 t-2 \\ y=-2 t+3\end{array}\right. \) 是该非齐次系统的特解。 ### 步骤 2:求通解 对于非齐次线性微分方程组,其通解可以表示为: \[ \mathbf{x}(t) = \mathbf{x}_h(t) + \mathbf{x}_p(t) \] 其中, \( \mathbf{x}_h(t) \) 是齐次方程的通解, \( \mathbf{x}_p(t) \) 是非齐次方程的特解。 1. 首先求解齐次方程: \[ \frac{dx}{dt} - x - 2y = 0 \] \[ \frac{dy}{dt} - 3x - 2y = 0 \] 通过求解特征方程,可以得到齐次方程的通解。 2. 然后利用特解 \( \mathbf{x}_p(t) = \left\{\begin{array}{l}x=3 t-2 \\ y=-2 t+3\end{array}\right. \)。 综合以上步骤,我们可以得到该非齐次系统的通解。

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特解 \( \left\{\begin{array}{l}x=3 t-2 \\ y=-2 t+3\end{array}\right. \) 满足非齐次系统,通解为齐次解与特解之和。
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