题目 Problem Find the particular solution of the differential equation d2xdt2+ dxdt−2x= 2t2+t−1,\begin{align*} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} +&\, \frac{\mathrm{d}x}{\mathrm{d}t} - 2x =&\, 2t^2 + t - 1, \end{align*}dt2d2x+dtdx−2x=2t2+t−1, given that, when t=0t = 0t=0, x=dxdt=0x = \frac{\mathrm{d}x}{\mathrm{d}t} = 0x=dtdx=0. [10] 题目中文翻译 求微分方程 d2xdt2+ dxdt−2x= 2t2+t−1,\begin{align*} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} +&\, \frac{\mathrm{d}x}{\mathrm{d}t} - 2x =&\, 2t^2 + t - 1, \end{align*}dt2d2x+dtdx−2x=2t2+t−1, 的特解，已知当 t=0t = 0t=0 时，x=dxdt=0x = \frac{\mathrm{d}x}{\mathrm{d}t} = 0x=dtdx=0。 解答
