题目 Problem Find the particular solution of the differential equation 6d2xdt2−5dxdt+x=t2+t+1,6\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 5\frac{\mathrm{d}x}{\mathrm{d}t} + x = t^2 + t + 1,6dt2d2x−5dtdx+x=t2+t+1, given that, when t=0t = 0t=0, x=12x = 12x=12 and dxdt=−6\frac{\mathrm{d}x}{\mathrm{d}t} = -6dtdx=−6. [10] 题目中文翻译 求微分方程 6d2xdt2−5dxdt+x=t2+t+16\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 5\frac{\mathrm{d}x}{\mathrm{d}t} + x = t^2 + t + 16dt2d2x−5dtdx+x=t2+t+1 的特解，已知当 t=0t = 0t=0 时，x=12x = 12x=12 且 dxdt=−6\frac{\mathrm{d}x}{\mathrm{d}t} = -6dtdx=−6。 解答
