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
