变量 $t$ 和 $x$ 满足微分方程

(a) 求 $x$ 关于 $t$ 的通解。

(b) 由此推导出当 $t$ 为很大的正数时，$\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}$ 的近似值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

辅助方程有一对共轭复根。右端是二次多项式，特解也试二次多项式，代入比较系数。

The auxiliary equation $m^2+m+1=0$ gives $m=-\frac12\pm\frac{\sqrt3}{2}\mathrm{i}$. Thus the complementary function is

For a particular integral, set $x_{\mathrm p}=pt^2+qt+r$. Then $x_{\mathrm p}'=2pt+q$ and $x_{\mathrm p}''=2p$, so

Matching coefficients gives $p=1$, $2p+q=0$ and $2p+q+r=1$, so $q=-2$ and $r=1$. Therefore

互补函数求两次导数后，仍是指数衰减因子乘有界的三角函数。二次特解求两次导数得到常数。

Differentiating the complementary function twice gives $e^{-t/2}$ times a fixed linear combination of sine and cosine, which tends to zero as $t\to+\infty$. The second derivative of $t^2-2t+1$ is $2$, so $\boxed{\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\approx2}$ for large positive $t$.