题目

Problem

Find the particular solution of the differential equation

6\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 5\frac{\mathrm{d}x}{\mathrm{d}t} + x = t^2 + t + 1,

given that, when $t = 0$ , $x = 12$ and $\frac{\mathrm{d}x}{\mathrm{d}t} = -6$ .

[10]

题目中文翻译

求微分方程

6\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 5\frac{\mathrm{d}x}{\mathrm{d}t} + x = t^2 + t + 1

的特解，已知当 $t = 0$ 时， $x = 12$ 且 $\frac{\mathrm{d}x}{\mathrm{d}t} = -6$ 。

(10)

原卷通用作答要求

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。

解答

解法一

思路

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解辅助方程得到两个指数项。右端是二次多项式，试同次数的特解，再代入两个初始条件。

答题过程

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The auxiliary equation is $(2m-1)(3m-1)=0$ , so the complementary function is $Ae^{t/2}+Be^{t/3}$ . For $x_p=pt^2+qt+r$ , we have $x_p'=2pt+q$ and $x_p''=2p$ . Substituting and collecting coefficients gives

p t^2+(q-10p)t+(12p-5q+r)=t^2+t+1.

Hence

p=1,\qquad -10p+q=1,\qquad 12p-5q+r=1.

Thus $q=11$ and $r=44$ . The general solution and its derivative are

\begin{align*} x=&\,Ae^{t/2}+Be^{t/3}+t^2+11t+44,\\[4mm] x'=&\,\frac A2e^{t/2}+\frac B3e^{t/3}+2t+11. \end{align*}

The initial conditions give

A+B+44=12,\qquad \frac A2+\frac B3+11=-6.

Thus $A+B=-32$ and $3A+2B=-102$ . Subtracting twice the first equation from the second gives $A=-38$ , and then $B=6$ . Therefore

x=-38e^{t/2}+6e^{t/3}+t^2+11t+44.

CIE 9231 2024 November Paper 21 Q5

题目

解答

解法一

思路

答题过程