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CIE 9231 2025 June Paper 23 Q4

A Level / CIE / FP2

CIE 9231 2025 June Paper 23 Paper · Question 4

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*}

given that, when t=0t = 0, x=dxdt=0x = \frac{\mathrm{d}x}{\mathrm{d}t} = 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*}

的特解，已知当 t=0t = 0 时，x=dxdt=0x = \frac{\mathrm{d}x}{\mathrm{d}t} = 0。

(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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先由辅助方程求互补函数。用二次多项式试一个特解，再由初始位置和初始导数求两个常数。

答题过程

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The auxiliary equation m2+m−2=0m^2+m-2=0 has roots 11 and −2-2, so the complementary function is Aet+Be−2tAe^t+Be^{-2t}. For a particular integral xp=pt2+qt+rx_p=pt^2+qt+r, we have xp′=2pt+qx_p'=2pt+q and xp′′=2px_p''=2p. Substitution gives

−2pt2+(2p−2q)t+(2p+q−2r)=2t2+t−1.-2pt^2+(2p-2q)t+(2p+q-2r)=2t^2+t-1.

Equating coefficients gives

−2p=2,2p−2q=1,2p+q−2r=−1.-2p=2,\qquad 2p-2q=1,\qquad 2p+q-2r=-1.

Thus p=−1p=-1, q=−32q=-\frac32 and r=−54r=-\frac54. The general solution is

x=Aet+Be−2t−t2−32t−54.x=Ae^t+Be^{-2t}-t^2-\frac32t-\frac54.

Differentiating gives

x′=Aet−2Be−2t−2t−32.x'=Ae^t-2Be^{-2t}-2t-\frac32.

The initial conditions imply A+B=54A+B=\frac54 and A−2B=32A-2B=\frac32. Hence A=43A=\frac43, B=−112B=-\frac1{12}, and

x=43et−112e−2t−t2−32t−54.x=\frac43 e^t-\frac1{12}e^{-2t}-t^2-\frac32t-\frac54.