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CIE 9231 2026 June Paper 23 Q4

A Level / CIE / FP2

CIE 9231 2026 June Paper 23 Paper · Question 4

题目

Problem

Find the particular solution of the differential equation

49d2xdt2+14dxdt+x=196e17t,49\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+14\frac{\mathrm{d}x}{\mathrm{d}t}+x=196\mathrm{e}^{\frac{1}{7}t},

given that, when t=0t=0, x=7x=7 and dxdt=0\dfrac{\mathrm{d}x}{\mathrm{d}t}=0.

(10)
题目中文翻译

求微分方程

49d2xdt2+14dxdt+x=196e17t49\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+14\frac{\mathrm{d}x}{\mathrm{d}t}+x=196\mathrm{e}^{\frac{1}{7}t}

满足当 t=0t=0 时,x=7x=7 且 dxdt=0\dfrac{\mathrm{d}x}{\mathrm{d}t}=0 的特解。

(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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辅助方程有重根 −1/7-1/7。互补函数要包含 te−t/7te^{-t/7},特解试作 ket/7ke^{t/7},最后代入初始条件。

答题过程

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The auxiliary equation is (7m+1)2=0(7m+1)^2=0, so the complementary function is (A+Bt)e−t/7(A+Bt)e^{-t/7}. For the trial xp=ket/7x_p=ke^{t/7}, its derivatives are xp′=ket/7/7x_p'=ke^{t/7}/7 and xp′′=ket/7/49x_p''=ke^{t/7}/49. Substituting gives (k+2k+k)et/7=196et/7(k+2k+k)e^{t/7}=196e^{t/7}, so k=49k=49. The general solution and its derivative are

x= (A+Bt)e−t/7+49et/7,x′= (B−A+Bt7)e−t/7+7et/7.\begin{align*} x=&\,(A+Bt)e^{-t/7}+49e^{t/7},\\ x'=&\,\bigg(B-\frac{A+Bt}{7}\bigg)e^{-t/7}+7e^{t/7}. \end{align*}

At t=0t=0, x(0)=A+49=7x(0)=A+49=7 and x′(0)=B−A/7+7=0x'(0)=B-A/7+7=0. Thus A=−42A=-42 and B=−13B=-13. Therefore

x=(−42−13t)e−t/7+49et/7.x=(-42-13t)e^{-t/7}+49e^{t/7}.