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CIE 9709 2025 March Paper 32 Q6

A Level / CIE / P3

CIE 9709 2025 March Paper 32 Paper · Question 6

题目

Problem

The variables xx and θ\theta satisfy the differential equation

dxdθ=(15x+1)sin⁡22θ,\frac{\mathrm{d}x}{\mathrm{d}\theta}=\left(\frac15x+1\right)\sin^22\theta,

and x=5x=5 when θ=0\theta=0.

Solve the differential equation and obtain an expression for xx in terms of θ\theta.

(7)
题目中文翻译

变量 xx 和 θ\theta 满足微分方程

dxdθ=(15x+1)sin⁡22θ,\frac{\mathrm{d}x}{\mathrm{d}\theta}=\left(\frac15x+1\right)\sin^22\theta,

且当 θ=0\theta=0 时,x=5x=5。

解此微分方程,并求出用 θ\theta 表示 xx 的表达式。

(7)
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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分离变量后,关于 xx 的积分产生对数,关于角的积分先降幂。初值确定常数后指数化。

答题过程

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Separating variables,

5x+5 dx=sin⁡22θ dθ.\frac5{x+5}\,\mathrm{d}x=\sin^22\theta\,\mathrm{d}\theta.

Using sin⁡22θ=(1−cos⁡4θ)/2\sin^22\theta=(1-\cos4\theta)/2,

∫sin⁡22θ dθ=θ2−18sin⁡4θ+C1,\int\sin^22\theta\,\mathrm{d}\theta =\frac\theta2-\frac18\sin4\theta+C_1,

where the factor 1/81/8 is (1/2)(1/4)(1/2)(1/4). Integrating both sides of the separated equation gives

5ln⁡∣x+5∣=θ2−18sin⁡4θ+C.5\ln|x+5|=\frac\theta2-\frac18\sin4\theta+C.

The initial condition x=5x=5, θ=0\theta=0 gives C=5ln⁡10C=5\ln10. On the initial branch x+5>0x+5>0, so

ln⁡x+510=θ10−140sin⁡4θ.\ln\frac{x+5}{10} =\frac{\theta}{10}-\frac1{40}\sin4\theta.

Exponentiating,

x=10exp⁡(θ10−sin⁡4θ40)−5.\boxed{x=10\exp\bigg(\frac{\theta}{10} -\frac{\sin4\theta}{40}\bigg)-5}.

The resulting expression has x+5>0x+5>0, consistent with the branch used.