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CIE 9709 2023 Nov Paper 33 Q8

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 8

题目

Problem

The variables xx and yy satisfy the differential equation

e4xdydx=cos⁡23y.\mathrm{e}^{4x}\frac{\mathrm{d}y}{\mathrm{d}x}=\cos^2 3y.

It is given that y=0y=0 when x=2x=2.

Solve the differential equation, obtaining an expression for yy in terms of xx.

(7)
题目中文翻译

变量 xx 和 yy 满足微分方程

e4xdydx=cos⁡23y.\mathrm{e}^{4x}\frac{\mathrm{d}y}{\mathrm{d}x}=\cos^2 3y.

已知当 x=2x=2 时,y=0y=0。

解此微分方程,得到用 xx 表示 yy 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,左侧用 sec⁡2(3y)\sec^2(3y) 积分,右侧积分指数函数。反正切有多个分支,要选通过初值 y=0y=0 的分支。

答题过程

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Separating variables on the branch through y=0y=0,

sec⁡2(3y) dy=e−4x dx.\sec^2(3y)\,\mathrm{d}y=e^{-4x}\,\mathrm{d}x.

Since d(tan⁡3y)/dy=3sec⁡2(3y)\mathrm{d}(\tan3y)/\mathrm{d}y=3\sec^2(3y) and d(e−4x)/dx=−4e−4x\mathrm{d}(e^{-4x})/\mathrm{d}x=-4e^{-4x}, integration gives

13tan⁡3y=−14e−4x+C.\frac13\tan3y=-\frac14e^{-4x}+C.

Using y=0y=0 when x=2x=2,

0= −14e−8+C,C= 14e−8.\begin{align*} 0=&\,-\frac14e^{-8}+C,\\[4mm] C=&\,\frac14e^{-8}. \end{align*}

Thus

tan⁡3y=34(e−8−e−4x).\tan3y=\frac34(e^{-8}-e^{-4x}).

The branch containing the initial point has −π/2<3y<π/2-\pi/2<3y<\pi/2. Hence

y=13tan⁡−1(34e−8−34e−4x).\boxed{y=\frac13\tan^{-1}\bigg(\frac34e^{-8}-\frac34e^{-4x}\bigg)}.