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CIE 9709 2024 March Paper 32 Q11

A Level / CIE / P3

CIE 9709 2024 March Paper 32 Paper · Question 11

题目

Problem

The variables yy and θ\theta satisfy the differential equation

(1+y)(1+cos⁡2θ)dydθ=e3y.(1+y)(1+\cos2\theta)\frac{\mathrm{d}y}{\mathrm{d}\theta}=\mathrm{e}^{3y}.

It is given that y=0y=0 when θ=14π\theta=\frac14\pi.

Solve the differential equation and find the exact value of tan⁡θ\tan\theta when y=1y=1.

(9)
题目中文翻译

变量 yy 和 θ\theta 满足微分方程

(1+y)(1+cos⁡2θ)dydθ=e3y.(1+y)(1+\cos2\theta)\frac{\mathrm{d}y}{\mathrm{d}\theta}=\mathrm{e}^{3y}.

已知当 θ=14π\theta=\frac14\pi 时，y=0y=0。

解此微分方程，并求 y=1y=1 时 tan⁡θ\tan\theta 的精确值。

(9)
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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先分离变量。含 yy 的一侧使用分部积分，含 θ\theta 的一侧先用 1+cos⁡2θ=2cos⁡2θ1+\cos2\theta=2\cos^2\theta。代入初值求常数，最后只需求 tan⁡θ\tan\theta。

答题过程

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Separating variables,

(1+y)e−3y dy=11+cos⁡2θ dθ.(1+y)e^{-3y}\,\mathrm{d}y =\frac{1}{1+\cos2\theta}\,\mathrm{d}\theta.

For the left side, integrate by parts with

u= 1+y,dv= e−3y dy,du= dy,v= −13e−3y.\begin{align*} u=&\,1+y,& \mathrm{d}v=&\,e^{-3y}\,\mathrm{d}y,\\[4mm] \mathrm{d}u=&\,\mathrm{d}y,& v=&\,-\frac13e^{-3y}. \end{align*}

Then

∫(1+y)e−3y dy= −13(1+y)e−3y+13∫e−3y dy= −13(1+y)e−3y−19e−3y+C1= −3y+49e−3y+C1.\begin{align*} \int(1+y)e^{-3y}\,\mathrm{d}y =&\,-\frac13(1+y)e^{-3y} +\frac13\int e^{-3y}\,\mathrm{d}y\\[4mm] =&\,-\frac13(1+y)e^{-3y}-\frac19e^{-3y}+C_1\\[4mm] =&\,-\frac{3y+4}{9}e^{-3y}+C_1. \end{align*}

For the right side,

∫11+cos⁡2θ dθ= 12∫sec⁡2θ dθ= 12tan⁡θ+C2.\begin{align*} \int\frac{1}{1+\cos2\theta}\,\mathrm{d}\theta =&\,\frac12\int\sec^2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac12\tan\theta+C_2. \end{align*}

Combining the two integration constants into a single constant CC gives

−3y+49e−3y=12tan⁡θ+C.-\frac{3y+4}{9}e^{-3y}=\frac12\tan\theta+C.

Using y=0y=0 when θ=π/4\theta=\pi/4 gives

−49=12+C,C=−1718.-\frac49=\frac12+C,\qquad C=-\frac{17}{18}.

The solution satisfying the initial condition can therefore be written

tan⁡θ=179−2(3y+4)9e−3y.\boxed{\tan\theta=\frac{17}{9} -\frac{2(3y+4)}9e^{-3y}}.

When y=1y=1,

tan⁡θ=179−149e−3.\boxed{\tan\theta=\frac{17}{9}-\frac{14}{9}e^{-3}}.