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CIE 9709 2025 June Paper 33 Q10

A Level / CIE / P3

CIE 9709 2025 June Paper 33 Paper · Question 10

题目

Problem

The variables xx and yy satisfy the differential equation

sin⁡4ydydx=xsin⁡2ysin⁡3x.\sin4y\frac{\mathrm{d}y}{\mathrm{d}x}=x\sin2y\sin3x.

It is given that y=112πy=\frac1{12}\pi when x=12πx=\frac12\pi.

(a) Solve the differential equation, obtaining a relation between xx and yy.

(8)

(b) Given that 0<y<12π0<y<\frac12\pi, find the values of yy when x=0x=0.

(2)
题目中文翻译

变量 xx 和 yy 满足微分方程

sin⁡4ydydx=xsin⁡2ysin⁡3x.\sin4y\frac{\mathrm{d}y}{\mathrm{d}x}=x\sin2y\sin3x.

已知当 x=12πx=\frac12\pi 时,y=112πy=\frac1{12}\pi。

(a) 解此微分方程,求出 xx 与 yy 之间的关系。

(8)

(b) 已知 0<y<12π0<y<\frac12\pi,求当 x=0x=0 时 yy 的值。

(2)
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

(a)

解法一

思路

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分离变量后用二倍角约去 sin⁡2y\sin2y,左边可直接积分,右边用分部积分。最后代入初值。

答题过程

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Near the initial point, sin⁡2y≠0\sin2y\ne0. Separating variables,

sin⁡4ysin⁡2y dy=xsin⁡3x dx.\frac{\sin4y}{\sin2y}\,\mathrm{d}y=x\sin3x\,\mathrm{d}x.

Since sin⁡4y=2sin⁡2ycos⁡2y\sin4y=2\sin2y\cos2y, the left-hand integral is

∫2cos⁡2y dy=sin⁡2y.\int2\cos2y\,\mathrm{d}y=\sin2y.

For the right-hand side, take u=xu=x and dv=sin⁡3x dx\mathrm{d}v=\sin3x\,\mathrm{d}x, so du=dx\mathrm{d}u=\mathrm{d}x and v=−cos⁡3x/3v=-\cos3x/3. Integration by parts gives

∫xsin⁡3x dx= −x3cos⁡3x+13∫cos⁡3x dx= −x3cos⁡3x+19sin⁡3x+C.\begin{align*} \int x\sin3x\,\mathrm{d}x =&\,-\frac{x}{3}\cos3x+\frac13\int\cos3x\,\mathrm{d}x\\ =&\,-\frac{x}{3}\cos3x+\frac19\sin3x+C. \end{align*}

Thus

sin⁡2y=−x3cos⁡3x+19sin⁡3x+C.\sin2y=-\frac{x}{3}\cos3x+\frac19\sin3x+C.

Using y=π/12y=\pi/12 at x=π/2x=\pi/2,

12=0−19+C⟹C=1118.\frac12=0-\frac19+C \quad\Longrightarrow\quad C=\frac{11}{18}.

Therefore the required relation is

sin⁡2y=−x3cos⁡3x+19sin⁡3x+1118.\boxed{\sin2y=-\frac{x}{3}\cos3x+\frac19\sin3x+\frac{11}{18}}.