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CIE 9709 2023 March Paper 32 Q9

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 9

题目

Problem

The variables xx and yy satisfy the differential equation

dydx=e3ysin⁡22x.\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}^{3y}\sin^2 2x.

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

Solve the differential equation and find the value of yy when x=12x=\frac12.

(7)
题目中文翻译

变量 xx 和 yy 满足微分方程

dydx=e3ysin⁡22x.\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{e}^{3y}\sin^2 2x.

已知当 x=0x=0 时,y=0y=0。

解此微分方程,并求 x=12x=\frac12 时 yy 的值。

(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,把 sin⁡22x\sin^22x 用二倍角公式化成 [1−cos⁡4x]/2[1-\cos4x]/2,再积分并使用初值。最后先写显式解,再代入 x=1/2x=1/2。

答题过程

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Separating variables,

e−3y dy=sin⁡22x dx=12(1−cos⁡4x) dx.e^{-3y}\,\mathrm{d}y=\sin^22x\,\mathrm{d}x =\frac12(1-\cos4x)\,\mathrm{d}x.

The derivatives of e−3ye^{-3y} and sin⁡4x\sin4x contribute factors −3-3 and 44 respectively. Integrating,

−13e−3y=x2−18sin⁡4x+C.-\frac13e^{-3y}=\frac x2-\frac18\sin4x+C.

Using x=0x=0, y=0y=0 gives C=−1/3C=-1/3. Therefore

e−3y=1−32x+38sin⁡4x,e^{-3y}=1-\frac32x+\frac38\sin4x,

so

y=−13ln⁡(1−32x+38sin⁡4x)\boxed{y=-\frac13\ln\bigg(1-\frac32x+\frac38\sin4x\bigg)}

on the branch containing the initial point, where the logarithm argument is positive.

When x=1/2x=1/2,

y=−13ln⁡(14+38sin⁡2).y=-\frac13\ln\bigg(\frac14+\frac38\sin2\bigg).

Hence y=0.175\boxed{y=0.175} to 3 significant figures.