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CIE 9709 2023 June Paper 32 Q8

A Level / CIE / P3

CIE 9709 2023 June Paper 32 Paper · Question 8

题目

Problem

(a) The variables xx and yy satisfy the differential equation

dydx=4+9y2e2x+1.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4+9y^2}{\mathrm{e}^{2x+1}}.

It is given that y=0y=0 when x=1x=1.

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

(7)

(b) State what happens to the value of yy as xx tends to infinity. Give your answer in an exact form.

(1)
题目中文翻译

(a) 变量 xx 和 yy 满足微分方程

dydx=4+9y2e2x+1.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4+9y^2}{\mathrm{e}^{2x+1}}.

已知当 x=1x=1 时,y=0y=0。

解此微分方程,求用 xx 表示 yy 的表达式。

(7)

(b) 说明当 xx 趋于无穷大时,yy 的值如何变化。答案用精确形式表示。

(1)
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

展开

分离变量后,左侧是反正切型积分,右侧是指数型积分。用初值求常数,再取正切写成显式解。

答题过程

展开

Separating variables,

14+9y2 dy=e−(2x+1) dx.\frac1{4+9y^2}\,\mathrm{d}y=e^{-(2x+1)}\,\mathrm{d}x.

The chain rule gives ddytan⁡−1(3y/2)=64+9y2\dfrac{\mathrm{d}}{\mathrm{d}y}\tan^{-1}(3y/2)=\dfrac6{4+9y^2}, and ddxe−(2x+1)=−2e−(2x+1)\dfrac{\mathrm{d}}{\mathrm{d}x}e^{-(2x+1)}=-2e^{-(2x+1)}. Thus integrating gives

16tan⁡−13y2=−12e−(2x+1)+C.\frac16\tan^{-1}\frac{3y}{2} =-\frac12e^{-(2x+1)}+C.

Using y=0y=0 at x=1x=1,

0= −12e−3+C,C= 12e−3.\begin{align*} 0=&\,-\frac12e^{-3}+C,\\[4mm] C=&\,\frac12e^{-3}. \end{align*}

Therefore

tan⁡−13y2=3e−3−3e−(2x+1).\tan^{-1}\frac{3y}{2} =3e^{-3}-3e^{-(2x+1)}.

Taking tangents,

y=23tan⁡(3e−3−3e−(2x+1)).\boxed{y=\frac23\tan\big(3e^{-3}-3e^{-(2x+1)}\big)}.

The branch through the initial point is taken where the argument of the tangent remains in (−π/2,π/2)(-\pi/2,\pi/2).

(b)

解法一

思路

展开

当 x→∞x\to\infty 时,随 xx 衰减的指数项趋于零,保留精确的极限表达式。

答题过程

展开

As x→∞x\to\infty, e−(2x+1)→0e^{-(2x+1)}\to0. Hence

y⟶23tan⁡(3e−3).\boxed{y\longrightarrow\frac23\tan(3e^{-3})}.