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CIE 9709 2023 Nov Paper 32 Q11

A Level / CIE / P3

CIE 9709 2023 Nov Paper 32 Paper · Question 11

题目

Problem

The variables xx and yy satisfy the differential equation

x2dydx+y2+y=0.x^2\frac{\mathrm{d}y}{\mathrm{d}x}+y^2+y=0.

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

(a) Solve the differential equation to obtain an expression for yy in terms of xx.

(8)

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

(1)
题目中文翻译

变量 xx 和 yy 满足微分方程

x2dydx+y2+y=0.x^2\frac{\mathrm{d}y}{\mathrm{d}x}+y^2+y=0.

已知当 y=1y=1 时,x=1x=1。

(a) 解此微分方程,得到用 xx 表示 yy 的表达式。

(8)

(b) 写出当 xx 趋于无穷大时 yy 的值的变化情况。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

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分离变量后,对 1/[y(y+1)]1/[y(y+1)] 做部分分式。初值所在分支有 y>0y>0,可合并对数;最后指数化并解出 yy,同时保留该分支的定义域。

答题过程

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Rearranging and separating variables,

1y(y+1) dy=−1x2 dx.\frac1{y(y+1)}\,\mathrm{d}y=-\frac1{x^2}\,\mathrm{d}x.

Since

1y−1y+1=(y+1)−yy(y+1)=1y(y+1),\frac1y-\frac1{y+1} =\frac{(y+1)-y}{y(y+1)} =\frac1{y(y+1)},

integration gives

ln⁡∣y∣−ln⁡∣y+1∣=1x+C.\ln|y|-\ln|y+1|=\frac1x+C.

On the branch through (1,1)(1,1), y>0y>0, so the initial condition gives

−ln⁡2=1+C,C=−1−ln⁡2.-\ln2=1+C,\qquad C=-1-\ln2.

Hence

ln⁡yy+1= 1x−1−ln⁡2,yy+1= 12e1/x−1.\begin{align*} \ln\frac{y}{y+1}=&\,\frac1x-1-\ln2,\\[4mm] \frac{y}{y+1}=&\,\frac12e^{1/x-1}. \end{align*}

Multiplying by 2(y+1)2(y+1) and collecting the terms in yy,

2y= (y+1)e1/x−1,y(2−e1/x−1)= e1/x−1.\begin{align*} 2y=&\,(y+1)e^{1/x-1},\\[4mm] y(2-e^{1/x-1})=&\,e^{1/x-1}. \end{align*}

Thus

y=e1/x−12−e1/x−1.\boxed{y=\frac{e^{1/x-1}}{2-e^{1/x-1}}}.

The branch containing the initial point has x>1/(1+ln⁡2)x>1/(1+\ln2), where the denominator is positive.

(b)

解法一

思路

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在初值分支内令 x→∞x\to\infty,指数中的 1/x1/x 趋于零。

答题过程

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As x→∞x\to\infty, e1/x−1→e−1e^{1/x-1}\to e^{-1}. Therefore

y⟶e−12−e−1=12e−1.\boxed{y\longrightarrow\frac{e^{-1}}{2-e^{-1}} =\frac1{2e-1}}.