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CIE 9709 2026 June Paper 35 Q6

A Level / CIE / P3

CIE 9709 2026 June Paper 35 Paper · Question 6

题目

Problem

A curve is such that the gradient of the curve at any point is proportional to the yy-coordinate at that point. The curve passes through the points (2,e2)(2,\mathrm{e}^2) and (5,e3)(5,\mathrm{e}^3).

Form and solve a differential equation to find yy in terms of xx.

(7)
题目中文翻译

一条曲线在任意一点处的斜率与该点的 yy 坐标成正比。曲线经过点 (2,e2)(2,\mathrm{e}^2) 和 (5,e3)(5,\mathrm{e}^3)。

建立并求解一个微分方程,求用 xx 表示的 yy。

(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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将“梯度与 yy 成正比”写成微分方程。分离变量积分,再用两点同时求比例常数和积分常数。

答题过程

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Let the proportionality constant be cc. Then

dydx=cy.\frac{\mathrm{d}y}{\mathrm{d}x}=cy.

On the branch through the given positive values, separate variables and integrate:

∫1y dy= ∫c dx,ln⁡y= cx+C.\begin{align*} \int\frac1y\,\mathrm{d}y=&\,\int c\,\mathrm{d}x,\\[4mm] \ln y=&\,cx+C. \end{align*}

Substituting the two points,

2= 2c+C,3= 5c+C.\begin{align*} 2=&\,2c+C,\\[4mm] 3=&\,5c+C. \end{align*}

Subtracting the first equation from the second gives 1=3c1=3c, so c=1/3c=1/3. Substituting back gives C=2−2/3=4/3C=2-2/3=4/3. The required solution is y=e(x+4)/3y=e^{(x+4)/3}.