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CIE 9231 2025 June Paper 23 Q7

A Level / CIE / FP2

CIE 9231 2025 June Paper 23 Paper · Question 7

题目

Problem

Find the solution of the differential equation

dydx−2x+6x2+6x+5y= 4,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{2x + 6}{x^2 + 6x + 5}y =&\, 4, \end{align*}

given that y=0y = 0 when x=0x = 0. Give your answer in an exact form.

[9]
题目中文翻译

求微分方程

dydx−2x+6x2+6x+5y= 4,\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{2x + 6}{x^2 + 6x + 5}y =&\, 4, \end{align*}

的解,已知当 x=0x = 0 时 y=0y = 0。答案需写成精确形式。

(9)
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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yy 的系数是 −Q′/Q-Q'/Q,其中 Q=x2+6x+5Q=x^2+6x+5。乘积分因子 1/Q1/Q 后,把右边拆成部分分式积分。

答题过程

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Let Q=(x+1)(x+5)Q=(x+1)(x+5). On the interval x>−1x>-1 containing the initial point, an integrating factor is 1/Q1/Q, since the coefficient of yy is −Q′/Q-Q'/Q. In full,

exp⁡(−∫Q′Q dx)=exp⁡(−ln⁡Q)=1Q,\exp\bigg(-\int\frac{Q'}Q\,\mathrm{d}x\bigg) =\exp(-\ln Q)=\frac1Q,

where Q>0Q>0 on this interval. Multiplying by this factor and using partial fractions gives

 ddx(yQ)= 4Q=1x+1−1x+5.\begin{align*} &\,\frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac yQ\bigg)\\[4mm] =&\,\frac4Q=\frac1{x+1}-\frac1{x+5}. \end{align*}

Integrating gives yQ=ln⁡(x+1x+5)+C\frac yQ=\ln\big(\frac{x+1}{x+5}\big)+C. At x=0x=0 we have y=0y=0, so C=ln⁡5C=\ln5. Therefore

y=(x2+6x+5)ln⁡(5(x+1)x+5).y=(x^2+6x+5)\ln\bigg(\frac{5(x+1)}{x+5}\bigg).