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CIE 9709 2022 March Paper 32 Q9

A Level / CIE / P3

CIE 9709 2022 March Paper 32 Paper · Question 9

题目

Problem

The variables xx and yy satisfy the differential equation

(x+1)(3x+1)dydx=y,(x+1)(3x+1)\frac{\mathrm{d}y}{\mathrm{d}x}=y,

and it is given that y=1y=1 when x=1x=1.

Solve the differential equation and find the exact value of yy when x=3x=3, giving your answer in a simplified form.

(9)
题目中文翻译

变量 xx 和 yy 满足微分方程

(x+1)(3x+1)dydx=y,(x+1)(3x+1)\frac{\mathrm{d}y}{\mathrm{d}x}=y,

且已知当 x=1x=1 时,y=1y=1。

解此微分方程,并求 x=3x=3 时 yy 的精确值,答案以化简后的形式表示。

(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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分离变量后对两个一次因子的倒数做部分分式。使用初值选正分支,最后代入 x=3x=3。

答题过程

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On the branch containing (1,1)(1,1), separate variables:

1y dy=1(x+1)(3x+1) dx.\frac1y\,\mathrm{d}y=\frac1{(x+1)(3x+1)}\,\mathrm{d}x.

Write

1(x+1)(3x+1)=Ax+1+B3x+1.\frac1{(x+1)(3x+1)}=\frac{A}{x+1}+\frac{B}{3x+1}.

Then 1=A(3x+1)+B(x+1)1=A(3x+1)+B(x+1). Comparing coefficients,

3A+B= 0,A+B= 1,\begin{align*} 3A+B=&\,0,\\[4mm] A+B=&\,1, \end{align*}

so A=−1/2A=-1/2, B=3/2B=3/2. For the second term, the factor 33 in the denominator gives

∫3/23x+1 dx=32⋅13ln⁡(3x+1)+C1.\int\frac{3/2}{3x+1}\,\mathrm{d}x=\frac32\cdot\frac13\ln(3x+1)+C_1.

Integrating on the interval x>−1/3x>-1/3 through the initial point,

ln⁡y=−12ln⁡(x+1)+12ln⁡(3x+1)+C.\ln y=-\frac12\ln(x+1)+\frac12\ln(3x+1)+C.

Using x=1x=1, y=1y=1,

0= −12ln⁡2+12ln⁡4+C,C= −12ln⁡2.\begin{align*} 0=&\,-\frac12\ln2+\frac12\ln4+C,\\[4mm] C=&\,-\frac12\ln2. \end{align*}

Therefore

ln⁡y=12ln⁡3x+12(x+1),\ln y=\frac12\ln\frac{3x+1}{2(x+1)},

so the initial-value solution is

y=3x+12(x+1),x>−13.\boxed{y=\sqrt{\frac{3x+1}{2(x+1)}},\qquad x>-\frac13}.

At x=3x=3,

y=108=52.\boxed{y=\sqrt{\frac{10}{8}}=\frac{\sqrt5}{2}}.