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CIE 9231 2026 June Paper 23 Q6

A Level / CIE / FP2

CIE 9231 2026 June Paper 23 Paper · Question 6

题目

Problem

(a) Show that an appropriate integrating factor for

(1−x2)dydx+y=(1−x2)1−x\big(1-x^2\big)\frac{\mathrm{d}y}{\mathrm{d}x}+y=\big(1-x^2\big)\sqrt{1-x}

is 1+x1−x\sqrt{\dfrac{1+x}{1-x}}.

(4)

(b) Hence find the solution of the differential equation

(1−x2)dydx+y=(1−x2)1−x\big(1-x^2\big)\frac{\mathrm{d}y}{\mathrm{d}x}+y=\big(1-x^2\big)\sqrt{1-x}

for which y=1y=1 when x=0x=0.

(6)
题目中文翻译

(a) 证明微分方程

(1−x2)dydx+y=(1−x2)1−x\big(1-x^2\big)\frac{\mathrm{d}y}{\mathrm{d}x}+y=\big(1-x^2\big)\sqrt{1-x}

的一个适当的积分因子为 1+x1−x\sqrt{\dfrac{1+x}{1-x}}。

(4)

(b) 由此求微分方程

(1−x2)dydx+y=(1−x2)1−x\big(1-x^2\big)\frac{\mathrm{d}y}{\mathrm{d}x}+y=\big(1-x^2\big)\sqrt{1-x}

满足当 x=0x=0 时 y=1y=1 的解。

(6)
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

在含初始点的 −1<x<1-1<x<1 区间,先把方程化为标准形式。yy 的系数是反双曲正切的导数,用对数形式写出积分因子。

答题过程

展开

Dividing by 1−x21-x^2 on −1<x<1-1<x<1 gives

y′+y1−x2=1−x.y'+\frac{y}{1-x^2}=\sqrt{1-x}.

An integrating factor is

μ= exp⁡(∫dx1−x2)= etanh⁡−1x= exp⁡(12ln⁡1+x1−x)= 1+x1−x.\begin{align*} \mu=&\,\exp\bigg(\int\frac{\mathrm{d}x}{1-x^2}\bigg)\\ =&\,e^{\tanh^{-1}x}\\ =&\,\exp\bigg(\frac12\ln\frac{1+x}{1-x}\bigg)\\ =&\,\sqrt{\frac{1+x}{1-x}}. \end{align*}

(b)

解法一

思路

展开

积分因子乘到右边后,1−x\sqrt{1-x} 抵消,只剩 1+x\sqrt{1+x}。积分后由初值确定常数。

答题过程

展开

Multiplying the standard equation by (1+x)/(1−x)\sqrt{(1+x)/(1-x)} gives

ddx(y1+x1−x)=1+x.\frac{\mathrm{d}}{\mathrm{d}x} \bigg(y\sqrt{\frac{1+x}{1-x}}\bigg)=\sqrt{1+x}.

Since ∫(1+x)1/2 dx=23(1+x)3/2+C\int(1+x)^{1/2}\,\mathrm{d}x=\frac23(1+x)^{3/2}+C, we have y(1+x)/(1−x)=23(1+x)3/2+Cy\sqrt{(1+x)/(1-x)}=\frac23(1+x)^{3/2}+C. At x=0x=0, y=1y=1, so C=13C=\frac13. Thus

y=1−x1+x[23(1+x)3/2+13],−1<x<1.y=\sqrt{\frac{1-x}{1+x}}\bigg[\frac23(1+x)^{3/2}+\frac13\bigg], \qquad -1<x<1.