题目

Problem

(a) Show that

\frac{\mathrm{d}}{\mathrm{d}x}\big(\ln(\tanh x)\big) = 2\operatorname{cosech}2x.

[3]

(b) Find the solution of the differential equation

\sinh 2x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = \sinh 2x

for which $y = 5$ when $x = \ln 2$ . Give your answer in an exact form.

[7]

题目中文翻译

(a) 证明

\frac{\mathrm{d}}{\mathrm{d}x}\big(\ln(\tanh x)\big) = 2\operatorname{cosech}2x

(3)

(b) 求微分方程

\sinh 2x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = \sinh 2x

的解，使得当 $x = \ln 2$ 时 $y = 5$ 。用精确形式给出答案。

(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

(a)

解法一

思路

展开

用对数的链式法则，再将 $\tanh x$ 和 $\operatorname{sech}x$ 改写成双曲正弦、双曲余弦。

答题过程

展开

For $x>0$ ,

\begin{align*} &\,\frac{\mathrm{d}}{\mathrm{d}x}\ln(\tanh x)\\[4mm] =&\,\frac{\operatorname{sech}^2x}{\tanh x}\\[4mm] =&\,\frac1{\sinh x\cosh x}\\[4mm] =&\,2\operatorname{csch}2x. \end{align*}

(b)

解法一

思路

展开

在含初始点的 $x>0$ 区间内，先除以 $\sinh 2x$ 。用 (a) 求出积分因子 $\tanh x$ ，积分后代入 $x=\ln2$ 。

答题过程

展开

On $x>0$ , the equation is $y'+2\operatorname{csch}(2x)y=1$ . By (a), an integrating factor is

\exp\bigg(\int2\operatorname{csch}(2x)\,\mathrm{d}x\bigg) =\exp(\ln(\tanh x))=\tanh x.

It gives

\frac{\mathrm{d}}{\mathrm{d}x}(y\tanh x)=\tanh x.

Integrating gives

y\tanh x=\ln(\cosh x)+C.

At $x=\ln2$ , the exponential definitions give

\cosh x=\frac{2+1/2}{2}=\frac54,\qquad \tanh x=\frac{2-1/2}{2+1/2}=\frac35.

Hence $5\cdot\frac35=\ln\frac54+C$ , and

y=\frac{\ln(\cosh x)+3-\ln(5/4)}{\tanh x}.

CIE 9231 2024 November Paper 22 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程