题目
Problem
(a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that
sinh2x=2sinhxcoshx.
[3]
(b) Using the substitution u=sinhx, find
∫sinh22xcoshxdx.
[4]
(c) Find the particular solution of the differential equation
dxdy+ytanhx=sinh22x,
given that y=4 when x=0. Give your answer in the form y=f(x).
[7]
题目中文翻译
(a) 从 cosh 和 sinh 关于指数函数的定义出发,证明
sinh2x=2sinhxcoshx
(3)
(b) 使用代换 u=sinhx,求
∫sinh22xcoshxdx
(4)
(c) 求微分方程
dxdy+ytanhx=sinh22x
的特解,已知当 x=0 时 y=4。将答案写成 y=f(x) 的形式。
(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
用指数定义把 2sinhxcoshx 写成两个括号的乘积,再用平方差展开。
答题过程
展开
The exponential definitions are
sinhx=2ex−e−x,coshx=2ex+e−x.
Therefore
2sinhxcoshx===21(ex−e−x)(ex+e−x)21(e2x−e−2x)sinh2x.
(b)
解法一
思路
展开
先用 (a) 展开平方,再用 cosh2x=1+sinh2x。代换 u=sinhx 后,只剩多项式。
答题过程
展开
By (a), sinh22x=4sinh2xcosh2x=4sinh2x(1+sinh2x). With u=sinhx, du=coshxdx, so
∫sinh22xcoshxdx===4∫(u2+u4)du34u3+54u5+C34sinh3x+54sinh5x+C.
(c)
解法一
思路
展开
由 ∫tanhxdx=ln(coshx)+C 求积分因子。乘上后,右边正好是 (b) 的被积函数。
答题过程
展开
The equation is already in standard linear form. An integrating factor is e∫tanhxdx=eln(coshx)=coshx. Thus
dxd(ycoshx)=sinh22xcoshx.
By (b), ycoshx=34sinh3x+54sinh5x+C. At x=0, y=4, so C=4. Hence
y=coshx4+34sinh3x+54sinh5x.