题目
Problem
The variables x and θ satisfy the differential equation
sin2θdθdx=(4x+3)cos2θ,
and x=0 when θ=121π.
Solve the differential equation and obtain an expression for x in terms of θ.
(7)
题目中文翻译
变量 x 和 θ 满足微分方程
sin2θdθdx=(4x+3)cos2θ,
且当 θ=121π 时,x=0。
解此微分方程,并求出 x 关于 θ 的表达式。
(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
解法一
思路
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分离变量后,左边是一次函数倒数,右边是对数正弦的导数。初值确定比例常数,再把 x 显式解出。
答题过程
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Separating variables,
4x+31dx=sin2θcos2θdθ.
The derivatives of 4x+3 and sin2θ are 4 and 2cos2θ, respectively. Thus the logarithmic integrals give
41ln∣4x+3∣=21ln∣sin2θ∣+C.
Using x=0 at θ=π/12,
41ln3=21ln21+C,
so C=41ln12.
On the interval 0<θ<π/2 containing the initial point, sin2θ>0 and the initial branch has 4x+3>0. Therefore
ln(4x+3)=2ln(sin2θ)+ln12.
Exponentiating,
4x+3=12sin22θ.
Hence
x=3sin22θ−43.