题目
Problem
(a) The variables x and y satisfy the differential equation
dxdy=e2x+14+9y2.
It is given that y=0 when x=1.
Solve the differential equation, obtaining an expression for y in terms of x.
(7)
(b) State what happens to the value of y as x tends to infinity. Give your answer in an exact form.
(1)
题目中文翻译
(a) 变量 x 和 y 满足微分方程
dxdy=e2x+14+9y2.
已知当 x=1 时,y=0。
解此微分方程,求用 x 表示 y 的表达式。
(7)
(b) 说明当 x 趋于无穷大时,y 的值如何变化。答案用精确形式表示。
(1)
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
分离变量后,左侧是反正切型积分,右侧是指数型积分。用初值求常数,再取正切写成显式解。
答题过程
展开
Separating variables,
4+9y21dy=e−(2x+1)dx.
The chain rule gives
dydtan−1(3y/2)=4+9y26,
and dxde−(2x+1)=−2e−(2x+1).
Thus integrating gives
61tan−123y=−21e−(2x+1)+C.
Using y=0 at x=1,
0=C=−21e−3+C,21e−3.
Therefore
tan−123y=3e−3−3e−(2x+1).
Taking tangents,
y=32tan(3e−3−3e−(2x+1)).
The branch through the initial point is taken where the argument of the tangent remains in (−π/2,π/2).
(b)
解法一
思路
展开
当 x→∞ 时,随 x 衰减的指数项趋于零,保留精确的极限表达式。
答题过程
展开
As x→∞, e−(2x+1)→0. Hence
y⟶32tan(3e−3).