题目
Problem
The variables x and y satisfy the differential equation
x2dxdy+y2+y=0.
It is given that x=1 when y=1.
(a) Solve the differential equation to obtain an expression for y in terms of x.
(8)
(b) State what happens to the value of y when x tends to infinity. Give your answer in an exact form.
(1)
题目中文翻译
变量 x 和 y 满足微分方程
x2dxdy+y2+y=0.
已知当 y=1 时,x=1。
(a) 解此微分方程,得到用 x 表示 y 的表达式。
(8)
(b) 写出当 x 趋于无穷大时 y 的值的变化情况。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
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分离变量后,对 1/[y(y+1)] 做部分分式。初值所在分支有 y>0,可合并对数;最后指数化并解出 y,同时保留该分支的定义域。
答题过程
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Rearranging and separating variables,
y(y+1)1dy=−x21dx.
Since
y1−y+11=y(y+1)(y+1)−y=y(y+1)1,
integration gives
ln∣y∣−ln∣y+1∣=x1+C.
On the branch through (1,1), y>0, so the initial condition gives
−ln2=1+C,C=−1−ln2.
Hence
lny+1y=y+1y=x1−1−ln2,21e1/x−1.
Multiplying by 2(y+1) and collecting the terms in y,
2y=y(2−e1/x−1)=(y+1)e1/x−1,e1/x−1.
Thus
y=2−e1/x−1e1/x−1.
The branch containing the initial point has x>1/(1+ln2), where the denominator is positive.
(b)
解法一
思路
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在初值分支内令 x→∞,指数中的 1/x 趋于零。
答题过程
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As x→∞, e1/x−1→e−1. Therefore
y⟶2−e−1e−1=2e−11.