题目
Problem
Find the solution of the differential equation
(4t2−1)dtdx+4x=4t2−1
for which x=3 when t=1. Give your answer in the form x=f(t).
(9)
题目中文翻译
求微分方程
(4t2−1)dtdx+4x=4t2−1
满足当 t=1 时 x=3 的解。将答案写成 x=f(t) 的形式。
(9)
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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在初值点所在的 t>21 区间内化成线性方程。将 x 的系数拆成部分分式来找积分因子。
答题过程
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For t>21, divide by 4t2−1 to obtain
dtdx+4t2−14x=1.
Since 4t2−14=2t−12−2t+12, we have
∫4t2−14dt=ln(2t−1)−ln(2t+1)+C.
An integrating factor is
μ=eln(2t−1)−ln(2t+1)=2t+12t−1.
Multiplying gives (μx)′=μ=1−2t+12. Integrating,
μx=∫(1−2t+12)dt=t−ln(2t+1)+C.
At t=1, x=3 and μ=1/3, giving 1=1−ln3+C, hence C=ln3. Therefore
x=2t−12t+1[t+ln(2t+13)].
