题目
Problem
Find the particular solution of the differential equation
49dt2d2x+14dtdx+x=196e71t,
given that, when t=0, x=7 and dtdx=0.
(10)
题目中文翻译
求微分方程
49dt2d2x+14dtdx+x=196e71t
满足当 t=0 时,x=7 且 dtdx=0 的特解。
(10)
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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辅助方程有重根 −1/7。互补函数要包含 te−t/7,特解试作 ket/7,最后代入初始条件。
答题过程
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The auxiliary equation is (7m+1)2=0, so the complementary function is (A+Bt)e−t/7. For the trial xp=ket/7, its derivatives are xp′=ket/7/7 and xp′′=ket/7/49. Substituting gives (k+2k+k)et/7=196et/7, so k=49. The general solution and its derivative are
x=x′=(A+Bt)e−t/7+49et/7,(B−7A+Bt)e−t/7+7et/7.
At t=0, x(0)=A+49=7 and x′(0)=B−A/7+7=0. Thus A=−42 and B=−13. Therefore
x=(−42−13t)e−t/7+49et/7.