题目
Problem
Find the particular solution of the differential equation
dt2d2x−12dtdx+36x=37sint,
given that, when t=0, x=dtdx=0.
(11)
题目中文翻译
求微分方程
dt2d2x−12dtdx+36x=37sint
满足当 t=0 时,x=dtdx=0 的特解。
(11)
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 6,所以互补函数含 te6t。特解同时试正弦和余弦,最后用两个零初值。
答题过程
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The auxiliary equation (m−6)2=0 gives the complementary function (A+Bt)e6t. For the particular integral, take xp=psint+qcost. Then xp′=pcost−qsint and xp′′=−psint−qcost. Substitution gives
(35p+12q)sint+(−12p+35q)cost=37sint.
Matching coefficients gives
35p+12q=37,−12p+35q=0.
The second equation gives q=12p/35. Substituting into the first gives 1369p/35=37, hence p=35/37 and q=12/37. Thus the general solution and its derivative are
x=x′=(A+Bt)e6t+3735sint+3712cost,[B+6(A+Bt)]e6t+3735cost−3712sint.
The initial conditions give
A+3712=0,6A+B+3735=0.
Thus A=−12/37 and B=1, so
x=(t−3712)e6t+3735sint+3712cost.