题目
Problem
Find the particular solution of the differential equation
dx2d2y+4dxdy+5y=13e3x
given that y=1 and dxdy=0 when x=0.
[10]
题目中文翻译
求微分方程
dx2d2y+4dxdy+5y=13e3x
的特解,已知当 x=0 时,y=1 且 dxdy=0。
(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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解辅助方程得到衰减振荡的互补函数。右边为 13e3x,特解试作 ke3x,最后用两个初始条件。
答题过程
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The auxiliary equation m2+4m+5=0 gives m=−2±i. The complementary function is e−2x(Acosx+Bsinx). For a particular integral yp=ke3x, the derivatives are yp′=3ke3x and yp′′=9ke3x. Substitution and division by e3x give (9+12+5)k=13, so the particular integral is 21e3x.
The general solution and its derivative are
y=y′=e−2x(Acosx+Bsinx)+21e3x,e−2x[(−2A+B)cosx+(−A−2B)sinx]+23e3x.
Now y(0)=1 gives A+21=1. The derivative at 0 gives B−2A+23=0, so A=21 and B=−21. Therefore
y=21e−2x(cosx−sinx)+21e3x.