题目
Problem
Find the particular solution of the differential equation
5dx2d2y+2dxdy+y=x2+5x+3,
given that, when x=0, y=dxdy=0.
[10]
题目中文翻译
求微分方程
5dx2d2y+2dxdy+y=x2+5x+3
的特解,已知当 x=0 时,y=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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由辅助方程求互补函数,用二次多项式试出一个特解。相加后用 y(0) 和 y′(0) 确定两个常数。
答题过程
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The auxiliary equation 5m2+2m+1=0 has roots −51±52i. The complementary function is
e−x/5(Acos52x+Bsin52x).
For a particular integral, substitute y=px2+qx+r, with y′=2px+q and y′′=2p. Equating coefficients gives
p=4p+q=10p+2q+r=1,5,3,
so p=q=1 and r=−9. The general solution is
y=x2+x−9+e−x/5(Acos52x+Bsin52x).
Differentiating,
y′=2x+1−51e−x/5(Acos52x+Bsin52x)+e−x/5(−52Asin52x+52Bcos52x).
Applying y(0)=y′(0)=0 gives
A−9=1−5A+52B=0,0.
Thus A=9, B=2 and
y=x2+x−9+e−x/5(9cos52x+2sin52x).