题目
Problem
Find the particular solution of the differential equation
5dx2d2y+4dxdy+y=x2+7x+1,
given that, when x=0, y=0 and dxdy=0.
(10)
题目中文翻译
求微分方程
5dx2d2y+4dxdy+y=x2+7x+1
满足当 x=0 时,y=0 且 dxdy=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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先解辅助方程。右端是二次多项式,试同次数的特解。最后用 y(0) 和 y′(0) 确定互补函数中的常数。
答题过程
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The auxiliary equation 5m2+4m+1=0 gives m=−52±51i. The complementary function is
e−2x/5(Acos5x+Bsin5x).
For the trial yp=px2+qx+r, the derivatives are yp′=2px+q and yp′′=2p. Substitution gives (p)x2+(8p+q)x+(10p+4q+r)=x2+7x+1. Comparing coefficients,
p=1,8p+q=7,10p+4q+r=1.
Thus q=−1 and r=−5. The general solution and its derivative are
y=y′=x2−x−5+e−2x/5(Acos5x+Bsin5x),2x−1−52e−2x/5(Acos5x+Bsin5x)+51e−2x/5(−Asin5x+Bcos5x).
Applying the two initial conditions gives
A−5=0,−52A+51B−1=0.
So A=5 and B=15, giving
y=x2−x−5+e−2x/5(5cos5x+15sin5x).