题目
Problem
Find the particular solution of the differential equation
2dx2d2y+2dxdy+y=4x2+3x+3,
given that, when x=0, y=dxdy=0.
(10)
题目中文翻译
求微分方程
2dx2d2y+2dxdy+y=4x2+3x+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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先解辅助方程得到互补函数。用二次多项式试特解,最后利用两个零初值确定常数。
答题过程
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The auxiliary equation 2m2+2m+1=0 gives m=−21±21i. The complementary function is
e−x/2(Acos2x+Bsin2x).
For the particular integral, take yp=px2+qx+r. Then yp′=2px+q and yp′′=2p, giving
2yp′′+2yp′+yp=px2+(4p+q)x+4p+2q+r.
Matching coefficients gives p=4, 4p+q=3 and 4p+2q+r=3, so q=−13 and r=13. The general solution and its derivative are
y=y′=4x2−13x+13+e−x/2(Acos2x+Bsin2x),8x−13−21e−x/2(Acos2x+Bsin2x)+21e−x/2(−Asin2x+Bcos2x).
At x=0, the two initial conditions give
A+13=0,−2A+2B−13=0.
Hence A=−13 and B=13, and the solution is
y=4x2−13x+13+13e−x/2(sin2x−cos2x).