题目
Problem
Find the particular solution of the differential equation
dx2d2y−2dxdy+y=4cosx,
given that, when x=0, y=−4 and dxdy=3.
[11]
题目中文翻译
求微分方程
dx2d2y−2dxdy+y=4cosx
的特解,已知当 x=0 时,y=−4 且 dxdy=3。
(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 1,所以互补函数包含 xex。特解同时试正弦和余弦,再用两个初始条件求常数。
答题过程
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The auxiliary equation (m−1)2=0 has a repeated root, so the complementary function is (A+Bx)ex. For a particular integral, take y=psinx+qcosx. Then yp′=pcosx−qsinx and yp′′=−psinx−qcosx. Since yp′′+yp=0 for this trial, the left-hand side reduces to
−2y′=−2pcosx+2qsinx.
Matching 4cosx gives p=−2 and q=0. The general solution and its derivative are
y=y′=(A+Bx)ex−2sinx,(A+B+Bx)ex−2cosx.
The initial conditions give A=−4 and A+B−2=3. Therefore
y=(9x−4)ex−2sinx.