Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2021 November Paper 21 Q5

A Level / CIE / FP2

CIE 9231 2021 Nov Paper 21 Paper · Question 5

题目

Problem

Find the particular solution of the differential equation

d2ydx2−2dydx+y=4cos⁡x,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4\cos x,

given that, when x=0x=0, y=−4y=-4 and dydx=3\frac{\mathrm{d}y}{\mathrm{d}x}=3.

[11]
题目中文翻译

求微分方程

d2ydx2−2dydx+y=4cos⁡x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4\cos x

的特解,已知当 x=0x=0 时,y=−4y=-4 且 dydx=3\frac{\mathrm{d}y}{\mathrm{d}x}=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

展开

辅助方程有重根 11,所以互补函数包含 xexxe^x。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

展开

The auxiliary equation (m−1)2=0(m-1)^2=0 has a repeated root, so the complementary function is (A+Bx)ex(A+Bx)e^x. For a particular integral, take y=psin⁡x+qcos⁡xy=p\sin x+q\cos x. Then yp′=pcos⁡x−qsin⁡xy_p'=p\cos x-q\sin x and yp′′=−psin⁡x−qcos⁡xy_p''=-p\sin x-q\cos x. Since yp′′+yp=0y_p''+y_p=0 for this trial, the left-hand side reduces to

−2y′=−2pcos⁡x+2qsin⁡x.-2y'=-2p\cos x+2q\sin x.

Matching 4cos⁡x4\cos x gives p=−2p=-2 and q=0q=0. The general solution and its derivative are

y= (A+Bx)ex−2sin⁡x,y′= (A+B+Bx)ex−2cos⁡x.\begin{align*} y=&\,(A+Bx)e^x-2\sin x,\\ y'=&\,(A+B+Bx)e^x-2\cos x. \end{align*}

The initial conditions give A=−4A=-4 and A+B−2=3A+B-2=3. Therefore

y=(9x−4)ex−2sin⁡x.y=(9x-4)e^x-2\sin x.