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CIE 9231 2025 Nov Paper 22 Q5

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 22 Paper · Question 5

题目

Problem

(a) Find the general solution of the differential equation

d2ydx2+4dydx+3y=5cos⁡x.\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\frac{\mathrm{d}y}{\mathrm{d}x} + 3y = 5\cos x.
[7]

(b) For large positive values of xx and for any initial conditions, show that the solution to part (a) can be approximated by

y≈Rsin⁡(x+ϕ),y \approx R\sin(x + \phi),

where the constants RR and ϕ\phi are to be determined.

[3]
题目中文翻译

(a) 求微分方程

d2ydx2+4dydx+3y=5cos⁡x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\frac{\mathrm{d}y}{\mathrm{d}x} + 3y = 5\cos x

的通解。

(7)

(b) 对于较大的正 xx 值以及任意初始条件,证明 (a) 中的解可近似为

y≈Rsin⁡(x+ϕ),y \approx R\sin(x + \phi),

其中常数 RR 和 ϕ\phi 需要确定。

(3)
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

(a)

解法一

思路

展开

由辅助方程求互补函数。右边是 cos⁡x\cos x,特解要同时试正弦和余弦,因为求导会互相转换。

答题过程

展开

The auxiliary equation (m+1)(m+3)=0(m+1)(m+3)=0 gives the complementary function Ae−x+Be−3xAe^{-x}+Be^{-3x}. For a particular integral, take y=psin⁡x+qcos⁡xy=p\sin x+q\cos x. Then

y′=pcos⁡x−qsin⁡x,y′′=−psin⁡x−qcos⁡x.y'=p\cos x-q\sin x,\qquad y''=-p\sin x-q\cos x.

Substitution gives

(2p−4q)sin⁡x+(4p+2q)cos⁡x=5cos⁡x.(2p-4q)\sin x+(4p+2q)\cos x=5\cos x.

Solving 2p−4q=02p-4q=0 and 4p+2q=54p+2q=5 gives p=1p=1 and q=12q=\frac12. The general solution is

y=Ae−x+Be−3x+sin⁡x+12cos⁡x.y=Ae^{-x}+Be^{-3x}+\sin x+\frac12\cos x.

(b)

解法一

思路

展开

初始条件只影响两个衰减的指数项。对留下的正弦和余弦比较系数,求振幅和相位。

答题过程

展开

For fixed A,BA,B, both Ae−xAe^{-x} and Be−3xBe^{-3x} tend to zero as x→+∞x\to+\infty. Matching sin⁡x+12cos⁡x\sin x+\frac12\cos x with Rsin⁡(x+ϕ)R\sin(x+\phi) gives

Rcos⁡ϕ=1,Rsin⁡ϕ=12.R\cos\phi=1,\qquad R\sin\phi=\frac12.

Taking R>0R>0, we have R2=12+(1/2)2=5/4R^2=1^2+(1/2)^2=5/4. Both Rcos⁡ϕR\cos\phi and Rsin⁡ϕR\sin\phi are positive, so we choose 0<ϕ<π/20<\phi<\pi/2. Hence

R=52,ϕ=arctan⁡12.R=\frac{\sqrt5}{2},\qquad \phi=\arctan\frac12.

Therefore, for large positive xx,

y≈52sin⁡(x+arctan⁡12).y\approx\frac{\sqrt5}{2}\sin\bigg(x+\arctan\frac12\bigg).