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CIE 9231 2025 Nov Paper 23 Q4

A Level / CIE / FP2

CIE 9231 2025 Nov Paper 23 Paper · Question 4

题目

Problem

Find the particular solution of the differential equation

5d2ydx2+2dydx+y=x2+5x+3,5\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + y = x^2 + 5x + 3,

given that, when x=0x = 0, y=dydx=0y = \frac{\mathrm{d}y}{\mathrm{d}x} = 0.

[10]
题目中文翻译

求微分方程

5d2ydx2+2dydx+y=x2+5x+35\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + y = x^2 + 5x + 3

的特解,已知当 x=0x = 0 时,y=dydx=0y = \frac{\mathrm{d}y}{\mathrm{d}x} = 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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由辅助方程求互补函数,用二次多项式试出一个特解。相加后用 y(0)y(0) 和 y′(0)y'(0) 确定两个常数。

答题过程

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The auxiliary equation 5m2+2m+1=05m^2+2m+1=0 has roots −15±25i-\frac15\pm\frac25\mathrm{i}. The complementary function is

e−x/5(Acos⁡2x5+Bsin⁡2x5).e^{-x/5}\bigg(A\cos\frac{2x}{5}+B\sin\frac{2x}{5}\bigg).

For a particular integral, substitute y=px2+qx+ry=px^2+qx+r, with y′=2px+qy'=2px+q and y′′=2py''=2p. Equating coefficients gives

p= 1,4p+q= 5,10p+2q+r= 3,\begin{align*} p=&\,1,\\ 4p+q=&\,5,\\ 10p+2q+r=&\,3, \end{align*}

so p=q=1p=q=1 and r=−9r=-9. The general solution is

y=x2+x−9+e−x/5(Acos⁡2x5+Bsin⁡2x5).y=x^2+x-9+e^{-x/5}\bigg(A\cos\frac{2x}{5}+B\sin\frac{2x}{5}\bigg).

Differentiating,

y′= 2x+1 −15e−x/5(Acos⁡2x5+Bsin⁡2x5) +e−x/5(−2A5sin⁡2x5+2B5cos⁡2x5).\begin{align*} y'=&\,2x+1\\ &\,\hspace{2pt}-\frac15e^{-x/5}\bigg(A\cos\frac{2x}{5}+B\sin\frac{2x}{5}\bigg)\\ &\,\hspace{4pt}+e^{-x/5}\bigg(-\frac{2A}{5}\sin\frac{2x}{5}+\frac{2B}{5}\cos\frac{2x}{5}\bigg). \end{align*}

Applying y(0)=y′(0)=0y(0)=y'(0)=0 gives

A−9= 0,1−A5+2B5= 0.\begin{align*} A-9=&\,0,\\ 1-\frac{A}{5}+\frac{2B}{5}=&\,0. \end{align*}

Thus A=9A=9, B=2B=2 and

y=x2+x−9+e−x/5(9cos⁡2x5+2sin⁡2x5).y=x^2+x-9+e^{-x/5}\bigg(9\cos\frac{2x}{5}+2\sin\frac{2x}{5}\bigg).