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CIE 9231 2025 June Paper 24 Q3

A Level / CIE / FP2

CIE 9231 2025 June Paper 24 Paper · Question 3

题目

Problem

Find the particular solution of the differential equation

d2ydx2+4dydx+5y=13e3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\frac{\mathrm{d}y}{\mathrm{d}x} + 5y = 13e^{3x}

given that y=1y = 1 and dydx=0\frac{\mathrm{d}y}{\mathrm{d}x} = 0 when x=0x = 0.

[10]
题目中文翻译

求微分方程

d2ydx2+4dydx+5y=13e3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4\frac{\mathrm{d}y}{\mathrm{d}x} + 5y = 13e^{3x}

的特解,已知当 x=0x = 0 时,y=1y = 1 且 dydx=0\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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解辅助方程得到衰减振荡的互补函数。右边为 13e3x13e^{3x},特解试作 ke3xke^{3x},最后用两个初始条件。

答题过程

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The auxiliary equation m2+4m+5=0m^2+4m+5=0 gives m=−2±im=-2\pm\mathrm{i}. The complementary function is e−2x(Acos⁡x+Bsin⁡x)e^{-2x}(A\cos x+B\sin x). For a particular integral yp=ke3xy_p=ke^{3x}, the derivatives are yp′=3ke3xy_p'=3ke^{3x} and yp′′=9ke3xy_p''=9ke^{3x}. Substitution and division by e3xe^{3x} give (9+12+5)k=13(9+12+5)k=13, so the particular integral is 12e3x\frac12e^{3x}.

The general solution and its derivative are

y= e−2x(Acos⁡x+Bsin⁡x)+12e3x,y′= e−2x[(−2A+B)cos⁡x+(−A−2B)sin⁡x] +32e3x.\begin{align*} y=&\,e^{-2x}(A\cos x+B\sin x)+\frac12e^{3x},\\[4mm] y'=&\,e^{-2x}\big[(-2A+B)\cos x+(-A-2B)\sin x\big]\\[4mm] &\,\hspace{2pt}+\frac32e^{3x}. \end{align*}

Now y(0)=1y(0)=1 gives A+12=1A+\frac12=1. The derivative at 00 gives B−2A+32=0B-2A+\frac32=0, so A=12A=\frac12 and B=−12B=-\frac12. Therefore

y=12e−2x(cos⁡x−sin⁡x)+12e3x.y=\frac12e^{-2x}(\cos x-\sin x)+\frac12e^{3x}.