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CIE 9231 2022 November Paper 23 Q5

A Level / CIE / FP2

CIE 9231 2022 Nov Paper 23 Paper · Question 5

题目

Problem

Find the particular solution of the differential equation

2d2ydx2+2dydx+y=4x2+3x+3,2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4x^2+3x+3,

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

(10)
题目中文翻译

求微分方程

2d2ydx2+2dydx+y=4x2+3x+32\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4x^2+3x+3

的特解,已知当 x=0x=0 时,y=dydx=0y=\dfrac{\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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先解辅助方程得到互补函数。用二次多项式试特解,最后利用两个零初值确定常数。

答题过程

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The auxiliary equation 2m2+2m+1=02m^2+2m+1=0 gives m=−12±12im=-\frac12\pm\frac12\mathrm{i}. The complementary function is

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

For the particular integral, take yp=px2+qx+ry_{\mathrm p}=px^2+qx+r. Then yp′=2px+qy_{\mathrm p}'=2px+q and yp′′=2py_{\mathrm p}''=2p, giving

2yp′′+2yp′+yp=px2+(4p+q)x+4p+2q+r.2y_{\mathrm p}''+2y_{\mathrm p}'+y_{\mathrm p} =px^2+(4p+q)x+4p+2q+r.

Matching coefficients gives p=4p=4, 4p+q=34p+q=3 and 4p+2q+r=34p+2q+r=3, so q=−13q=-13 and r=13r=13. The general solution and its derivative are

y= 4x2−13x+13+e−x/2(Acos⁡x2+Bsin⁡x2),y′= 8x−13−12e−x/2(Acos⁡x2+Bsin⁡x2) +12e−x/2(−Asin⁡x2+Bcos⁡x2).\begin{align*} y=&\,4x^2-13x+13+e^{-x/2}\bigg(A\cos\frac x2+B\sin\frac x2\bigg),\\ y'=&\,8x-13-\frac12e^{-x/2}\bigg(A\cos\frac x2+B\sin\frac x2\bigg)\\ &\,\hspace{2pt}+\frac12e^{-x/2}\bigg(-A\sin\frac x2+B\cos\frac x2\bigg). \end{align*}

At x=0x=0, the two initial conditions give

A+13=0,−A2+B2−13=0.A+13=0,\qquad -\frac A2+\frac B2-13=0.

Hence A=−13A=-13 and B=13B=13, and the solution is

y=4x2−13x+13+13e−x/2(sin⁡x2−cos⁡x2).y=4x^2-13x+13+13e^{-x/2}\bigg(\sin\frac x2-\cos\frac x2\bigg).