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CIE 9231 2026 June Paper 22 Q6

A Level / CIE / FP2

CIE 9231 2026 June Paper 22 Paper · Question 6

题目

Problem

Find the particular solution of the differential equation

5d2ydx2+4dydx+y=x2+7x+1,5\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+4\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2+7x+1,

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

(10)
题目中文翻译

求微分方程

5d2ydx2+4dydx+y=x2+7x+15\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+4\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2+7x+1

满足当 x=0x=0 时,y=0y=0 且 dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 的特解。

(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。右端是二次多项式,试同次数的特解。最后用 y(0)y(0) 和 y′(0)y'(0) 确定互补函数中的常数。

答题过程

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The auxiliary equation 5m2+4m+1=05m^2+4m+1=0 gives m=−25±15im=-\frac25\pm\frac15\mathrm{i}. The complementary function is

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

For the trial yp=px2+qx+ry_p=px^2+qx+r, the derivatives are yp′=2px+qy_p'=2px+q and yp′′=2py_p''=2p. Substitution gives (p)x2+(8p+q)x+(10p+4q+r)=x2+7x+1(p)x^2+(8p+q)x+(10p+4q+r)=x^2+7x+1. Comparing coefficients,

p=1,8p+q=7,10p+4q+r=1.p=1,\qquad 8p+q=7,\qquad 10p+4q+r=1.

Thus q=−1q=-1 and r=−5r=-5. The general solution and its derivative are

y= x2−x−5+e−2x/5(Acos⁡x5+Bsin⁡x5),y′= 2x−1−25e−2x/5(Acos⁡x5+Bsin⁡x5) +15e−2x/5(−Asin⁡x5+Bcos⁡x5).\begin{align*} y=&\,x^2-x-5+e^{-2x/5}\bigg(A\cos\frac x5+B\sin\frac x5\bigg),\\ y'=&\,2x-1-\frac25e^{-2x/5}\bigg(A\cos\frac x5+B\sin\frac x5\bigg)\\ &\,\hspace{2pt}+\frac15e^{-2x/5}\bigg(-A\sin\frac x5+B\cos\frac x5\bigg). \end{align*}

Applying the two initial conditions gives

A−5=0,−25A+15B−1=0.A-5=0,\qquad -\frac25A+\frac15B-1=0.

So A=5A=5 and B=15B=15, giving

y=x2−x−5+e−2x/5(5cos⁡x5+15sin⁡x5).y=x^2-x-5+e^{-2x/5}\bigg(5\cos\frac x5+15\sin\frac x5\bigg).