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CIE 9231 2024 November Paper 22 Q5

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 22 Paper · Question 5

题目

Problem

Find the particular solution of the differential equation

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

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

3d2ydx2+2dydx+y=x23\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + y = x^2

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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辅助方程有一对共轭复根。用二次多项式试特解,再用 y(0)y(0) 和 y′(0)y'(0) 解出互补函数的两个常数。

答题过程

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The auxiliary equation 3m2+2m+1=03m^2+2m+1=0 gives m=−13±23im=-\frac13\pm\frac{\sqrt2}{3}\mathrm{i}. The complementary function is

e−x/3(Acos⁡2x3+Bsin⁡2x3).e^{-x/3}\bigg(A\cos\frac{\sqrt2x}{3}+B\sin\frac{\sqrt2x}{3}\bigg).

For a particular integral yp=px2+qx+ry_p=px^2+qx+r, we have yp′=2px+qy_p'=2px+q and yp′′=2py_p''=2p. Substituting and comparing coefficients gives p=1p=1, 4p+q=04p+q=0 and 6p+2q+r=06p+2q+r=0. Hence the particular integral is x2−4x+2x^2-4x+2. Write w=2/3w=\sqrt2/3. The general solution and its derivative are

y= x2−4x+2+e−x/3(Acos⁡wx+Bsin⁡wx),y′= 2x−4+e−x/3[(−A3+wB)cos⁡wx +(−wA−B3)sin⁡wx].\begin{align*} y=&\,x^2-4x+2+e^{-x/3}(A\cos wx+B\sin wx),\\[4mm] y'=&\,2x-4+e^{-x/3}\bigg[\bigg(-\frac A3+wB\bigg)\cos wx\\[4mm] &\,\hspace{2pt}+\bigg(-wA-\frac B3\bigg)\sin wx\bigg]. \end{align*}

From the initial conditions,

A+2=0,−A3+23B−4=0.A+2=0,\qquad -\frac A3+\frac{\sqrt2}3B-4=0.

The first equation gives A=−2A=-2. Then 2/3+2B/3−4=02/3+\sqrt2 B/3-4=0, so 2B=10\sqrt2 B=10 and B=52B=5\sqrt2. The solution is

y= x2−4x+2 +e−x/3(−2cos⁡2x3+52sin⁡2x3).\begin{align*} y=&\,x^2-4x+2\\[4mm] &\,\hspace{2pt}+e^{-x/3}\bigg(-2\cos\frac{\sqrt2x}{3}+5\sqrt2\sin\frac{\sqrt2x}{3}\bigg). \end{align*}