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CIE 9231 2024 June Paper 21 Q6

A Level / CIE / FP2

CIE 9231 2024 June Paper 21 Paper · Question 6

题目

Problem

(a) Show that

(cosh⁡x+sinh⁡x)12=e12x.\left(\cosh x+\sinh x\right)^{\frac{1}{2}}=e^{\frac{1}{2}x}.
[2]

(b) Find the particular solution of the differential equation

d2ydx2+dydx+3y=5(cosh⁡x+sinh⁡x)12,\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}+3y=5\left(\cosh x+\sinh x\right)^{\frac{1}{2}},

given that, when x=0x=0, y=1y=1 and dydx=43\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4}{3}.

[10]
题目中文翻译

(a) 证明

(cosh⁡x+sinh⁡x)12=e12x\left(\cosh x+\sinh x\right)^{\frac{1}{2}}=e^{\frac{1}{2}x}
(2)

(b) 求微分方程

d2ydx2+dydx+3y=5(cosh⁡x+sinh⁡x)12\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\frac{\mathrm{d}y}{\mathrm{d}x}+3y=5\left(\cosh x+\sinh x\right)^{\frac{1}{2}}

的特解,已知当 x=0x=0 时,y=1y=1 且 dydx=43\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4}{3}。

(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

(a)

解法一

思路

展开

代入双曲正弦和余弦的指数定义。相加后,含 e−xe^{-x} 的项抵消。

答题过程

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The exponential definitions give

cosh⁡x+sinh⁡x=ex+e−x2+ex−e−x2=ex.\cosh x+\sinh x=\frac{e^x+e^{-x}}2+\frac{e^x-e^{-x}}2=e^x.

Taking the positive square root gives (cosh⁡x+sinh⁡x)1/2=ex/2(\cosh x+\sinh x)^{1/2}=e^{x/2}.

(b)

解法一

思路

展开

用 (a) 把右边化成指数。先由辅助方程求互补函数,再试 kex/2ke^{x/2} 求特解,最后使用两个初始条件。

答题过程

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The auxiliary equation m2+m+3=0m^2+m+3=0 has roots −12±112i-\frac12\pm\frac{\sqrt{11}}2\mathrm{i}. With ω=112\omega=\frac{\sqrt{11}}2, the complementary function is e−x/2(Acos⁡ωx+Bsin⁡ωx)e^{-x/2}(A\cos\omega x+B\sin\omega x). For the particular integral yp=kex/2y_p=ke^{x/2}, we have yp′=k2ex/2y_p'=\frac{k}{2}e^{x/2} and yp′′=k4ex/2y_p''=\frac{k}{4}e^{x/2}. Substituting gives

(14+12+3)k=5,k=43.\bigg(\frac14+\frac12+3\bigg)k=5, \qquad k=\frac43.

The general solution and its derivative are

y= e−x/2(Acos⁡ωx+Bsin⁡ωx)+43ex/2,y′= −12e−x/2(Acos⁡ωx+Bsin⁡ωx) +ωe−x/2(−Asin⁡ωx+Bcos⁡ωx) +23ex/2.\begin{align*} y=&\,e^{-x/2}(A\cos\omega x+B\sin\omega x)+\frac43e^{x/2},\\ y'=&\,-\frac12 e^{-x/2}(A\cos\omega x+B\sin\omega x)\\ &\,\hspace{2pt}+\omega e^{-x/2}(-A\sin\omega x+B\cos\omega x)\\ &\,\hspace{4pt}+\frac23e^{x/2}. \end{align*}

The initial conditions give

A+43=1,−A2+ωB+23=43.A+\frac43=1,\qquad -\frac A2+\omega B+\frac23=\frac43.

The first equation gives A=−13A=-\frac13. The second then gives ωB=12\omega B=\frac12, so B=1/11B=1/\sqrt{11}. Therefore

y= e−x/2(−13cos⁡11 x2 +111sin⁡11 x2)+43ex/2.\begin{align*} y=&\,e^{-x/2}\bigg(-\frac13\cos\frac{\sqrt{11}\,x}{2}\\ &\,\hspace{2pt}+\frac1{\sqrt{11}}\sin\frac{\sqrt{11}\,x}{2}\bigg) +\frac43e^{x/2}. \end{align*}