Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2023 June Paper 21 Q6

A Level / CIE / FP2

CIE 9231 2023 June Paper 21 Paper · Question 6

题目

Problem

Find the particular solution of the differential equation

d2xdt2−12dxdt+36x=37sin⁡t,\frac{\mathrm{d}^2x}{\mathrm{d}t^2}-12\frac{\mathrm{d}x}{\mathrm{d}t}+36x=37\sin t,

given that, when t=0t=0, x=dxdt=0x=\dfrac{\mathrm{d}x}{\mathrm{d}t}=0.

(11)
题目中文翻译

求微分方程

d2xdt2−12dxdt+36x=37sin⁡t\frac{\mathrm{d}^2x}{\mathrm{d}t^2}-12\frac{\mathrm{d}x}{\mathrm{d}t}+36x=37\sin t

满足当 t=0t=0 时,x=dxdt=0x=\dfrac{\mathrm{d}x}{\mathrm{d}t}=0 的特解。

(11)
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

展开

辅助方程有重根 66,所以互补函数含 te6tte^{6t}。特解同时试正弦和余弦,最后用两个零初值。

答题过程

展开

The auxiliary equation (m−6)2=0(m-6)^2=0 gives the complementary function (A+Bt)e6t(A+Bt)e^{6t}. For the particular integral, take xp=psin⁡t+qcos⁡tx_{\mathrm p}=p\sin t+q\cos t. Then xp′=pcos⁡t−qsin⁡tx_{\mathrm p}'=p\cos t-q\sin t and xp′′=−psin⁡t−qcos⁡tx_{\mathrm p}''=-p\sin t-q\cos t. Substitution gives

(35p+12q)sin⁡t+(−12p+35q)cos⁡t=37sin⁡t.(35p+12q)\sin t+(-12p+35q)\cos t=37\sin t.

Matching coefficients gives

35p+12q=37,−12p+35q=0.35p+12q=37,\qquad -12p+35q=0.

The second equation gives q=12p/35q=12p/35. Substituting into the first gives 1369p/35=371369p/35=37, hence p=35/37p=35/37 and q=12/37q=12/37. Thus the general solution and its derivative are

x= (A+Bt)e6t+3537sin⁡t+1237cos⁡t,x′= [B+6(A+Bt)]e6t +3537cos⁡t−1237sin⁡t.\begin{align*} x=&\,(A+Bt)e^{6t}+\frac{35}{37}\sin t+\frac{12}{37}\cos t,\\ x'=&\,\big[B+6(A+Bt)\big]e^{6t}\\ &\,\hspace{2pt}+\frac{35}{37}\cos t-\frac{12}{37}\sin t. \end{align*}

The initial conditions give

A+1237=0,6A+B+3537=0.A+\frac{12}{37}=0,\qquad 6A+B+\frac{35}{37}=0.

Thus A=−12/37A=-12/37 and B=1B=1, so

x=(t−1237)e6t+3537sin⁡t+1237cos⁡t.x=\bigg(t-\frac{12}{37}\bigg)e^{6t} +\frac{35}{37}\sin t+\frac{12}{37}\cos t.