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FP2 2024 Jan Q6

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 6

题目

Problem

6.

The differential equation

d2xdt2+6dxdt+13x= 8e−3tt⩾0\begin{align*} \frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 6\frac{\mathrm{d}x}{\mathrm{d}t} + 13x =&\, 8\mathrm{e}^{-3t} \qquad t \geqslant 0 \end{align*}

describes the motion of a particle along the xx-axis.

(a) Determine the general solution of this differential equation.

(6)

Given that the motion of the particle satisfies x=12x = \frac{1}{2} and dxdt=12\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2} when t=0t = 0

(b) determine the particular solution for the motion of the particle.

(4)

On the graph of the particular solution found in part (b), the first turning point for t>0t > 0 occurs at x=ax = a.

(c) Determine, to 3 significant figures, the value of aa.

[Solutions relying entirely on calculator technology are not acceptable.]

(4)
题目中文翻译

微分方程

d2xdt2+6dxdt+13x= 8e−3t,t⩾0 \begin{align*} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} +6\frac{\mathrm{d}x}{\mathrm{d}t}+13x =&\,8\mathrm{e}^{-3t},\qquad t\geqslant0 \end{align*}

描述一个质点沿 xx 轴的运动。

(a) 求该微分方程的通解。

(6)

已知质点的运动满足：t=0t=0 时，x=12x=\frac12 且 dxdt=12\frac{\mathrm{d}x}{\mathrm{d}t}=\frac12。

(b) 求描述该质点运动的特解。

(4)

在 (b) 中所得特解的图像上，t>0t>0 时的第一个转折点满足 x=ax=a。

(c) 求 aa 的值，保留三位有效数字。

[完全依赖计算器技术的解答不予接受。]

(4)

解答

(a)

解法一

思路

展开

这是二阶常系数非齐次微分方程。先用辅助方程求 complementary function；右边是 8e−3t8\mathrm{e}^{-3t}，而 −3-3 不是辅助方程的根，所以 particular integral 可设为 λe−3t\lambda\mathrm{e}^{-3t}。

答题过程

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The auxiliary equation is

m2+6m+13=0\begin{align*} m^2+6m+13=0 \end{align*}

So

m= −6±36−522= −3±2i\begin{align*} m =&\, \frac{-6\pm\sqrt{36-52}}{2}\\[4mm] =&\, -3\pm2\mathrm{i} \end{align*}

Hence the complementary function is

xc=e−3t(Acos⁡2t+Bsin⁡2t)\begin{align*} x_c=\mathrm{e}^{-3t}(A\cos2t+B\sin2t) \end{align*}

For the particular integral, let

xp=λe−3t\begin{align*} x_p=\lambda\mathrm{e}^{-3t} \end{align*}

Then

xp′= −3λe−3txp′′= 9λe−3t\begin{align*} x_p'=&\,-3\lambda\mathrm{e}^{-3t}\\[2mm] x_p''=&\,9\lambda\mathrm{e}^{-3t} \end{align*}

Substitute into the differential equation:

9λe−3t+6(−3λe−3t)+13λe−3t= 8e−3t(9λ−18λ+13λ)e−3t= 8e−3t4λ= 8λ= 2\begin{align*} 9\lambda\mathrm{e}^{-3t} +6(-3\lambda\mathrm{e}^{-3t}) +13\lambda\mathrm{e}^{-3t} =&\, 8\mathrm{e}^{-3t}\\[4mm] (9\lambda-18\lambda+13\lambda)\mathrm{e}^{-3t} =&\, 8\mathrm{e}^{-3t}\\[4mm] 4\lambda=&\,8\\[2mm] \lambda=&\,2 \end{align*}

Therefore

x=e−3t(Acos⁡2t+Bsin⁡2t)+2e−3t\begin{align*} \boxed{ x=\mathrm{e}^{-3t}(A\cos2t+B\sin2t)+2\mathrm{e}^{-3t} } \end{align*}

(b)

解法一

思路

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先用 x(0)=12x(0)=\frac12 求 AA，再对通解求导，用 x′(0)=12x'(0)=\frac12 求 BB。求导时指数因子和括号都要处理。

答题过程

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Using x=12x=\frac12 at t=0t=0,

12= A+2A= −32\begin{align*} \frac12 =&\, A+2\\[2mm] A=&\,-\frac32 \end{align*}

Differentiate the general solution:

dxdt= e−3t(−2Asin⁡2t+2Bcos⁡2t) −3e−3t(Acos⁡2t+Bsin⁡2t)−6e−3t\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\, \mathrm{e}^{-3t}(-2A\sin2t+2B\cos2t)\\[2mm] &\,\hspace{2pt}-3\mathrm{e}^{-3t}(A\cos2t+B\sin2t) -6\mathrm{e}^{-3t} \end{align*}

Using dxdt=12\frac{\mathrm{d}x}{\mathrm{d}t}=\frac12 at t=0t=0,

12= 2B−3A−612= 2B−3(−32)−612= 2B−322B= 2B= 1\begin{align*} \frac12 =&\, 2B-3A-6\\[4mm] \frac12 =&\, 2B-3\bigg(-\frac32\bigg)-6\\[4mm] \frac12 =&\, 2B-\frac32\\[4mm] 2B=&\,2\\[2mm] B=&\,1 \end{align*}

The particular solution is

x=e−3t(−32cos⁡2t+sin⁡2t+2)\begin{align*} \boxed{ x=\mathrm{e}^{-3t} \bigg(-\frac32\cos2t+\sin2t+2\bigg) } \end{align*}

(c)

解法一

思路

展开

Turning point 需要 dxdt=0\frac{\mathrm{d}x}{\mathrm{d}t}=0。先把 (b) 的导数化简，会得到只含 cos⁡2t\cos2t 的方程。然后用这个 tt 代回 xx，但不能只给计算器黑箱答案。

答题过程

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From the derivative in part (b), with A=−32A=-\frac32 and B=1B=1,

dxdt= e−3t(3sin⁡2t+2cos⁡2t−3(−32cos⁡2t+sin⁡2t)−6)= e−3t(132cos⁡2t−6)\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\, \mathrm{e}^{-3t} \bigg( 3\sin2t+2\cos2t -3\bigg(-\frac32\cos2t+\sin2t\bigg)-6 \bigg)\\[4mm] =&\, \mathrm{e}^{-3t} \bigg( \frac{13}{2}\cos2t-6 \bigg) \end{align*}

At a turning point,

e−3t(132cos⁡2t−6)= 0\begin{align*} \mathrm{e}^{-3t} \bigg( \frac{13}{2}\cos2t-6 \bigg) =&\,0 \end{align*}

Since e−3t≠0\mathrm{e}^{-3t}\ne0,

132cos⁡2t−6= 0cos⁡2t= 1213\begin{align*} \frac{13}{2}\cos2t-6=&\,0\\[2mm] \cos2t=&\,\frac{12}{13} \end{align*}

For the first turning point with t>0t>0,

t= 12arccos⁡(1213)\begin{align*} t =&\, \frac12\arccos\bigg(\frac{12}{13}\bigg) \end{align*}

At this point, sin⁡2t=513\sin2t=\frac{5}{13}, so

x= e−3t(−32⋅1213+513+2)= e−3t(−1813+513+2613)= e−3t\begin{align*} x =&\, \mathrm{e}^{-3t} \bigg( -\frac32\cdot\frac{12}{13} +\frac{5}{13}+2 \bigg)\\[4mm] =&\, \mathrm{e}^{-3t} \bigg( -\frac{18}{13}+\frac{5}{13}+\frac{26}{13} \bigg)\\[4mm] =&\, \mathrm{e}^{-3t} \end{align*}

Therefore

a= e−32arccos⁡(1213)= 0.553116…\begin{align*} a =&\, \mathrm{e}^{-\frac32\arccos\bigg(\frac{12}{13}\bigg)}\\[4mm] =&\, 0.553116\ldots \end{align*}

Hence

a=0.553 to 3 significant figures\begin{align*} \boxed{a=0.553\text{ to 3 significant figures}} \end{align*}