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IAL 2022 Oct Q1

A Level / Edexcel / FP2

IAL 2022 Oct Paper · Question 1

题目

Problem

Given that

d2ydx2+3xdydx= 2cos⁡x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 3x\frac{\mathrm{d}y}{\mathrm{d}x} =\,& 2\cos x\\[2mm] \end{align*}

(a) Express d3ydx3\frac{\mathrm{d}^3y}{\mathrm{d}x^3} in terms of xx , dydx\frac{\mathrm{d}y}{\mathrm{d}x} and d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}

(3)

At x=0x = 0 , y=2y = 2 and dydx=5\frac{\mathrm{d}y}{\mathrm{d}x} = 5

(b) Determine the value of d3ydx3\frac{\mathrm{d}^3y}{\mathrm{d}x^3} at x=0x = 0

(1)

(c) Express yy as a series in ascending powers of xx , up to and including the term in x3x^3

(3)

解答

(a)

解法一

思路

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对原方程两边求导即可。左边有乘积 3xy′3xy'，所以要用乘积法则：

ddx(3xy′)=3y′+3xy′′.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}(3xy')=3y'+3xy''. \end{align*}

最后把 y′′′y''' 单独放在等号左边。

答题过程

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Given

d2ydx2+3xdydx=2cos⁡x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +3x\frac{\mathrm{d}y}{\mathrm{d}x} =2\cos x. \end{align*}

Differentiate both sides with respect to xx:

d3ydx3+3dydx+3xd2ydx2= −2sin⁡x.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} +3\frac{\mathrm{d}y}{\mathrm{d}x} +3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,-2\sin x. \end{align*}

Therefore

d3ydx3=−2sin⁡x−3dydx−3xd2ydx2.\begin{align*} \boxed{ \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =-2\sin x -3\frac{\mathrm{d}y}{\mathrm{d}x} -3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} }. \end{align*}

(b)

解法一

思路

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把 x=0x=0 和 y′=5y'=5 代入 (a) 的结果。因为含 y′′y'' 的项前面有 xx，所以这一项在 x=0x=0 时为 00。

答题过程

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At x=0x=0,

dydx=5.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=5. \end{align*}

Using part (a),

d3ydx3= −2sin⁡0−3(5)−3(0)d2ydx2= −15.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,-2\sin0-3(5)-3(0)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] =&\,-15. \end{align*}

Thus

d3ydx3∣x=0=−15.\begin{align*} \boxed{\left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=0}=-15}. \end{align*}

(c)

解法一

思路

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Taylor 展开到 x3x^3 需要 y(0)y(0)、y′(0)y'(0)、y′′(0)y''(0) 和 y′′′(0)y'''(0)。

题目给了 y(0)y(0) 和 y′(0)y'(0)，(b) 给了 y′′′(0)y'''(0)。还缺 y′′(0)y''(0)，可以直接把 x=0x=0 代入原微分方程求出。

答题过程

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At x=0x=0,

y=2,dydx=5.\begin{align*} y=2,\qquad \frac{\mathrm{d}y}{\mathrm{d}x}=5. \end{align*}

Use the original differential equation:

d2ydx2+3xdydx= 2cos⁡x.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +3x\frac{\mathrm{d}y}{\mathrm{d}x} =&\,2\cos x. \end{align*}

Substitute x=0x=0:

d2ydx2∣x=0+3(0)(5)= 2cos⁡0d2ydx2∣x=0= 2.\begin{align*} \left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} +3(0)(5)=&\,2\cos0\\[2mm] \left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} =&\,2. \end{align*}

Also, from part (b),

d3ydx3∣x=0=−15.\begin{align*} \left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=0}=-15. \end{align*}

Therefore

y= y(0)+xy′(0)+x22!y′′(0)+x33!y′′′(0)+⋯= 2+5x+x22(2)+x36(−15)+⋯= 2+5x+x2−52x3+⋯ .\begin{align*} y=&\,y(0)+xy'(0)+\frac{x^2}{2!}y''(0) +\frac{x^3}{3!}y'''(0)+\cdots\\[2mm] =&\,2+5x+\frac{x^2}{2}(2) +\frac{x^3}{6}(-15)+\cdots\\[2mm] =&\,2+5x+x^2-\frac{5}{2}x^3+\cdots. \end{align*}

So, up to and including the term in x3x^3,

y=2+5x+x2−52x3.\begin{align*} \boxed{y=2+5x+x^2-\frac{5}{2}x^3}. \end{align*}