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IAL 2020 Oct Q1

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 1

题目

Problem

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*}

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

(3)

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

(b) Determine the value of d3ydx3\dfrac{\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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对原方程两边关于 xx 求导。左边的 3xdydx3x\dfrac{\mathrm{d}y}{\mathrm{d}x} 要用乘积法则，所以会产生两项。

#### 答题过程
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Differentiate both sides:

ddx(d2ydx2+3xdydx)= ddx(2cos⁡x)d3ydx3+3dydx+3xd2ydx2= −2sin⁡x\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +3x\frac{\mathrm{d}y}{\mathrm{d}x} \right) =&\,\frac{\mathrm{d}}{\mathrm{d}x}(2\cos x)\\[4mm] \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*} \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 dydx=5\dfrac{\mathrm{d}y}{\mathrm{d}x}=5 代入 (a)。含有 xx 的那一项会直接变成 00。

答题过程

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At x=0x=0,

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}\\[4mm] =&\,-15 \end{align*}

(c)

解法一

思路

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Maclaurin 展开到 x3x^3 需要 y(0),y′(0),y′′(0),y′′′(0)y(0),y'(0),y''(0),y'''(0)。前两个题目已给，y′′′(0)y'''(0) 来自 (b)，还要从原微分方程求 y′′(0)y''(0)。

答题过程

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From the original differential equation,

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

At x=0x=0,

y′′(0)= 2cos⁡0−3(0)(5)= 2\begin{align*} y''(0) =&\,2\cos0-3(0)(5)\\[4mm] =&\,2 \end{align*}

Using the Maclaurin expansion,

y= y(0)+y′(0)x+y′′(0)2!x2+y′′′(0)3!x3+⋯= 2+5x+22x2+−156x3+⋯= 2+5x+x2−52x3+⋯\begin{align*} y =&\,y(0)+y'(0)x+\frac{y''(0)}{2!}x^2 +\frac{y'''(0)}{3!}x^3+\cdots\\[4mm] =&\,2+5x+\frac{2}{2}x^2+\frac{-15}{6}x^3+\cdots\\[4mm] =&\,2+5x+x^2-\frac52x^3+\cdots \end{align*}

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

y=2+5x+x2−52x3\begin{align*} y=2+5x+x^2-\frac52x^3 \end{align*}