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IAL 2021 Jan Q5

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 5

题目

Problem

Given that

(2−x2)d2ydx2+5x(dydx)2=3y\begin{align*} (2 - x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 3y \end{align*}

(a) show that

d3ydx3= 1(2−x2)(2xd2ydx2(1−5dydx)−5(dydx)2+3dydx)\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, \frac{1}{(2 - x^2)} \bigg( 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \left(1 - 5\frac{\mathrm{d}y}{\mathrm{d}x}\right) - 5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 3\frac{\mathrm{d}y}{\mathrm{d}x} \bigg) \end{align*}
(5)

Given also that y=3y = 3 and dydx=14\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{4} at x=0x = 0

(b) obtain a series solution for yy in ascending powers of xx with simplified coefficients, up to and including the term in x3x^3

(4)

解答

(a)

解法一

思路

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这题的关键是逐项对原方程求导，然后整理出 d3ydx3\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}。最容易漏的是对

5x(dydx)2\begin{align*} 5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \end{align*}

同时用 product rule 和 chain rule。这个方法最贴近原方程，不需要先重排。

答题过程

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Differentiate

(2−x2)d2ydx2+5x(dydx)2=3y\begin{align*} (2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 =3y \end{align*}

with respect to xx:

−2xd2ydx2+(2−x2)d3ydx3+5(dydx)2+10xdydxd2ydx2=3dydx.\begin{align*} -2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +(2-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} +5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +10x\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =3\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Rearranging,

(2−x2)d3ydx3= 2xd2ydx2−10xdydxd2ydx2−5(dydx)2+3dydx= 2xd2ydx2(1−5dydx)−5(dydx)2+3dydx.\begin{align*} (2-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -10x\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \left(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\right) -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Therefore,

d3ydx3= 12−x2(2xd2ydx2(1−5dydx)−5(dydx)2+3dydx).\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{1}{2-x^2}\bigg( 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \left(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\right) -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \bigg). \end{align*}

This is the required result.

解法二

思路

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另一种做法是先把原方程重排成一个商：

d2ydx2=3y−5x(dydx)22−x2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{3y-5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} {2-x^2}. \end{align*}

然后对整个商使用 quotient rule。这个方法的好处是直接从 y′′y'' 求 y′′′y'''，但分子较长，符号要特别小心。

答题过程

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From the given equation,

d2ydx2= 3y−5x(dydx)22−x2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{3y-5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} {2-x^2}. \end{align*}

Differentiate both sides with respect to xx:

d3ydx3= [3dydx−5(dydx)2](2−x2)(2−x2)2 +[−10xdydxd2ydx2](2−x2)(2−x2)2 +2x[3y−5x(dydx)2](2−x2)2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{ \left[ 3\frac{\mathrm{d}y}{\mathrm{d}x} -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right](2-x^2) }{(2-x^2)^2}\\[4mm] &\,+\frac{ \left[ -10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right](2-x^2) }{(2-x^2)^2}\\[4mm] &\,+\frac{ 2x\left[ 3y-5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] }{(2-x^2)^2}. \end{align*}

From the original equation,

3y−5x(dydx)2= (2−x2)d2ydx2.\begin{align*} 3y-5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 =&\,(2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Substitute this into the last term:

d3ydx3= 12−x2[3dydx−5(dydx)2 −10xdydxd2ydx2+2xd2ydx2].\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{1}{2-x^2} \Bigg[ 3\frac{\mathrm{d}y}{\mathrm{d}x} -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\\[4mm] &\,\hspace{22pt} -10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Bigg]. \end{align*}

Collecting the d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} terms gives

d3ydx3= 12−x2(2xd2ydx2(1−5dydx)−5(dydx)2+3dydx).\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{1}{2-x^2} \bigg( 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \left(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\right) -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \bigg). \end{align*}

This is the required result.

解法三

思路

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也可以先把 y′′y'' 拆成两个商：

d2ydx2=3y2−x2−5x(dydx)22−x2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{3y}{2-x^2} -\frac{5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{2-x^2}. \end{align*}

然后分别用 quotient rule。它和解法二本质接近，但每次求导的对象更小，适合担心大分子出错的学生。

答题过程

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Rearrange the original equation as

d2ydx2= 3y2−x2−5x(dydx)22−x2.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac{3y}{2-x^2} -\frac{5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{2-x^2}. \end{align*}

Differentiate term by term:

d3ydx3= 3dydx(2−x2)+6xy(2−x2)2 −[5(dydx)2+10xdydxd2ydx2](2−x2)(2−x2)2 −10x2(dydx)2(2−x2)2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{ 3\frac{\mathrm{d}y}{\mathrm{d}x}(2-x^2)+6xy }{(2-x^2)^2}\\[4mm] &\,-\frac{ \left[ 5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right](2-x^2) }{(2-x^2)^2}\\[4mm] &\,-\frac{ 10x^2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 }{(2-x^2)^2}. \end{align*}

From the original equation,

3y= (2−x2)d2ydx2+5x(dydx)2.\begin{align*} 3y =&\,(2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +5x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2. \end{align*}

So

6xy= 2x(3y)= 2x(2−x2)d2ydx2+10x2(dydx)2.\begin{align*} 6xy =&\,2x(3y)\\[4mm] =&\,2x(2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +10x^2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2. \end{align*}

Substitute this expression for 6xy6xy. The 10x2(dydx)210x^2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 terms cancel, leaving

d3ydx3= 12−x2[3dydx+2xd2ydx2 −5(dydx)2−10xdydxd2ydx2]= 12−x2(2xd2ydx2(1−5dydx)−5(dydx)2+3dydx).\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{1}{2-x^2} \Bigg[ 3\frac{\mathrm{d}y}{\mathrm{d}x} +2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{22pt} -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 -10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Bigg]\\[4mm] =&\,\frac{1}{2-x^2} \bigg( 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \left(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\right) -5\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \bigg). \end{align*}

This is the required result.

(b)

解法一

思路

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Taylor series 到 x3x^3 需要 y(0)y(0)、y′(0)y'(0)、y′′(0)y''(0)、y′′′(0)y'''(0)。其中 y′′(0)y''(0) 用原方程求，y′′′(0)y'''(0) 用 (a) 的结果求。

答题过程

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At x=0x=0,

2d2ydx2= 3y= 9.\begin{align*} 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,3y\\[4mm] =&\,9. \end{align*}

So

d2ydx2∣x=0=92.\begin{align*} \left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0} =\frac92. \end{align*}

Using the result from part (a),

d3ydx3∣x=0= 12(−5(14)2+3(14))= 12(−516+34)= 732.\begin{align*} \left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=0} =&\,\frac12\bigg( -5\left(\frac14\right)^2+3\left(\frac14\right) \bigg)\\[4mm] =&\,\frac12\left(-\frac5{16}+\frac34\right)\\[4mm] =&\,\frac7{32}. \end{align*}

Therefore,

y= y(0)+y′(0)x+y′′(0)2!x2+y′′′(0)3!x3= 3+14x+94x2+7192x3.\begin{align*} y =&\,y(0)+y'(0)x+\frac{y''(0)}{2!}x^2+\frac{y'''(0)}{3!}x^3\\[4mm] =&\,3+\frac14x+\frac94x^2+\frac7{192}x^3. \end{align*}