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FP2 2024 June Q2

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 2

题目

Problem

2.

xdydx−y3=4\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} - y^3 = 4 \end{align*}

(a) Show that

xd3ydx3= ay(dydx)2+(by2+c)d2ydx2\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, ay\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 + (by^2 + c)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

where aa , bb and cc are integers to be determined.

(4)

Given that y=1y = 1 at x=2x = 2

(b) determine the Taylor series expansion for yy in ascending powers of (x−2)(x - 2) , up to and including the term in (x−2)3(x - 2)^3 , giving each coefficient in simplest form.

(3)
题目中文翻译

已知

xdydx−y3=4.\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x}-y^3=4. \end{align*}

(a) 证明

xd3ydx3= ay(dydx)2 +(by2+c)d2ydx2,\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,ay\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[2mm] &\,\hspace{2pt}+(by^2+c)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}, \end{align*}

并求出整数 aa、bb、cc 的值。

(4)

已知 x=2x=2 时 y=1y=1。

(b) 求 yy 按 (x−2)(x-2) 的升幂排列的泰勒展开式，直到并包括 (x−2)3(x-2)^3 项，各项系数须写成最简形式。

(3)

解答

(a)

解法一

思路

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题目要三阶导，所以对原方程连续隐式求导两次。第一次求导会得到含 y′′y'' 的式子，第二次再求导时注意 3y2y′3y^2y' 要用乘积法则。

答题过程

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Starting from

xdydx−y3=4\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x}-y^3=4 \end{align*}

differentiate with respect to xx:

dydx+xd2ydx2−3y2dydx= 0\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} +x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} -3y^2\frac{\mathrm{d}y}{\mathrm{d}x} =&\,0 \end{align*}

Differentiate again:

d2ydx2+(d2ydx2+xd3ydx3)−(6y(dydx)2+3y2d2ydx2)= 02d2ydx2+xd3ydx3−6y(dydx)2−3y2d2ydx2= 0\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\bigg( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} \bigg) -\bigg( 6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 +3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \bigg) =&\,0\\[4mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} -6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 -3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,0 \end{align*}

Therefore

xd3ydx3= 6y(dydx)2+(3y2−2)d2ydx2\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, 6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 +(3y^2-2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

So

a=6,b=3,c=−2\begin{align*} \boxed{a=6,\qquad b=3,\qquad c=-2} \end{align*}

解法二

思路

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将原微分方程整理为一阶导数 dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} 的显式表达，即 y′=y3+4xy' = \dfrac{y^3+4}{x}。然后使用商的求导法则（Quotient Rule）进行连续求导：先求出二阶导数并代入一阶导数化简，再求三阶导数，即可求得所需的代数等式。

答题过程

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Rearrange the given differential equation to make y′y' the subject:

xdydx−y3= 4dydx= y3+4x\begin{align*} x\frac{\mathrm{d}y}{\mathrm{d}x} - y^3 =&\, 4\\[3mm] \frac{\mathrm{d}y}{\mathrm{d}x} =&\, \frac{y^3+4}{x} \end{align*}

Differentiate with respect to xx using the Quotient Rule:

d2ydx2= (3y2dydx)x−(y3+4)x2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\bigg(3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)x - (y^3+4)}{x^2} \end{align*}

Substitute y3+4=xdydxy^3+4 = x\dfrac{\mathrm{d}y}{\mathrm{d}x} into the numerator:

d2ydx2= 3xy2dydx−xdydxx2= dydx(3y2−1)x\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{3xy^2\frac{\mathrm{d}y}{\mathrm{d}x} - x\frac{\mathrm{d}y}{\mathrm{d}x}}{x^2}\\[4mm] =&\, \frac{\frac{\mathrm{d}y}{\mathrm{d}x}(3y^2-1)}{x} \end{align*}

which can be rewritten as:

xd2ydx2= dydx(3y2−1)\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\, \frac{\mathrm{d}y}{\mathrm{d}x}(3y^2-1) \end{align*}

Differentiate both sides with respect to xx using the Product Rule:

d2ydx2+xd3ydx3= d2ydx2(3y2−1)+dydx(6ydydx)= (3y2−1)d2ydx2+6y(dydx)2\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, \frac{\mathrm{d}^2y}{\mathrm{d}x^2}(3y^2-1) + \frac{\mathrm{d}y}{\mathrm{d}x}\bigg( 6y\frac{\mathrm{d}y}{\mathrm{d}x} \bigg)\\[4mm] =&\, (3y^2-1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 \end{align*}

Rearrange to make xd3ydx3x\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} the subject:

xd3ydx3= 6y(dydx)2+(3y2−1)d2ydx2−d2ydx2= 6y(dydx)2+(3y2−2)d2ydx2\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, 6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 + (3y^2-1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] =&\, 6y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 + (3y^2-2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

Comparing this with the given form:

xd3ydx3= ay(dydx)2+(by2+c)d2ydx2\begin{align*} x\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\, ay\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 + (by^2+c)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{align*}

we get:

a=6,b=3,c=−2\begin{align*} \boxed{a=6,\qquad b=3,\qquad c=-2} \end{align*}

(b)

解法一

思路

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Taylor series 关于 x=2x=2，需要 y(2)y(2)、y′(2)y'(2)、y′′(2)y''(2)、y′′′(2)y'''(2)。先用原方程求 y′(2)y'(2)，再用 (a) 过程中得到的一阶导关系求 y′′(2)y''(2)，最后用 (a) 的结果求 y′′′(2)y'''(2)。

答题过程

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At x=2x=2, y=1y=1. From the original equation,

2y′−13= 42y′= 5y′= 52\begin{align*} 2y'-1^3=&\,4\\[2mm] 2y'=&\,5\\[2mm] y'=&\,\frac52 \end{align*}

Use

y′+xy′′−3y2y′=0\begin{align*} y'+xy''-3y^2y'=0 \end{align*}

At x=2x=2, y=1y=1, y′=52y'=\frac52:

52+2y′′−3⋅12⋅52= 02y′′= 5y′′= 52\begin{align*} \frac52+2y''-3\cdot1^2\cdot\frac52 =&\,0\\[4mm] 2y''=&\,5\\[2mm] y''=&\,\frac52 \end{align*}

From part (a),

xy′′′= 6y(y′)2+(3y2−2)y′′\begin{align*} x y''' =&\, 6y(y')^2+(3y^2-2)y'' \end{align*}

Substitute x=2x=2, y=1y=1, y′=52y'=\frac52, y′′=52y''=\frac52:

2y′′′= 6(52)2+(3−2)52= 1504+52= 752+52= 40\begin{align*} 2y''' =&\, 6\bigg(\frac52\bigg)^2+(3-2)\frac52\\[4mm] =&\, \frac{150}{4}+\frac52\\[4mm] =&\, \frac{75}{2}+\frac52\\[4mm] =&\,40 \end{align*}

Thus

y′′′=20\begin{align*} y'''=20 \end{align*}

The Taylor series is

y= 1+52(x−2)+522!(x−2)2+203!(x−2)3+⋯= 1+52(x−2)+54(x−2)2+103(x−2)3+⋯\begin{align*} y =&\, 1+\frac52(x-2) +\frac{\frac52}{2!}(x-2)^2 +\frac{20}{3!}(x-2)^3+\cdots\\[4mm] =&\, \boxed{ 1+\frac52(x-2)+\frac54(x-2)^2+\frac{10}{3}(x-2)^3+\cdots } \end{align*}