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IAL 2022 June Q5

A Level / Edexcel / FP2

IAL 2022 June Paper · Question 5

题目

Problem

Given that

yd2ydx2+2(dydx)2−2y= 0y>0\begin{align*} y \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2 \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 - 2y =\,& 0 \qquad y > 0\\[2mm] \end{align*}

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

(4)

Given that y=2y = 2 and dydx=1\frac{\mathrm{d}y}{\mathrm{d}x} = 1 at x=0x = 0

(b) determine a series solution for yy in ascending powers of xx , up to and including the term in x3x^3 , giving each coefficient in its simplest form.

(4)

解答

(a)

解法一

思路

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先把方程整理成 y′′y'' 的表达式，再对它求导。因为题目给出 y>0y>0，所以除以 yy 是合法的。

求导时要特别小心：(y′)2y\dfrac{(y')^2}{y} 是商或乘积形式，会同时产生 y′′y'' 和 (y′)3(y')^3。

答题过程

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The given equation is

yd2ydx2+2(dydx)2−2y=0.\begin{align*} y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 -2y=0. \end{align*}

Since y>0y>0, divide by yy:

d2ydx2+2y(dydx)2−2= 0.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{2}{y}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 -2=&\,0. \end{align*}

Hence

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

Differentiate both sides with respect to xx:

d3ydx3= −2ddx{y−1(dydx)2}= −2{−y−2dydx(dydx)2 +y−1⋅2dydxd2ydx2}= 2y2(dydx)3−4ydydxd2ydx2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,-2\frac{\mathrm{d}}{\mathrm{d}x} \left\{ y^{-1}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right\}\\[2mm] =&\,-2\left\{ -y^{-2}\frac{\mathrm{d}y}{\mathrm{d}x} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right.\\[2mm] &\,\hspace{2pt}\left. +y^{-1}\cdot 2\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \right\}\\[2mm] =&\,\frac{2}{y^2}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 -\frac{4}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Therefore

d3ydx3=−4ydydxd2ydx2+2y2(dydx)3.\begin{align*} \boxed{ \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =-\frac{4}{y}\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{2}{y^2}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3 }. \end{align*}

解法二

思路

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也可以直接对原方程求导。这样会先得到一个含 y′′′y''' 的方程，再把 y′′′y''' 解出来。

这条路线的优点是运算较短，但要注意乘积求导：

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

答题过程

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Differentiate

yd2ydx2+2(dydx)2−2y=0\begin{align*} y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 -2y=0 \end{align*}

with respect to xx:

dydxd2ydx2+yd3ydx3+4dydxd2ydx2 −2dydx=0.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} &+4\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] &\,\hspace{2pt}-2\frac{\mathrm{d}y}{\mathrm{d}x}=0. \end{align*}

So

yd3ydx3+5dydxd2ydx2−2dydx= 0.\begin{align*} y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} +5\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} =&\,0. \end{align*}

Rearranging,

yd3ydx3= 2dydx−5dydxd2ydx2= (2−5d2ydx2)dydx.\begin{align*} y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,2\frac{\mathrm{d}y}{\mathrm{d}x} -5\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[2mm] =&\,\left(2-5\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) \frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Since y>0y>0,

d3ydx3=1y(2−5d2ydx2)dydx.\begin{align*} \boxed{ \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =\frac{1}{y} \left(2-5\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) \frac{\mathrm{d}y}{\mathrm{d}x} }. \end{align*}

(b)

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)。

题目已经给了前两个。先用原微分方程求 y′′(0)y''(0)，再用 (a) 的结果求 y′′′(0)y'''(0)，最后代入 Taylor 公式。

答题过程

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At x=0x=0,

y=2,dydx=1.\begin{align*} y=2,\qquad \frac{\mathrm{d}y}{\mathrm{d}x}=1. \end{align*}

Use the original differential equation:

yd2ydx2+2(dydx)2−2y= 0.\begin{align*} y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 -2y=&\,0. \end{align*}

Substituting y=2y=2 and y′=1y'=1,

2d2ydx2+2(1)2−2(2)= 02d2ydx2−2= 0.\begin{align*} 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2(1)^2-2(2)=&\,0\\[2mm] 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}-2=&\,0. \end{align*}

So

d2ydx2∣x=0=1.\begin{align*} \left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=0}=1. \end{align*}

Using

d3ydx3=1y(2−5d2ydx2)dydx,\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =\frac{1}{y} \left(2-5\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) \frac{\mathrm{d}y}{\mathrm{d}x}, \end{align*}

we get

d3ydx3∣x=0= 12(2−5⋅1)(1)= −32.\begin{align*} \left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=0} =&\,\frac{1}{2}(2-5\cdot 1)(1)\\[2mm] =&\,-\frac{3}{2}. \end{align*}

Therefore

y= y(0)+xy′(0)+x22!y′′(0)+x33!y′′′(0)+⋯= 2+x+x22+x36(−32)+⋯= 2+x+12x2−14x3+⋯ .\begin{align*} y=&\,y(0)+xy'(0)+\frac{x^2}{2!}y''(0) +\frac{x^3}{3!}y'''(0)+\cdots\\[2mm] =&\,2+x+\frac{x^2}{2} +\frac{x^3}{6}\left(-\frac{3}{2}\right)+\cdots\\[2mm] =&\,2+x+\frac{1}{2}x^2-\frac{1}{4}x^3+\cdots. \end{align*}

Thus the series solution up to and including the term in x3x^3 is

y=2+x+12x2−14x3.\begin{align*} \boxed{y=2+x+\frac{1}{2}x^2-\frac{1}{4}x^3}. \end{align*}