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FP2 2021 June Q4

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 4

题目

Problem

4. Given that

 yd2ydx2 −4(dydx)2+3y= 0.\begin{align*} &\,y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,-4\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2+3y\\[4mm] =&\,0. \end{align*}

(a) show that

 d3ydx3= 28y2(dydx)3 −24ydydx.\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac{28}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,-\frac{24}{y}\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}
(5)

Given also that y=8y = 8 and dydx=1\frac{\mathrm{d}y}{\mathrm{d}x} = 1 at x=0x = 0

(b) find a series solution for yy in ascending powers of xx, up to and including the term in x3x^3, simplifying the coefficients where possible.

(4)
题目中文翻译

4. 已知

 yd2ydx2 −4(dydx)2+3y= 0.\begin{align*} &\,y\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,-4\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2+3y\\[4mm] =&\,0. \end{align*}

(a) 证明

 d3ydx3= 28y2(dydx)3 −24ydydx.\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac{28}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,-\frac{24}{y}\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}
(5)

又已知，当 x=0x=0 时，y=8y=8 且 dydx=1\frac{\mathrm{d}y}{\mathrm{d}x}=1。

(b) 求 yy 按 xx 的升幂展开的级数解，展开至 x3x^3 项（含该项），并尽可能化简各项系数。

(4)

解答

(a)

解法一

思路

展开

先把原方程除以 yy，解出二阶导数，再使用乘积法则和链式法则求导。最后把二阶导数代回去，目标式就只含 yy 与一阶导数。

答题过程

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Rearranging the given equation and dividing by yy gives

d2ydx2= 4y(dydx)2−3.\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\frac4y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2-3. \end{align*}

Differentiate using the product and chain rules:

 d3ydx3= 4ddx[y−1(dydx)2]= 4[−y−2(dydx)3 +2y−1dydxd2ydx2]= −4y2(dydx)3 +8ydydxd2ydx2.\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,4\frac{\mathrm{d}}{\mathrm{d}x} \bigg[y^{-1}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\bigg]\\[4mm] =&\,4\bigg[-y^{-2} \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}+2y^{-1}\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg]\\[4mm] =&\,-\frac4{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}+\frac8y\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Substitute the expression for the second derivative:

 d3ydx3= −4y2(dydx)3 +8ydydx[ 4y(dydx)2−3]= −4y2(dydx)3 +32y2(dydx)3 −24ydydx= 28y2(dydx)3 −24ydydx.\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,-\frac4{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}+\frac8y\frac{\mathrm{d}y}{\mathrm{d}x}\bigg[\\[4mm] &\,\hspace{4pt}\frac4y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2-3\bigg]\\[4mm] =&\,-\frac4{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}+\frac{32}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{4pt}-\frac{24}y\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,\frac{28}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}-\frac{24}y\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

This is the required result.

解法二

思路

展开

也可以先直接微分原方程。对含 yy 与二阶导数的乘积使用乘积法则，对一阶导数的平方使用链式法则；合并后把三阶导数单独留下，再用原方程消去二阶导数。

答题过程

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Differentiate the original equation directly:

 dydxd2ydx2 +yd3ydx3 −8dydxd2ydx2 +3dydx= 0.\begin{align*} &\,\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+y\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] &\,\hspace{4pt}-8\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{6pt}+3\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,0. \end{align*}

Collecting the first and third terms and rearranging,

yd3ydx3= 7dydxd2ydx2 −3dydx.\begin{align*} y\frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,7\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}-3\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

The original equation gives

d2ydx2=4y(dydx)2−3.\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac4y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2-3.

Substituting this and dividing by yy,

 d3ydx3= 7ydydx[ 4y(dydx)2−3] −3ydydx= 28y2(dydx)3 −21ydydx −3ydydx= 28y2(dydx)3 −24ydydx.\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac7y\frac{\mathrm{d}y}{\mathrm{d}x}\bigg[\\[4mm] &\,\hspace{2pt}\frac4y\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2-3\bigg]\\[4mm] &\,\hspace{4pt}-\frac3y\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,\frac{28}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}-\frac{21}y\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{4pt}-\frac3y\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] =&\,\frac{28}{y^2}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^3\\[4mm] &\,\hspace{2pt}-\frac{24}y\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

This is the required result.

(b)

解法一

思路

展开

展开到 x3x^3 需要函数及前三阶导数在 x=0x=0 处的值。二阶导数由原方程求出，三阶导数用 (a) 的结果求出，再代入 Maclaurin 公式。

答题过程

展开

At x=0x=0, we have y=8y=8 and dydx=1\dfrac{\mathrm{d}y}{\mathrm{d}x}=1. The original equation gives

8d2ydx2∣x=0= 4(1)2−3(8)= −20.\begin{align*} 8\bigg.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=0} =&\,4(1)^2-3(8)\\[4mm] =&\,-20. \end{align*}

Hence

d2ydx2∣x=0=−52.\bigg.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=0} =-\frac52.

Using part (a),

 d3ydx3∣x=0= 2882(1)3−248(1)= 716−3= −4116.\begin{align*} &\,\bigg.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\bigg|_{x=0}\\[4mm] =&\,\frac{28}{8^2}(1)^3-\frac{24}8(1)\\[4mm] =&\,\frac7{16}-3\\[4mm] =&\,-\frac{41}{16}. \end{align*}

The Maclaurin expansion is

y= y(0)+y′(0)x +y′′(0)2!x2 +y′′′(0)3!x3.\begin{align*} y=&\,y(0)+y'(0)x\\[4mm] &\,\hspace{2pt}+\frac{y''(0)}{2!}x^2\\[4mm] &\,\hspace{4pt}+\frac{y'''(0)}{3!}x^3. \end{align*}

Substituting the derivative values,

y= 8+x−52⋅2!x2 −4116⋅3!x3= 8+x−54x2 −4196x3.\begin{align*} y=&\,8+x-\frac5{2\cdot2!}x^2\\[4mm] &\,\hspace{2pt}-\frac{41}{16\cdot3!}x^3\\[4mm] =&\,8+x-\frac54x^2\\[4mm] &\,\hspace{2pt}-\frac{41}{96}x^3. \end{align*}