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FP2 2021 Jan Q5

A Level / Edexcel / FP2

IAL 2021 Jan Paper · Question 5

题目

Problem

5. Given that

 (2−x2)d2ydx2 +5x(dydx)2= 3y.\begin{align*} &\,(2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+5x\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] =&\,3y. \end{align*}

(a) show that

 d3ydx3= 12−x2[ 2xd2ydx2 ×(1−5dydx) −5(dydx)2 +3dydx].\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac1{2-x^2}\bigg[\\[4mm] &\,\hspace{2pt}2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{4pt}\times\bigg(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{6pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{8pt}+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}=\dfrac14 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)
题目中文翻译

5. 已知

 (2−x2)d2ydx2 +5x(dydx)2= 3y.\begin{align*} &\,(2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+5x\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] =&\,3y. \end{align*}

(a) 证明

 d3ydx3= 12−x2[ 2xd2ydx2 ×(1−5dydx) −5(dydx)2 +3dydx].\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac1{2-x^2}\bigg[\\[4mm] &\,\hspace{2pt}2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{4pt}\times\bigg(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{6pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{8pt}+3\frac{\mathrm{d}y}{\mathrm{d}x}\bigg]. \end{align*}
(5)

另已知在 x=0x=0 时，y=3y=3 且 dydx=14\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac14。

(b) 求 yy 按 xx 的升幂排列的级数解，写至并包括 x3x^3 项，各项系数须化简。

(4)

解答

(a)

解法一

思路

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直接对原方程求导。第一项使用乘积法则；第二项同时使用乘积法则和链式法则。把三阶导数单独留下，再合并其余项。

答题过程

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Differentiate the given equation with respect to xx:

 −2xd2ydx2 +(2−x2)d3ydx3 +5(dydx)2 +10xdydxd2ydx2= 3dydx.\begin{align*} &\,-2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}+(2-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] &\,\hspace{4pt}+5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{6pt}+10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] =&\,3\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Rearranging,

 (2−x2)d3ydx3= 2xd2ydx2 −10xdydxd2ydx2 −5(dydx)2 +3dydx.\begin{align*} &\,(2-x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}-10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{4pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{6pt}+3\frac{\mathrm{d}y}{\mathrm{d}x}. \end{align*}

Factor the first two terms and divide by 2−x22-x^2:

 d3ydx3= 12−x2[ 2xd2ydx2(1−5dydx) −5(dydx)2+3dydx].\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac{1}{2-x^2}\bigg[\\[4mm] &\,\hspace{2pt}2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \bigg(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{2pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^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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先把二阶导数写成一个商，再使用商法则。用 DD 和 NN 分别表示分母与分子，只是为了让长式子更易读；求导后利用原方程中的 N=Dd2ydx2N=D\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} 简化。

答题过程

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Let

D= 2−x2,N= 3y−5x(dydx)2.\begin{align*} D=&\,2-x^2,\\[4mm] N=&\,3y-5x\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2. \end{align*}

The given equation becomes

d2ydx2=ND.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{N}{D}.

Using the product and chain rules,

dDdx= −2x,dNdx= 3dydx −5(dydx)2 −10xdydxd2ydx2.\begin{align*} \frac{\mathrm{d}D}{\mathrm{d}x}=&\,-2x,\\[4mm] \frac{\mathrm{d}N}{\mathrm{d}x}=&\,3\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{4pt}-10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

By the quotient rule,

d3ydx3= DdNdx−NdDdxD2= 1DdNdx+2xND2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{D\frac{\mathrm{d}N}{\mathrm{d}x} -N\frac{\mathrm{d}D}{\mathrm{d}x}}{D^2}\\[4mm] =&\,\frac{1}{D}\frac{\mathrm{d}N}{\mathrm{d}x} +\frac{2xN}{D^2}. \end{align*}

Since N=Dd2ydx2N=D\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2},

2xND2= 2xDd2ydx2.\begin{align*} \frac{2xN}{D^2} =&\,\frac{2x}{D}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Substituting and collecting terms gives

 Dd3ydx3= 3dydx−5(dydx)2 −10xdydxd2ydx2 +2xd2ydx2.\begin{align*} &\,D\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,3\frac{\mathrm{d}y}{\mathrm{d}x} -5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{2pt}-10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{4pt}+2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Hence, with D=2−x2D=2-x^2,

 d3ydx3= 12−x2[ 2xd2ydx2(1−5dydx) −5(dydx)2+3dydx].\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac{1}{2-x^2}\bigg[\\[4mm] &\,\hspace{2pt}2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \bigg(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{2pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^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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也可以先拆成两个商，分别使用商法则。为看清消项过程，用 UU、VV 表示两个分子，DD 表示共同分母。原方程给出 U−V=Dd2ydx2U-V=D\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}，可用来合并分母为 D2D^2 的两项。

答题过程

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Let

D= 2−x2,U= 3y,V= 5x(dydx)2.\begin{align*} D=&\,2-x^2,\\[4mm] U=&\,3y,\\[4mm] V=&\,5x\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2. \end{align*}

Then

d2ydx2=UD−VD.\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =\frac{U}{D}-\frac{V}{D}.

The derivatives of the numerators are

dUdx= 3dydx,dVdx= 5(dydx)2 +10xdydxd2ydx2.\begin{align*} \frac{\mathrm{d}U}{\mathrm{d}x} =&\,3\frac{\mathrm{d}y}{\mathrm{d}x},\\[4mm] \frac{\mathrm{d}V}{\mathrm{d}x} =&\,5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{2pt}+10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Since dDdx=−2x\dfrac{\mathrm{d}D}{\mathrm{d}x}=-2x, the quotient rule on each fraction gives

ddx(UD)= DdUdx+2xUD2,ddx(VD)= DdVdx+2xVD2.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{U}{D}\bigg) =&\,\frac{D\frac{\mathrm{d}U}{\mathrm{d}x}+2xU}{D^2},\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{V}{D}\bigg) =&\,\frac{D\frac{\mathrm{d}V}{\mathrm{d}x}+2xV}{D^2}. \end{align*}

Subtracting these expressions,

d3ydx3= 1DdUdx+2xUD2 −1DdVdx−2xVD2.\begin{align*} \frac{\mathrm{d}^3y}{\mathrm{d}x^3} =&\,\frac{1}{D}\frac{\mathrm{d}U}{\mathrm{d}x} +\frac{2xU}{D^2}\\[4mm] &\,\hspace{2pt}-\frac{1}{D}\frac{\mathrm{d}V}{\mathrm{d}x} -\frac{2xV}{D^2}. \end{align*}

From the original equation,

U=V+Dd2ydx2.U=V+D\frac{\mathrm{d}^2y}{\mathrm{d}x^2}.

Thus

2xUD2= 2xVD2+2xDd2ydx2.\begin{align*} \frac{2xU}{D^2} =&\,\frac{2xV}{D^2} +\frac{2x}{D}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

The 2xVD2\dfrac{2xV}{D^2} terms cancel. Substituting the numerator derivatives gives

 Dd3ydx3= 3dydx+2xd2ydx2 −5(dydx)2 −10xdydxd2ydx2.\begin{align*} &\,D\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,3\frac{\mathrm{d}y}{\mathrm{d}x} +2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\\[4mm] &\,\hspace{2pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2\\[4mm] &\,\hspace{4pt}-10x\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2}. \end{align*}

Collecting the second-derivative terms and dividing by D=2−x2D=2-x^2,

 d3ydx3= 12−x2[ 2xd2ydx2(1−5dydx) −5(dydx)2+3dydx].\begin{align*} &\,\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\\[4mm] =&\,\frac{1}{2-x^2}\bigg[\\[4mm] &\,\hspace{2pt}2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \bigg(1-5\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\\[4mm] &\,\hspace{2pt}-5\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^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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展开到 x3x^3 需要前三阶导数在 x=0x=0 处的值。先由原方程求二阶导数，再用 (a) 求三阶导数，最后代入 Maclaurin 公式。

答题过程

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At x=0x=0, the original equation gives

2d2ydx2= 3(3)=9.\begin{align*} 2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,3(3)=9. \end{align*}

Hence

d2ydx2∣x=0=92.\bigg.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg|_{x=0} =\frac92.

Using part (a),

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

Therefore,

y= 3+14x+94x2 +7192x3.\begin{align*} y=&\,3+\frac14x+\frac94x^2\\[4mm] &\,\hspace{2pt}+\frac7{192}x^3. \end{align*}