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FP2 2026 Jan A Q5

A Level / Edexcel / FP2

IAL 2026 Jan A 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} +5x\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 =&\, 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} =&\, \frac{1}{2-x^2} \Big[ 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Big( 1-5\frac{\mathrm{d}y}{\mathrm{d}x} \Big) -5\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \Big] \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

已知

(2−x2)d2ydx2+5x(dydx)2= 3y\begin{align*} (2-x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +5x\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 =&\, 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} =&\, \frac{1}{2-x^2} \Big[ 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Big( 1-5\frac{\mathrm{d}y}{\mathrm{d}x} \Big) -5\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \Big] \end{align*}
(5)

又已知当 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，直至并包括 x3x^3 项，并将各项系数化简。

(4)

解答

(a)

解法一

思路

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直接对原方程两边求导，然后把 d3ydx3\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} 单独整理出来。将含有 xy′′xy'' 的两项提出公因式，就能得到目标形式。

答题过程

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Differentiate

(2−x2)y′′+5x(y′)2=3y\begin{align*} (2-x^2)y''+5x(y')^2=3y \end{align*}

with respect to xx:

−2xy′′+(2−x2)y′′′+5(y′)2+10xy′y′′= 3y′\begin{align*} -2xy''+(2-x^2)y''' +5(y')^2+10xy'y'' =&\, 3y' \end{align*}

Rearrange to make y′′′y''' the subject:

(2−x2)y′′′= 2xy′′−5(y′)2−10xy′y′′+3y′= 2xy′′(1−5y′)−5(y′)2+3y′\begin{align*} (2-x^2)y''' =&\, 2xy''-5(y')^2-10xy'y''+3y'\\[3mm] =&\, 2xy''(1-5y')-5(y')^2+3y' \end{align*}

Therefore

y′′′= 12−x2[2xy′′(1−5y′)−5(y′)2+3y′]\begin{align*} y''' =&\, \frac{1}{2-x^2} \Big[ 2xy''(1-5y')-5(y')^2+3y' \Big] \end{align*}

That is,

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} \Big[ 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Big( 1-5\frac{\mathrm{d}y}{\mathrm{d}x} \Big) -5\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \Big] \end{align*}

解法二

思路

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先将原方程变形，将二阶导数 d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} 独立表达为关于其他项的商式，随后利用商的求导法则（Quotient Rule）求得三阶导数，再将分子中的原方程关系代回简化即可。

答题过程

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

y′′= 3y−5x(y′)22−x2\begin{align*} y'' =&\, \frac{3y-5x(y')^2}{2-x^2} \end{align*}

Differentiate with respect to xx using the Quotient Rule:

y′′′= ddx[3y−5x(y′)22−x2]= 3y′−5(y′)2−10xy′y′′2−x2 +2x[3y−5x(y′)2](2−x2)2\begin{align*} y''' =&\, \frac{\mathrm{d}}{\mathrm{d}x}\Big[ \frac{3y-5x(y')^2}{2-x^2} \Big]\\[4mm] =&\, \frac{3y'-5(y')^2-10xy'y''}{2-x^2}\\[3mm] &\,\hspace{2pt}+\frac{2x\big[3y-5x(y')^2\big]}{(2-x^2)^2} \end{align*}

Substitute 3y−5x(y′)2=(2−x2)y′′3y-5x(y')^2 = (2-x^2)y'' into the second fraction:

y′′′= 3y′−5(y′)2−10xy′y′′2−x2 +2x(2−x2)y′′(2−x2)2= 3y′−5(y′)2−10xy′y′′+2xy′′2−x2= 12−x2[2xy′′(1−5y′)−5(y′)2+3y′]\begin{align*} y''' =&\, \frac{3y'-5(y')^2-10xy'y''}{2-x^2}\\[2mm] &\,\hspace{2pt}+\frac{2x(2-x^2)y''}{(2-x^2)^2}\\[4mm] =&\, \frac{3y'-5(y')^2-10xy'y''+2xy''}{2-x^2}\\[4mm] =&\, \frac{1}{2-x^2}\Big[ 2xy''(1-5y') - 5(y')^2 + 3y' \Big] \end{align*}

Replacing yy, y′y', y′′y'', and y′′′y''' with standard notation:

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} \Big[ 2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} \Big( 1-5\frac{\mathrm{d}y}{\mathrm{d}x} \Big) -5\Big(\frac{\mathrm{d}y}{\mathrm{d}x}\Big)^2 +3\frac{\mathrm{d}y}{\mathrm{d}x} \Big] \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)、y′′′(0)y'''(0)。前两个已知，y′′(0)y''(0) 用原方程算，y′′′(0)y'''(0) 用 (a) 的式子算。

答题过程

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At x=0x=0,

y=3,y′=14\begin{align*} y=3, \qquad y'=\frac14 \end{align*}

Using

(2−x2)y′′+5x(y′)2=3y,\begin{align*} (2-x^2)y''+5x(y')^2=3y, \end{align*}

at x=0x=0:

2y′′(0)=9⟹y′′(0)=92\begin{align*} 2y''(0)=9 \quad\Longrightarrow\quad y''(0)=\frac92 \end{align*}

Using part (a),

y′′′(0)= 12[0−5(14)2+3(14)]= 12(−516+34)= 12⋅716= 732\begin{align*} y'''(0) =&\, \frac12 \Big[ 0 -5\Big(\frac14\Big)^2 +3\Big(\frac14\Big) \Big]\\[3mm] =&\, \frac12 \Big( -\frac{5}{16}+\frac34 \Big)\\[3mm] =&\, \frac12\cdot\frac{7}{16}\\[3mm] =&\, \frac{7}{32} \end{align*}

Therefore the Taylor series is

y= 3+14x+9/22!x2+7/323!x3= 3+14x+94x2+7192x3\begin{align*} y =&\, 3+\frac14x+\frac{9/2}{2!}x^2 +\frac{7/32}{3!}x^3\\[3mm] =&\, 3+\frac14x+\frac94x^2+\frac{7}{192}x^3 \end{align*}