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FP2 2025 June Q6

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 6

题目

Problem

6.

(a) Show that the substitution z=1y2z=\frac{1}{y^2} transforms the differential equation

2dydx+(cot⁡x)y+(tan⁡xsec⁡x)y3=00<x<π2(I)\begin{align*} 2\frac{\mathrm{d}y}{\mathrm{d}x} +(\cot x)y +(\tan x\sec x)y^3=0 \qquad 0<x<\frac{\pi}{2} \qquad \text{(I)} \end{align*}

into the differential equation

dzdx−(cot⁡x)z= tan⁡xsec⁡x0<x<π2(II)\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} -(\cot x)z =&\, \tan x\sec x \qquad 0<x<\frac{\pi}{2} \qquad \text{(II)} \end{align*}
(3)

(b) Hence determine the general solution of differential equation (I), giving your answer in the form y2=f(x)y^2=f(x)

(5)

Given that y2=433y^2=\frac{4\sqrt3}{3} when x=π6x=\frac{\pi}{6}

(c) determine the exact values of yy when x=π3x=\frac{\pi}{3}

(3)
题目中文翻译

(a) 证明代换 z=1y2z=\frac1{y^2} 能将微分方程

2dydx+(cot⁡x)y+(tan⁡xsec⁡x)y3= 0, 0<x<π2(I)\begin{align*} 2\frac{\mathrm{d}y}{\mathrm{d}x} +(\cot x)y+(\tan x\sec x)y^3=&\,0,\\[3mm] &\,0<x<\frac\pi2 \qquad\text{(I)} \end{align*}

变为微分方程

dzdx−(cot⁡x)z= tan⁡xsec⁡x, 0<x<π2(II).\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-(\cot x)z =&\,\tan x\sec x,\\[3mm] &\,0<x<\frac\pi2 \qquad\text{(II)}. \end{align*}
(3)

(b) 由此求微分方程 (I) 的通解，答案写成 y2=f(x)y^2=f(x) 的形式。

(5)

已知 x=π6x=\frac\pi6 时 y2=433y^2=\frac{4\sqrt3}{3}。

(c) 求 x=π3x=\frac\pi3 时 yy 的精确值。

(3)

解答

(a)

解法一

思路

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由 z=y−2z=y^{-2} 得 z′=−2y−3y′z'=-2y^{-3}y'。原方程中也有 2y′2y'，所以把 2y′2y' 改写成 −y3z′-y^3z' 后，整式除以 y3y^3 就能得到关于 zz 的线性方程。

答题过程

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Since

z=1y2=y−2,\begin{align*} z=\frac1{y^2}=y^{-2}, \end{align*}

we have

dzdx= −2y−3dydx\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} =&\, -2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Therefore

2dydx= −y3dzdx\begin{align*} 2\frac{\mathrm{d}y}{\mathrm{d}x} =&\, -y^3\frac{\mathrm{d}z}{\mathrm{d}x} \end{align*}

Substitute into (I):

−y3dzdx+(cot⁡x)y+(tan⁡xsec⁡x)y3= 0\begin{align*} -y^3\frac{\mathrm{d}z}{\mathrm{d}x} +(\cot x)y +(\tan x\sec x)y^3 =&\, 0 \end{align*}

Divide by −y3-y^3:

dzdx−(cot⁡x)y−2−tan⁡xsec⁡x= 0\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} -(\cot x)y^{-2} -\tan x\sec x =&\, 0 \end{align*}

Since y−2=zy^{-2}=z,

dzdx−(cot⁡x)z= tan⁡xsec⁡x\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} -(\cot x)z =&\, \tan x\sec x \end{align*}

This is (II).

解法二

思路

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原方程是典型的伯努利微分方程（Bernoulli Equation）。我们先将方程 (I) 全体除以 y3y^3，整理出项 y−3dydxy^{-3} \dfrac{\mathrm{d}y}{\mathrm{d}x} 和 y−2y^{-2}。再求代换式 z=y−2z = y^{-2} 的导数，并将相应的项直接代入，即可非常自然地化简为关于 zz 的一阶线性微分方程。

答题过程

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Divide the original differential equation (I) by y3y^3 (since y≠0y \ne 0):

2y−3dydx+(cot⁡x)y−2+tan⁡xsec⁡x= 0\begin{align*} 2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} + (\cot x)y^{-2} + \tan x\sec x =&\, 0 \end{align*}

Since z=1y2=y−2z = \dfrac{1}{y^2} = y^{-2}, differentiating with respect to xx using the Chain Rule gives:

dzdx= −2y−3dydx\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} =&\, -2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

which implies:

2y−3dydx= −dzdx\begin{align*} 2y^{-3}\frac{\mathrm{d}y}{\mathrm{d}x} =&\, -\frac{\mathrm{d}z}{\mathrm{d}x} \end{align*}

Substitute 2y−3dydx=−dzdx2y^{-3}\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{\mathrm{d}z}{\mathrm{d}x} and y−2=zy^{-2} = z into the divided equation:

−dzdx+(cot⁡x)z+tan⁡xsec⁡x= 0dzdx−(cot⁡x)z= tan⁡xsec⁡x\begin{align*} -\frac{\mathrm{d}z}{\mathrm{d}x} + (\cot x)z + \tan x\sec x =&\, 0\\[4mm] \frac{\mathrm{d}z}{\mathrm{d}x} - (\cot x)z =&\, \tan x\sec x \end{align*}

This is equation (II).

(b)

解法一

思路

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(II) 是一阶线性方程。积分因子为 e∫−cot⁡x dx=1sin⁡x\mathrm{e}^{\int-\cot x\,\mathrm{d}x}=\frac1{\sin x}，右边乘上积分因子后会化成 sec⁡2x\sec^2x。

答题过程

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The differential equation is

dzdx−(cot⁡x)z=tan⁡xsec⁡x\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x}-(\cot x)z=\tan x\sec x \end{align*}

The integrating factor is

I.F.= e∫−cot⁡x dx= e−ln⁡(sin⁡x)= 1sin⁡x\begin{align*} \mathrm{I.F.} =&\, \mathrm{e}^{\int-\cot x\,\mathrm{d}x}\\[3mm] =&\, \mathrm{e}^{-\ln(\sin x)}\\[3mm] =&\, \frac1{\sin x} \end{align*}

This is valid since 0<x<π20<x<\frac{\pi}{2}, so sin⁡x>0\sin x>0.

Multiply by 1sin⁡x\frac1{\sin x}:

ddx(zsin⁡x)= 1sin⁡xtan⁡xsec⁡x= 1sin⁡x⋅sin⁡xcos⁡x⋅1cos⁡x= sec⁡2x\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \bigg(\frac{z}{\sin x}\bigg) =&\, \frac1{\sin x}\tan x\sec x\\[3mm] =&\, \frac1{\sin x} \cdot \frac{\sin x}{\cos x} \cdot \frac1{\cos x}\\[3mm] =&\, \sec^2x \end{align*}

Integrate:

zsin⁡x= ∫sec⁡2x dx= tan⁡x+C\begin{align*} \frac{z}{\sin x} =&\, \int\sec^2x\,\mathrm{d}x\\[2mm] =&\, \tan x+C \end{align*}

Thus

z= sin⁡x(tan⁡x+C)\begin{align*} z =&\, \sin x(\tan x+C) \end{align*}

Since z=1y2z=\frac1{y^2},

1y2= sin⁡x(tan⁡x+C)\begin{align*} \frac1{y^2} =&\, \sin x(\tan x+C) \end{align*}

Therefore

y2=1sin⁡x(tan⁡x+C)\begin{align*} \boxed{ y^2=\frac{1}{\sin x(\tan x+C)} } \end{align*}

(c)

解法一

思路

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先用 x=π6x=\frac{\pi}{6} 和 y2=433y^2=\frac{4\sqrt3}{3} 求常数 CC，再代入 x=π3x=\frac{\pi}{3} 求 y2y^2。题目问 yy，所以正负平方根都要给出。

答题过程

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From part (b),

y2=1sin⁡x(tan⁡x+C)\begin{align*} y^2=\frac{1}{\sin x(\tan x+C)} \end{align*}

Use y2=433y^2=\frac{4\sqrt3}{3} when x=π6x=\frac{\pi}{6}:

433= 1sin⁡π6(tan⁡π6+C)433= 112(13+C)\begin{align*} \frac{4\sqrt3}{3} =&\, \frac1{ \sin\frac{\pi}{6} \bigg( \tan\frac{\pi}{6}+C \bigg) }\\[4mm] \frac{4\sqrt3}{3} =&\, \frac1{ \frac12\bigg(\frac1{\sqrt3}+C\bigg) } \end{align*}

So

12(13+C)= 34313+C= 323C= 123=36\begin{align*} \frac12\bigg(\frac1{\sqrt3}+C\bigg) =&\, \frac{3}{4\sqrt3}\\[3mm] \frac1{\sqrt3}+C =&\, \frac{3}{2\sqrt3}\\[3mm] C =&\, \frac{1}{2\sqrt3} =\frac{\sqrt3}{6} \end{align*}

At x=π3x=\frac{\pi}{3},

y2= 1sin⁡π3(tan⁡π3+36)= 132(3+36)= 132⋅736= 17/4= 47\begin{align*} y^2 =&\, \frac1{ \sin\frac{\pi}{3} \bigg( \tan\frac{\pi}{3}+\frac{\sqrt3}{6} \bigg) }\\[4mm] =&\, \frac1{ \frac{\sqrt3}{2} \bigg( \sqrt3+\frac{\sqrt3}{6} \bigg) }\\[4mm] =&\, \frac1{ \frac{\sqrt3}{2}\cdot\frac{7\sqrt3}{6} }\\[4mm] =&\, \frac1{7/4}\\[2mm] =&\, \frac47 \end{align*}

Therefore

y=±27\begin{align*} \boxed{ y=\pm\frac{2}{\sqrt7} } \end{align*}