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P4 2022 Jan Q9

A Level / Edexcel / P4

IAL 2022 Jan Paper · Question 9

题目

Problem

(a) Find the derivative with respect to yy of

1(1+2ln⁡y)2\frac{1}{(1+2\ln y)^2}
(2)

(b) Hence find a general solution to the differential equation

 3cosec⁡(2x)dydx= y(1+2ln⁡y)3, y>0, −π2<x<π2.\begin{align*} &\,3\cosec(2x)\frac{\mathrm{d}y}{\mathrm{d}x}\\ =&\,y(1+2\ln y)^3,\\ &\,y>0,\\ &\,-\frac{\pi}{2}<x<\frac{\pi}{2}. \end{align*}
(4)

(c) Show that the particular solution of this differential equation for which y=1y=1 at x=π6x=\dfrac{\pi}{6} is given by

y=eAsec⁡x−12y=e^{A\sec x-\frac{1}{2}}

where AA is an irrational number to be found.

(5)
题目中文翻译

(a) 求

1(1+2ln⁡y)2\frac{1}{(1+2\ln y)^2}

关于 yy 的导数。（2 分）

(b) 进而求微分方程

 3cosec⁡(2x)dydx= y(1+2ln⁡y)3, y>0, −π2<x<π2.\begin{align*} &\,3\cosec(2x)\frac{\mathrm{d}y}{\mathrm{d}x}\\ =&\,y(1+2\ln y)^3,\\ &\,y>0,\\ &\,-\frac{\pi}{2}<x<\frac{\pi}{2}. \end{align*}

的通解。（4 分）

(c) 证明满足 x=π6x=\dfrac{\pi}{6} 时 y=1y=1 的该微分方程特解为

y=eAsec⁡x−12y=e^{A\sec x-\frac{1}{2}}

其中 AA 是一个待求的无理数。（5 分）

解答

(a)

解法一

思路

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把原式写成 (1+2ln⁡y)−2(1+2\ln y)^{-2}，再使用链式法则。外层幂函数先求导，内层 1+2ln⁡y1+2\ln y 的导数为 2y\frac2y。

答题过程

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Using the chain rule,

 ddy(1+2ln⁡y)−2= −2(1+2ln⁡y)−3×(2y)= −4y(1+2ln⁡y)3.\begin{align*} &\,\frac{\mathrm{d}}{\mathrm{d}y} \big(1+2\ln y\big)^{-2}\\ =&\,-2\big(1+2\ln y\big)^{-3} \\ &\quad\times\bigg(\frac2y\bigg)\\ =&\,\boxed{ -\frac{4}{y(1+2\ln y)^3} }. \end{align*}

(b)

解法一

思路

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先把变量分离。左边的积分可以直接利用 (a)：所需被积函数恰好是 (a) 中导数的 −14-\frac14；右边把 cosec⁡(2x)\cosec(2x) 移到分母后化为 sin⁡(2x)\sin(2x)。也可以用 sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x 积分，得到含 sin⁡2x\sin^2x 的等价通解，两式只差积分常数的写法。

答题过程

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Separating the variables gives

 1y(1+2ln⁡y)3 dy= 13sin⁡(2x) dx.\begin{align*} &\,\frac{1}{y(1+2\ln y)^3}\,\mathrm{d}y\\ =&\,\frac13\sin(2x)\,\mathrm{d}x. \end{align*}

From part (a),

 ∫1y(1+2ln⁡y)3 dy= −14(1+2ln⁡y)2.\begin{align*} &\,\int\frac{1}{y(1+2\ln y)^3}\,\mathrm{d}y\\ =&\,-\frac{1}{4(1+2\ln y)^2}. \end{align*}

Also,

 13∫sin⁡(2x) dx= −16cos⁡(2x).\begin{align*} &\,\frac13\int\sin(2x)\,\mathrm{d}x\\ =&\,-\frac16\cos(2x). \end{align*}

Therefore, a general solution is

 −34(1+2ln⁡y)2= −12cos⁡(2x)+C.\boxed{ \begin{aligned} &\,-\frac{3}{4(1+2\ln y)^2}\\ =&\,-\frac12\cos(2x)+C \end{aligned} }.

Alternatively, using sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x gives

13∫sin⁡(2x) dx=13sin⁡2x.\frac13\int\sin(2x)\,\mathrm{d}x =\frac13\sin^2x.

Hence the same general solution can be written as

 −34(1+2ln⁡y)2= sin⁡2x+D.\begin{align*} &\,-\frac{3}{4(1+2\ln y)^2}\\ =&\,\sin^2x+D. \end{align*}

Since −12cos⁡(2x)=sin⁡2x−12-\frac12\cos(2x)=\sin^2x-\frac12, the constants satisfy D=C−12D=C-\frac12.

(c)

解法一

思路

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承接 (b)，先代入初值求积分常数。随后使用 1+cos⁡(2x)=2cos⁡2x1+\cos(2x)=2\cos^2x，把关于 1+2ln⁡y1+2\ln y 的平方关系化简。开平方时结合给定区间及初值选择正号，最后解出 yy。

答题过程

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Using y=1y=1 when x=π6x=\frac{\pi}{6} in the general solution from part (b),

 −34(1+2ln⁡1)2= −12cos⁡(π3)+C−34= −14+C.\begin{align*} &\,-\frac{3}{4(1+2\ln1)^2}\\ =&\,-\frac12\cos\bigg(\frac{\pi}{3}\bigg)+C\\ -\frac34=&\,-\frac14+C. \end{align*}

Hence

C=−12.C=-\frac12.

The particular solution therefore satisfies

 −34(1+2ln⁡y)2= −12cos⁡(2x)−12.\begin{align*} &\,-\frac{3}{4(1+2\ln y)^2}\\ =&\,-\frac12\cos(2x)-\frac12. \end{align*}

Thus

 34(1+2ln⁡y)2= 12(1+cos⁡(2x))= cos⁡2x.\begin{align*} &\,\frac{3}{4(1+2\ln y)^2}\\ =&\,\frac12\big(1+\cos(2x)\big)\\ =&\,\cos^2x. \end{align*}

Since −π2<x<π2-\frac{\pi}{2}<x<\frac{\pi}{2}, we have sec⁡x>0\sec x>0. The initial condition gives 1+2ln⁡y=1>01+2\ln y=1>0, so the positive square root is required:

1+2ln⁡y=32sec⁡x.1+2\ln y=\frac{\sqrt3}{2}\sec x.

Therefore,

ln⁡y= 34sec⁡x−12,y= e34sec⁡x−12.\begin{align*} \ln y =&\,\frac{\sqrt3}{4}\sec x-\frac12,\\ y =&\,\boxed{ e^{\frac{\sqrt3}{4}\sec x-\frac12} }. \end{align*}

Hence the required irrational number is

A=34.\boxed{A=\frac{\sqrt3}{4}}.