题目

Problem

(a) For all the values of $x$ where the identity is defined, prove that

\begin{align*} \cot 2x + \tan x \equiv \cosec 2x \end{align*}

(3)

(b) Show that the substitution $y^2 = w\sin 2x$ , where $w$ is a function of $x$ , transforms the differential equation

\begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 \tan x ={}& \sin x \qquad 0 < x < \frac{\pi}{2} \qquad \text{(I)} \end{align*}

into the differential equation

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} + 2w\cosec 2x ={}& \sec x \qquad 0 < x < \frac{\pi}{2} \qquad \text{(II)} \end{align*}

(4)

(c) By solving differential equation (II) , determine a general solution of differential equation (I) in the form $y^2 = f(x)$ , where $f(x)$ is a function in terms of $\cos x$

You may use without proof

\Bigg[ \int \cosec 2x \,\mathrm{d}x = \frac{1}{2}\ln |\tan x| + (\text{constant})\Bigg]

(6)

解答

(a)

解法一

思路

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把 $\cot2x$ 和 $\tan x$ 都写成 $\sin,\cos$ ，通分后分子会出现 $\cos2x\cos x+\sin2x\sin x$ ，可以用 $\cos(A-B)$ 化简。

答题过程

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\begin{align*} \cot2x+\tan x ={}& \frac{\cos2x}{\sin2x}+\frac{\sin x}{\cos x}\\[4mm] ={}& \frac{\cos2x\cos x+\sin2x\sin x} {\sin2x\cos x}\\[4mm] ={}& \frac{\cos(2x-x)}{\sin2x\cos x}\\[4mm] ={}& \frac{\cos x}{\sin2x\cos x}\\[4mm] ={}& \frac{1}{\sin2x}\\[4mm] ={}& \cosec2x \end{align*}

Therefore

\begin{align*} \cot2x+\tan x\equiv \cosec2x \end{align*}

(b)

解法一

思路

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先对 $y^2=w\sin2x$ 两边求导，得到 $2y\frac{\mathrm{d}y}{\mathrm{d}x}$ 的表达式。再把原方程中的 $y\frac{\mathrm{d}y}{\mathrm{d}x}$ 和 $y^2$ 都换成 $w$ ，最后用 (a) 的恒等式化简。

答题过程

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Given

\begin{align*} y^2=w\sin2x \end{align*}

differentiate both sides with respect to $x$ :

\begin{align*} 2y\frac{\mathrm{d}y}{\mathrm{d}x} ={}& \frac{\mathrm{d}w}{\mathrm{d}x}\sin2x+2w\cos2x \end{align*}

Hence

\begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x} ={}& \frac12\frac{\mathrm{d}w}{\mathrm{d}x}\sin2x+w\cos2x \end{align*}

Substitute this and $y^2=w\sin2x$ into (I):

\begin{align*} \frac12\frac{\mathrm{d}w}{\mathrm{d}x}\sin2x +w\cos2x +w\sin2x\tan x ={}& \sin x \end{align*}

Multiply by $2$ :

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x}\sin2x +2w\cos2x +2w\sin2x\tan x ={}& 2\sin x \end{align*}

Since $0<x<\frac{\pi}{2}$ , $\sin2x\ne0$ , so divide by $\sin2x$ :

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\frac{\cos2x}{\sin2x} +2w\tan x ={}& \frac{2\sin x}{\sin2x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} +2w(\cot2x+\tan x) ={}& \frac{2\sin x}{2\sin x\cos x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} +2w(\cot2x+\tan x) ={}& \sec x \end{align*}

Using the identity from part (a),

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\cosec2x ={}& \sec x \end{align*}

This is differential equation (II).

(c)

解法一

思路

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(II) 是一阶线性微分方程。积分因子是 $\mathrm{e}^{\int 2\cosec2x\,\mathrm{d}x}$ ，题目已经给出所需积分。最后要把 $w$ 换回 $y^2$ ，并整理成只含 $\cos x$ 的形式。

答题过程

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The differential equation is

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} +2w\cosec2x ={}& \sec x \end{align*}

The integrating factor is

\begin{align*} \mathrm{I.F.} ={}& \mathrm{e}^{\int 2\cosec2x\,\mathrm{d}x}\\[4mm] ={}& \mathrm{e}^{\ln|\tan x|}\\[4mm] ={}& \tan x \end{align*}

For $0<x<\frac{\pi}{2}$ , $\tan x>0$ , so the absolute value causes no difficulty.

Multiply (II) by $\tan x$ :

\begin{align*} \tan x\frac{\mathrm{d}w}{\mathrm{d}x} +2w\tan x\cosec2x ={}& \sec x\tan x\\[4mm] \frac{\mathrm{d}}{\mathrm{d}x}(w\tan x) ={}& \sec x\tan x \end{align*}

Integrate:

\begin{align*} w\tan x ={}& \int \sec x\tan x\,\mathrm{d}x\\[4mm] ={}& \sec x+C \end{align*}

Since $y^2=w\sin2x$ ,

\begin{align*} w={}&\frac{y^2}{\sin2x} \end{align*}

Substitute into $w\tan x=\sec x+C$ :

\begin{align*} \frac{y^2}{\sin2x}\tan x ={}& \sec x+C\\[4mm] \frac{y^2}{2\sin x\cos x}\cdot\frac{\sin x}{\cos x} ={}& \frac{1}{\cos x}+C\\[4mm] \frac{y^2}{2\cos^2x} ={}& \frac{1}{\cos x}+C \end{align*}

Multiply by $2\cos^2x$ :

\begin{align*} y^2 ={}& 2\cos x+2C\cos^2x \end{align*}

Writing the arbitrary constant as $A$ , the general solution is

\begin{align*} \boxed{y^2=2\cos x+A\cos^2x} \end{align*}

IAL 2024 Jan Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程