IAL 2024 Jan Q8 | 國際數學站

(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} \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} \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)

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解答

(a)

To prove the identity:

\begin{align*} \cot 2x + \tan x = &\,\frac{\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x} \end{align*}

Find a common denominator:

\begin{align*} \frac{\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x} = &\,\frac{\cos 2x \cos x + \sin 2x \sin x}{\sin 2x \cos x} \end{align*}

Using the trigonometric addition formula $\cos(A-B) = \cos A\cos B + \sin A\sin B$ on the numerator:

\begin{align*} = &\,\frac{\cos(2x - x)}{\sin 2x \cos x}\\[4mm] = &\,\frac{\cos x}{\sin 2x \cos x}\\[4mm] = &\,\frac{1}{\sin 2x}\\[4mm] = &\,\csc 2x \end{align*}

Thus, $\cot 2x + \tan x \equiv \csc 2x$ .

(b)

Given $y^2 = w\sin 2x$ . Differentiate both sides with respect to $x$ using the product rule:

\begin{align*} 2y\frac{\mathrm{d}y}{\mathrm{d}x} = &\,\frac{\mathrm{d}w}{\mathrm{d}x}\sin 2x + 2w\cos 2x \end{align*}

We are given differential equation (I):

\begin{align*} y\frac{\mathrm{d}y}{\mathrm{d}x} + y^2\tan x = &\,\sin x \end{align*}

Substitute $y\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\left( \frac{\mathrm{d}w}{\mathrm{d}x}\sin 2x + 2w\cos 2x \right)$ and $y^2 = w\sin 2x$ into (I):

\begin{align*} \frac{1}{2}\left( \frac{\mathrm{d}w}{\mathrm{d}x}\sin 2x + 2w\cos 2x \right) + (w\sin 2x)\tan x = &\,\sin x \end{align*}

Multiply the entire equation by 2:

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

Divide through by $\sin 2x$ :

\begin{align*} \frac{\mathrm{d}w}{\mathrm{d}x} + 2w\frac{\cos 2x}{\sin 2x} + 2w\tan x = &\,\frac{2\sin x}{2\sin x \cos x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} + 2w(\cot 2x + \tan x) = &\,\frac{1}{\cos x}\\[4mm] \frac{\mathrm{d}w}{\mathrm{d}x} + 2w(\cot 2x + \tan x) = &\,\sec x \end{align*}

Using the identity proven in part (a), $\cot 2x + \tan x = \csc 2x$ :

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

This is the required differential equation (II).

(c)

To solve (II), this is a first-order linear differential equation. Find the integrating factor (IF):

\begin{align*} \text{IF} = &\,\mathrm{e}^{\int 2\csc 2x \,\mathrm{d}x} \end{align*}

We are given $\int \csc 2x \,\mathrm{d}x = \frac{1}{2}\ln|\tan x|$ , so:

\begin{align*} \text{IF} = &\,\mathrm{e}^{2\left(\frac{1}{2}\ln|\tan x|\right)}\\[4mm] = &\,\mathrm{e}^{\ln(\tan x)}\\[4mm] = &\,\tan x \end{align*}

Multiply equation (II) by the IF:

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

Integrate both sides with respect to $x$ :

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

Now substitute $w$ back in terms of $y$ . We know $y^2 = w\sin 2x$ , so $w = \frac{y^2}{\sin 2x}$ :

\begin{align*} \left(\frac{y^2}{\sin 2x}\right) \tan x = &\,\sec x + c\\[4mm] \left(\frac{y^2}{2\sin x \cos x}\right) \left(\frac{\sin x}{\cos x}\right) = &\,\frac{1}{\cos x} + c\\[4mm] \frac{y^2}{2\cos^2 x} = &\,\frac{1 + c\cos x}{\cos x}\\[4mm] y^2 = &\,2\cos^2 x \left(\frac{1 + c\cos x}{\cos x}\right)\\[4mm] y^2 = &\,2\cos x (1 + c\cos x)\\[4mm] y^2 = &\,2\cos x + 2c\cos^2 x \end{align*}

Letting $A = 2c$ , the general solution in the required form is:

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