IAL 2025 June Q6 | 國際數學站

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

\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

\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 $y^2 = f(x)$

(5)

Given that $y^2 = \dfrac{4\sqrt{3}}{3}$ when $x = \dfrac{\pi}{6}$

(3)

解答

(a)

Given $z = \frac{1}{y^2} = y^{-2}$ :

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

So, $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y^3}{2} \frac{\mathrm{d}z}{\mathrm{d}x}$ .

Substitute this into the original differential equation:

\begin{align*} 2\left(-\frac{y^3}{2} \frac{\mathrm{d}z}{\mathrm{d}x}\right) + (\cot x)y + (\tan x \sec x)y^3 = &\,0\\[4mm] -y^3 \frac{\mathrm{d}z}{\mathrm{d}x} + (\cot x)y + (\tan x \sec x)y^3 = &\,0 \end{align*}

Divide the entire equation by $-y^3$ :

\begin{align*} \frac{\mathrm{d}z}{\mathrm{d}x} - (\cot x)y^{-2} - \tan x \sec x = &\,0\\[4mm] \frac{\mathrm{d}z}{\mathrm{d}x} - (\cot x)z = &\,\tan x \sec x \end{align*}

(b)

This is a linear differential equation of the form $\frac{\mathrm{d}z}{\mathrm{d}x} + P(x)z = Q(x)$ . The integrating factor (IF) is:

\begin{align*} \text{IF} = &\,\mathrm{e}^{\int -\cot x \,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{-\ln|\sin x|}\\[4mm] = &\,\frac{1}{\sin x} \end{align*}

Multiply both sides by the integrating factor:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left( z \cdot \frac{1}{\sin x} \right) = &\,\frac{1}{\sin x} \tan x \sec x\\[4mm] = &\,\frac{1}{\sin x} \frac{\sin x}{\cos x} \frac{1}{\cos x}\\[4mm] = &\,\sec^2 x \end{align*}

Integrate both sides:

\begin{align*} \frac{z}{\sin x} = &\,\int \sec^2 x \,\mathrm{d}x\\[4mm] \frac{z}{\sin x} = &\,\tan x + C\\[4mm] z = &\,\sin x (\tan x + C) \end{align*}

Substitute back $z = \frac{1}{y^2}$ :

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

(c)

Given $y^2 = \frac{4\sqrt{3}}{3}$ when $x = \frac{\pi}{6}$ :

\begin{align*} \frac{4\sqrt{3}}{3} = &\,\frac{1}{\sin\left(\frac{\pi}{6}\right) \left(\tan\left(\frac{\pi}{6}\right) + C\right)}\\[4mm] \frac{4\sqrt{3}}{3} = &\,\frac{1}{\frac{1}{2} \left(\frac{\sqrt{3}}{3} + C\right)}\\[4mm] \frac{1}{2} \left(\frac{\sqrt{3}}{3} + C\right) = &\,\frac{3}{4\sqrt{3}}\\[4mm] \frac{\sqrt{3}}{3} + C = &\,\frac{6}{4\sqrt{3}}\\[4mm] \frac{\sqrt{3}}{3} + C = &\,\frac{\sqrt{3}}{2}\\[4mm] C = &\,\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{6} \end{align*}

Now determine $y$ when $x = \frac{\pi}{3}$ :

\begin{align*} y^2 = &\,\frac{1}{\sin\left(\frac{\pi}{3}\right) \left(\tan\left(\frac{\pi}{3}\right) + \frac{\sqrt{3}}{6}\right)}\\[4mm] = &\,\frac{1}{\frac{\sqrt{3}}{2} \left(\sqrt{3} + \frac{\sqrt{3}}{6}\right)}\\[4mm] = &\,\frac{1}{\frac{\sqrt{3}}{2} \left(\frac{7\sqrt{3}}{6}\right)}\\[4mm] = &\,\frac{1}{\frac{21}{12}}\\[4mm] = &\,\frac{12}{21} = \frac{4}{7} \end{align*}

Hence, the exact values of $y$ are:

\begin{align*} y = \pm \frac{2}{\sqrt{7}} = \pm \frac{2\sqrt{7}}{7} \end{align*}