IAL 2025 Jan Q1 | 國際數學站

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} - 3y\tan x = \sec^2 x \end{align*}

(a) Show that an integrating factor for this differential equation is given by

\begin{align*} p(x) = \cos^3 x \end{align*}

(2)

Given that $y = 4$ when $x = \frac{\pi}{4}$

(b) determine the particular solution of the differential equation.

Give your answer in the form $y = f(x)$ .

(4)

解答

(a)

The integrating factor is given by

\begin{align*} I(x) = &\,\exp\left(\int -3 \tan x \,\mathrm{d}x\right)\\[4mm] = &\,\exp\left(3 \ln (\cos x)\right)\\[4mm] = &\,\cos^3 x \end{align*}

(b)

Multiplying the differential equation by the integrating factor:

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

Given that $y = 4$ when $x = \frac{\pi}{4}$ :

\begin{align*} 4 \cos^3 \left(\frac{\pi}{4}\right) = &\,\sin \left(\frac{\pi}{4}\right) + c\\[4mm] 4 \left(\frac{\sqrt{2}}{2}\right)^3 = &\,\frac{\sqrt{2}}{2} + c\\[4mm] 4 \left(\frac{2\sqrt{2}}{8}\right) = &\,\frac{\sqrt{2}}{2} + c\\[4mm] \sqrt{2} = &\,\frac{\sqrt{2}}{2} + c\\[4mm] c = &\,\frac{\sqrt{2}}{2} \end{align*}

So the particular solution is

\begin{align*} y \cos^3 x = &\,\sin x + \frac{\sqrt{2}}{2}\\[4mm] y = &\,\frac{\sin x}{\cos^3 x} + \frac{\sqrt{2}}{2\cos^3 x}\\[4mm] = &\,\tan x \sec^2 x + \frac{\sqrt{2}}{2}\sec^3 x \end{align*}