IAL 2020 Oct Q6 | 國際數學站

Obtain the general solution of the equation

\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} + (x\cot x + 2)xy = 4\sin x \qquad 0 < x < \pi \end{align*}

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

(8)

解答

\begin{align*} x^2\frac{\mathrm{d}y}{\mathrm{d}x} + (x\cot x + 2)xy = 4\sin x \end{align*}

Divide through by $x^2$ :

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + \left(\cot x + \frac{2}{x}\right)y = &\,\frac{4\sin x}{x^2} \end{align*}

The integrating factor is

\begin{align*} &\,\mathrm{e}^{\int (\cot x + \frac{2}{x})\,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{\int \cot x\,\mathrm{d}x + \int \frac{2}{x}\,\mathrm{d}x}\\[4mm] = &\,\mathrm{e}^{\ln(\sin x) + 2\ln x}\\[4mm] = &\,x^2\sin x \end{align*}

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

Hence

\begin{align*} x^2\sin x\, y = &\,\int 4\sin^2 x\,\mathrm{d}x\\[4mm] = &\,\int 2(1 - \cos 2x)\,\mathrm{d}x\\[4mm] = &\,2x - \sin 2x + C \end{align*}

Therefore

\begin{align*} y = &\,\frac{2x - \sin 2x + C}{x^2\sin x} \end{align*}