==問題==
設$y = y(x)$是微分方程
$$\sin x \frac{dy}{dx} + y \cos x = 4x, \quad x \in (0, \pi)$$
的解。若$\displaystyle y \left( \frac{\pi}{2} \right) = 0$，則$\displaystyle y \left( \frac{\pi}{6} \right)=$？
(1) $\displaystyle \frac{4}{9 \sqrt{3}} \pi^2$
(2) $\displaystyle \frac{-8}{9 \sqrt{3}} \pi^2$
(3) $\displaystyle -\frac{8}{9} \pi^2$
(4) $\displaystyle -\frac{4}{9} \pi^2$
==答案==
(3)==解答==
將題目的微分方程式稍微改寫為$$\frac{dy}{dx} \sin x + y \cos x = 4x,$$
然後
$$\frac{dy}{dx} \sin x + y \frac{d \sin x}{dx} = 4x.$$
聯想乘積的求導公式
$$\frac{d}{dx} fg = f'g + fg'.$$
得到
$$\frac{d}{dx} (y \sin x) = 4x.$$
左右同時對$x$積分，
$$\int \frac{d}{dx} (y \sin x) \, {\rm d}x = \int 4x \, {\rm d}x.$$
於是
$$y \sin x = 2x^2 + C.$$
這裡$C$為待定常數。
將初始條件$\displaystyle y \left( \frac{\pi}{2} \right) = 0$代入上式，得
$$0 = \frac{\pi^2}{2} + C.$$
解出$C = -\frac{\pi^2}{2}$。所以
$$y(x) = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}.$$
因此所求
$$y \left( \frac{\pi}{6} \right) = \frac{2 \left( \frac{\pi}{6} \right)^2 - \frac{\pi^2}{2}}{1/2} = -\frac{8}{9}\pi^2.$$
(解答結束)
