Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 1 / b / Solution

Recognize the left-hand side as a product rule derivative, and use $\frac{1}{2}\sin 2x=\sin x\cos x$:At $x=\pi /6$, both $y\sin x$ and $\frac{1}{2}{\sin }^{2}x$ equal $1/8$, so $C=0$. ThereforeDividing by $\sin x$ first obtains this solution on $(0,\pi )$, the interval containing the initial condition. The explicit expression is smooth on the entire real line and satisfies the original, undivided differential equation even where $\sin x=0$.