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(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.This can be written as $\frac{d}{dx}(y \sin x) = 4x$.Integrating both si

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$-\frac{8}{9} \pi^2$

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(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.This can be written as $\frac{d}{dx}(y \sin x) = 4x$.Integrating both sides with respect to $x$,we get $y \sin x = \int 4x \, dx = 2x^2 + C$.Given that $y(\frac{\pi}{2}) = 0$,we substitute $x = \frac{\pi}{2}$ and $y = 0$ into the equation:$0 \cdot \sin(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 + C \Rightarrow 0 = 2(\frac{\pi^2}{4}) + C \Rightarrow C = -\frac{\pi^2}{2}$.Thus,the solution is $y \sin x = 2x^2 - \frac{\pi^2}{2}$.Now,we find $y(\frac{\pi}{6})$ by substituting $x = \frac{\pi}{6}$:$y \sin(\frac{\pi}{6}) = 2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}$.$y(\frac{1}{2}) = 2(\frac{\pi^2}{36}) - \frac{\pi^2}{2} = \frac{\pi^2}{18} - \frac{\pi^2}{2} = \frac{\pi^2 - 9\pi^2}{18} = -\frac{8\pi^2}{18} = -\frac{4\pi^2}{9}$.Therefore,$y = 2 	imes (-\frac{4\pi^2}{9}) = -\frac{8\pi^2}{9}$.

Let $y = y(x)$ be the solution of the differential equation $\sin x \frac{dy}{dx} + y \cos x = 4x$,where $x \in (0, \pi)$. If $y(\frac{\pi}{2}) = 0$,then find the value of $y(\frac{\pi}{6})$.

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