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(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$. Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin

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

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(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$. Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = \frac{4x}{\sin x}$. The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. Substituting the values,$y \sin x = \int \frac{4x}{\sin x} \cdot \sin x dx + C = \int 4x dx + C = 2x^2 + C$. Thus,$y = \frac{2x^2 + C}{\sin x}$. Given $y(\frac{\pi}{2}) = 0$,we have $0 = \frac{2(\frac{\pi}{2})^2 + C}{\sin(\frac{\pi}{2})} = \frac{\pi^2}{2} + C$,so $C = -\frac{\pi^2}{2}$. The particular solution is $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$. Evaluating at $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}}{\sin(\frac{\pi}{6})} = \frac{\frac{2\pi^2}{36} - \frac{\pi^2}{2}}{1/2} = 2 \left( \frac{\pi^2}{18} - \frac{\pi^2}{2} \right) = 2 \left( \frac{\pi^2 - 9\pi^2}{18} \right) = 2 \left( \frac{-8\pi^2}{18} \right) = -\frac{8}{9} \pi^2$.

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

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