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(C) Given the differential equation: $2 \cos x \frac{d y}{d x} = \sin 2x - 4y \sin x$. Dividing by $2 \cos x$,we get: $\frac{d y}{d x} = \frac{2 \sin

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(C) Given the differential equation: $2 \cos x \frac{d y}{d x} = \sin 2x - 4y \sin x$. Dividing by $2 \cos x$,we get: $\frac{d y}{d x} = \frac{2 \sin x \cos x}{2 \cos x} - \frac{4y \sin x}{2 \cos x} = \sin x - 2y 	an x$. Rearranging gives the linear differential equation: $\frac{d y}{d x} + 2y 	an x = \sin x$. The integrating factor $I.F. = e^{\int 2 	an x \, dx} = e^{2 \ln |\sec x|} = \sec^2 x$. Multiplying by $I.F.$,we have: $y \sec^2 x = \int \sin x \sec^2 x \, dx = \int 	an x \sec x \, dx = \sec x + C$. Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$: $0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + C \implies 0 = 2 + C \implies C = -2$. Thus,$y \sec^2 x = \sec x - 2$,which simplifies to $y = \cos x - 2 \cos^2 x$. Now,$y(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) - 2 \cos^2(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} - 1$. Next,$y^{\prime}(x) = -\sin x - 4 \cos x(-\sin x) = -\sin x + 4 \sin x \cos x = -\sin x + 2 \sin 2x$. $y^{\prime}(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) + 2 \sin(\frac{\pi}{2}) = -\frac{1}{\sqrt{2}} + 2$. Finally,$y^{\prime}(\frac{\pi}{4}) + y(\frac{\pi}{4}) = (-\frac{1}{\sqrt{2}} + 2) + (\frac{1}{\sqrt{2}} - 1) = 1$.

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

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