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(C) The given differential equation is $x\frac{dy}{dx}-y=x^2 \cot x$. Dividing by $x^2$,we get $\frac{x \frac{dy}{dx}-y}{x^2} = \cot x$. This can be w

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$-\pi$

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(C) The given differential equation is $x\frac{dy}{dx}-y=x^2 \cot x$. Dividing by $x^2$,we get $\frac{x \frac{dy}{dx}-y}{x^2} = \cot x$. This can be written as $\frac{d}{dx}\left(\frac{y}{x}\right) = \cot x$. Integrating both sides with respect to $x$,we get $\frac{y}{x} = \int \cot x \, dx = \ln|\sin x| + C$. Given $y(\frac{\pi}{2}) = \frac{\pi}{2}$,we substitute $x = \frac{\pi}{2}$ and $y = \frac{\pi}{2}$: $\frac{\pi/2}{\pi/2} = \ln(\sin \frac{\pi}{2}) + C \implies 1 = \ln(1) + C \implies C = 1$. Thus,the solution is $y = x(\ln(\sin x) + 1)$. Now,calculate $y(\frac{\pi}{6}) = \frac{\pi}{6}(\ln(\sin \frac{\pi}{6}) + 1) = \frac{\pi}{6}(\ln(\frac{1}{2}) + 1) = \frac{\pi}{6}(1 - \ln 2)$. Calculate $y(\frac{\pi}{4}) = \frac{\pi}{4}(\ln(\sin \frac{\pi}{4}) + 1) = \frac{\pi}{4}(\ln(\frac{1}{\sqrt{2}}) + 1) = \frac{\pi}{4}(1 - \frac{1}{2}\ln 2)$. Finally,$6y(\frac{\pi}{6}) - 8y(\frac{\pi}{4}) = 6[\frac{\pi}{6}(1 - \ln 2)] - 8[\frac{\pi}{4}(1 - \frac{1}{2}\ln 2)]$. $= \pi(1 - \ln 2) - 2\pi(1 - \frac{1}{2}\ln 2) = \pi - \pi \ln 2 - 2\pi + \pi \ln 2 = -\pi$.

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-y=x^{2}\cot x, x\in(0,\pi)$. If $y(\frac{\pi}{2})=\frac{\pi}{2}$,then $6y(\frac{\pi}{6})-8y(\frac{\pi}{4})$ is equal to :

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