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(D) The given differential equation is $(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0$.Dividing by $dx$ and rearranging,we get $(x \cos x) \frac{d

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$2$

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(D) The given differential equation is $(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0$.Dividing by $dx$ and rearranging,we get $(x \cos x) \frac{dy}{dx} + (x \sin x + \cos x) y = 1$.Dividing by $x \cos x$,we have $\frac{dy}{dx} + (	an x + \frac{1}{x}) y = \frac{1}{x \cos x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 	an x + \frac{1}{x}$ and $Q(x) = \frac{1}{x \cos x}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int (	an x + \frac{1}{x}) dx} = e^{\ln(\sec x) + \ln x} = x \sec x$.The solution is $y(x \sec x) = \int (x \sec x) \frac{1}{x \cos x} dx + C = \int \sec^2 x dx + C = 	an x + C$.Given $\frac{\pi}{3} y(\frac{\pi}{3}) = \sqrt{3}$,we have $y(\frac{\pi}{3}) = \frac{3\sqrt{3}}{\pi}$.Substituting $x = \frac{\pi}{3}$ into $y(x \sec x) = 	an x + C$,we get $\frac{3\sqrt{3}}{\pi} \cdot \frac{\pi}{3} \cdot 2 = 	an(\frac{\pi}{3}) + C \implies 2\sqrt{3} = \sqrt{3} + C \implies C = \sqrt{3}$.Thus,$y = \frac{	an x + \sqrt{3}}{x \sec x} = \frac{\sin x + \sqrt{3} \cos x}{x}$.Calculating $y'$ and $y''$ at $x = \frac{\pi}{6}$: $y' = \frac{x(\cos x - \sqrt{3} \sin x) - (\sin x + \sqrt{3} \cos x)}{x^2}$.At $x = \frac{\pi}{6}$,$y' = \frac{\frac{\pi}{6}(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}) - (\frac{1}{2} + \frac{3}{2})}{(\pi/6)^2} = \frac{-2}{\pi^2/36} = -\frac{72}{\pi^2}$.Evaluating the expression $|\frac{\pi}{6} y''(\frac{\pi}{6}) + 2 y'(\frac{\pi}{6})|$,we find it equals $2$.

Let $y=y(x)$ be a solution of the differential equation $(x \cos x) dy + (xy \sin x + y \cos x - 1) dx = 0$,$0 < x < \frac{\pi}{2}$. If $\frac{\pi}{3} y(\frac{\pi}{3}) = \sqrt{3}$,then $|\frac{\pi}{6} y''(\frac{\pi}{6}) + 2 y'(\frac{\pi}{6})|$ is equal to $.........$.

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