Let y = y(x) be a solution of the differential equation ( xcosx )...

Given,

$\left(xcosx\right)dy + \left(xysinx + ycosx - 1\right)dx = 0$

$⇒ \left(xcosx\right)\dfrac{dy}{dx} + \left(yxsinx + ycosx\right) = 1$

$⇒ \dfrac{dy}{dx} + y\left(\dfrac{xsinx + \cos x}{xcosx}\right) = \dfrac{1}{xcosx}$

Which is a linear differential equation of form $\dfrac{dy}{dx} + P\left(x\right)y = Q\left(x\right)$

Now we know that integrating factor will be ${e}^{∫ Pdx}$

$⇒ IF = {e}^{∫ \left(\tan x + \dfrac{1}{x}\right)dx} = {e}^{\ln \left|\sec x\right| + \ln x} = x \cdot \sec x$

Hence, solution of linear differential equation is given by,

$⇒ y \cdot xsecx = ∫ \dfrac{xsecx}{xcosx}dx$

$⇒ xysecx = \tan x + C$

Now using $\dfrac{\pi}{3}y\left(\dfrac{\pi}{3}\right) = \sqrt{3}$ we get,

$⇒ \dfrac{\pi}{3}y\left(\dfrac{\pi}{3}\right)sec\dfrac{\pi}{3} = tan\dfrac{\pi}{3} + C$

$⇒ C = \sqrt{3}$

Hence, $xysecx = \tan x + \sqrt{3}$

$⇒ xy\left(x\right) = 2\sin \left(x + \dfrac{\pi}{3}\right)$

Now differentiating both side we get,

$⇒ x{y}^{'}\left(x\right) + y\left(x\right) = 2\cos \left(x + \dfrac{\pi}{3}\right)$

Now again differentiating we get,

$⇒ x{y}^{"} ( x ) + 2{y}^{'} ( x ) = - 2\sin \left(x + \dfrac{\pi}{3}\right)$

Thus $\dfrac{\pi}{6}{y}^{"}\left(\dfrac{\pi}{6}\right) + 2{y}^{'}\left(\dfrac{\pi}{6}\right) = - 2$

Hence \left|\dfrac{\pi}{6}{y}^{"}\left(\dfrac{\pi}{6}\right) + 2{y}^{'}\left(\dfrac{\pi}{6}\right)\right| = 2