Let y = y x be a solution of the differential equation ( x   cos x ) d y + ( x y   sin x + y &#160…

Given,

$\left(x \cos x\right)dy+\left(xy \sin x+y \cos x-1\right)dx=0$

$\Rightarrow \left(x \cos x\right)\frac{dy}{dx}+\left(yx \sin x+y \cos x\right)=1$

$\Rightarrow \frac{dy}{dx}+y\left(\frac{x \sin x+\cos x}{x \cos x}\right)=\frac{1}{x \cos x}$

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

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

$\Rightarrow IF=e^{\int \left(\tan x+\frac{1}{x}\right)dx}=e^{\ln \left|\sec x\right|+\ln x}=x·\sec x$

Hence, solution of linear differential equation is given by,

$\Rightarrow y·x \sec x=\int \frac{x \sec x}{x \cos x}dx$

$\Rightarrow xy \sec x=\tan x+C$

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

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

$\Rightarrow C=\sqrt{3}$

Hence, $xy \sec x=\tan x+\sqrt{3}$

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

Now differentiating both side we get,

$\Rightarrow x{y}^{\text{'}}\left(x\right)+y\left(x\right)=2\cos \left(x+\frac{\pi }{3}\right)$

Now again differentiating we get,

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

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

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