JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Differential Equations

Dividing by $x\cos x$ gives the linear form $\frac{dy}{dx}+\left(\tan x+\frac{1}{x}\right)y=\frac{1}{x\cos x}$. The integrating factor is $e^{\int (\tan x+1/x)dx}=e^{-\ln \cos x+\ln x}=x\sec x$, so $\frac{d}{dx}\left(xy\sec x\right)=\frac{x\sec x}{x\cos x}={\sec }^{2}x$ and $xy\sec x=\tan x+C$. The condition $\frac{\pi }{3}y\left(\frac{\pi }{3}\right)=\sqrt{3}$ makes $xy\sec x$ at $x=\frac{\pi }{3}$ equal to $\sqrt{3}\cdot \sec \frac{\pi }{3}=2\sqrt{3}$, while $\tan \frac{\pi }{3}=\sqrt{3}$, so $C=\sqrt{3}$. Hence $xy=\sin x+\sqrt{3}\cos x$. Writing $g(x)=xy$, differentiation gives $g^{\prime \prime }=x{y}^{\prime \prime }+2y^{\prime }$, and $g^{\prime \prime }(x)=-\sin x-\sqrt{3}\cos x=-g(x)$. At $x=\frac{\pi }{6}$, $g\left(\frac{\pi }{6}\right)=\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}=2$, so $\frac{\pi }{6}{y}^{\prime \prime }\left(\frac{\pi }{6}\right)+2y^{\prime }\left(\frac{\pi }{6}\right)=-2$, whose absolute value is $2$.