Question

If $\cos x\frac{dy}{dx}-y\sin x=6x,(0<x<\frac{\pi }{2})$ and $y\left(\frac{\pi }{3}\right)=0$ then $y\left(\frac{\pi }{6}\right)$ is equal to:

• A. $-\frac{{\pi }^2}{4\sqrt{3}}$
• B. $-\frac{{\pi }^2}{2}$
• C. $-\frac{{\pi }^2}{2\sqrt{3}}$
• D. $\frac{{\pi }^2}{2\sqrt{3}}$

Answer 1

The correct option is C $-\frac{{\pi }^2}{2\sqrt{3}}$

$\frac{dy}{dx}-y\tan x=6x\sec x$

$y\left(\frac{\pi }{3}\right)=0;y\left(\frac{\pi }{6}\right)=7$

$e^{\int pdx}=e^{-\int \tan xdx}=e^{\ln \cos x}=\cos x$

$y.\cos x=\int 6x\sec x\cos xdx$

$y.\cos x=\frac{6x^2}{2}+C$

$y=3x^2\sec x+C\sec x$

$0=3.\frac{{\pi }^2}{9}.\left(2\right)+C\left(2\right)$

$2C=\frac{-2{\pi }^2}{3}\Rightarrow C=-\frac{2{\pi }^2}{3}$

$y\left(\pi /6\right)=3.\frac{{\pi }^2}{36}.\left(\frac{2}{\sqrt{3}}\right)+\left(\frac{2}{\sqrt{3}}\right).(-\frac{{\pi }^2}{3})\Rightarrow y=-\frac{{\pi }^2}{2\sqrt{3}}$