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

Answer: (C)

$\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^{\ell n \cos x} = \cos x $
$ y.\cos x = \int6x \sec x \cos x dx $
$ 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{\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) . \left(- \frac{\pi^{2}}{3}\right) $
$ \Rightarrow y = - \frac{\pi^{2}}{2\sqrt{3}} $