If   \cos x \frac{dy}{dx} - y \sin x = 6x, \left ( 0,<x<\frac{\pi}{2} \right )

and  y \left ( \frac{\pi}{3} \right )=0 , then  y \left ( \frac{\pi}{6} \right )  is equal to :

  • Option 1)

    \frac{\pi^{2}}{2\sqrt{3}}

  • Option 2)

    -\frac{\pi^{2}}{2}

  • Option 3)

    -\frac{\pi^{2}}{2\sqrt{3}}

     

  • Option 4)

    -\frac{\pi^{2}}{4\sqrt{3}}

 

Answers (1)

\\ \cos x \frac{dy}{dx}-y\:\sin x= 6x\\\\\: y (\frac{\pi}{3}) = 0 \\\\\: \frac{dy}{dx} - y \tan x=\frac{6x}{\cos x}\\\\\: I.F.=e^{\int \tan x \:dx}= \cos x\\\\\:y\: \cos x = \int \frac{6x}{\cos x}.\cos x \:dx

\\y.\cos x =3x^{2}+c\\\\\: 0.\cos \left ( \pi/3 \right )=3\left ( \pi/3 \right )^{2}+c\\\\\:c=-\frac{\pi^{2}}{3}\\\\\:y.\cos x =3x^{2}-\frac{\pi^{2}}{3}\\\\\:y \left ( \pi/6 \right )=\frac{3\left ( \pi/6 \right )^{2}-\frac{\pi^{2}}{3}}{\cos \left ( \frac{\pi}{6} \right )}=\frac{3\frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}}{\sqrt{3}/2}

\frac{\frac{-3\pi^{2}}{12}}{\sqrt{3}/2}=\frac{-6\pi^{2}}{12\sqrt{3}}\\\\\:\:\: =\frac{-\pi^{2}}{2\sqrt{3}}


Option 1)

\frac{\pi^{2}}{2\sqrt{3}}

Option 2)

-\frac{\pi^{2}}{2}

Option 3)

-\frac{\pi^{2}}{2\sqrt{3}}

 

Option 4)

-\frac{\pi^{2}}{4\sqrt{3}}

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