if cos x


If $\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0

$y\left(\frac{\pi}{6}\right)$ is equal to:


  1. (1) $\frac{\pi^{2}}{2 \sqrt{3}}$

  2. (2) $-\frac{\pi^{2}}{2}$

  3. (3) $-\frac{\pi^{2}}{2 \sqrt{3}}$

  4. (4) $-\frac{\pi^{2}}{4 \sqrt{3}}$

Correct Option: , 3


$\cos x d y-(\sin x) y d x=6 x d x$

$\Rightarrow \int d(y \cos x)=\int 6 x d x \Rightarrow y \cos x=3 x^{2}+C \ldots(1)$

Given, $y\left(\frac{\pi}{3}\right)=0$

Putting $x=\frac{\pi}{3}$ and $y=0$ in eq. (1), we get

$(10) \times\left(\frac{1}{2}\right)=\frac{3 \pi^{2}}{9}+C \Rightarrow C \frac{-\pi^{2}}{3}$

So, from (1) $y \cos x=3 x^{2}-\frac{\pi^{2}}{3}$

Now, put $\quad-$ in the above equation,

$y \frac{\sqrt{3}}{2}=\frac{3 \pi^{2}}{36}-\frac{\pi^{2}}{3} \Rightarrow \frac{\sqrt{3} y}{2}=\frac{-3 \pi^{2}}{12} \Rightarrow y=\frac{-\pi^{2}}{2 \sqrt{3}}$

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