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

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

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

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

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Differential Equations

cos x \frac{d y}{d x} - y sin x = 6 x, (0 < x < \frac{π}{2})

and

y (\frac{π}{3}) = 0

, then

y (\frac{π}{6})

is equal to:

Answer: (D)

Dividing through by

cos x

puts the equation in the linear form

\frac{d y}{d x} - y tan x = 6 x sec x

. The integrating factor is

e^{- \int t a n x d x} = e^{l n c o s x} = cos x

, and multiplying gives

\frac{d}{d x} (y cos x) = 6 x sec x cos x = 6 x

. Integrating,

y cos x = 3 x^{2} + C

. The condition

y (\frac{π}{3}) = 0

gives

0 = 3 \cdot \frac{π ^{2}}{9} + C

, so

C = - \frac{π ^{2}}{3}

. At

x = \frac{π}{6}

y cos \frac{π}{6} = 3 \cdot \frac{π ^{2}}{36} - \frac{π ^{2}}{3} = \frac{π ^{2}}{12} - \frac{π ^{2}}{3} = - \frac{π ^{2}}{4}

, so

y = - \frac{π ^{2}}{4} \cdot \frac{2}{3} = - \frac{π ^{2}}{2 3}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option