Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=0$, then $y\left(\frac{\pi}{6}\right)$ is equal to :

(B) $\frac{4}{9 \sqrt{3}} \pi^{2}$

(C) $\frac{-8}{9 \sqrt{3}} \pi^{2}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Differential Equations

Let

y = y (x)

be the solution of the differential equation

sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)

. If

y (\frac{π}{2}) = 0

, then

y (\frac{π}{6})

is equal to :

Answer: (D)

The left side is an exact derivative,

\frac{d}{d x} (y sin x) = 4 x

, so integrating gives

y sin x = 2 x^{2} + c

. The condition

y (\frac{π}{2}) = 0

gives

0 = 2 (\frac{π}{2})^{2} + c

, so

c = - \frac{π ^{2}}{2}

and

y sin x = 2 x^{2} - \frac{π ^{2}}{2}

. At

x = \frac{π}{6}

, where

sin x = \frac{1}{2}

\frac{y}{2} = 2 (\frac{π}{6})^{2} - \frac{π ^{2}}{2} = \frac{π ^{2}}{18} - \frac{π ^{2}}{2} = - \frac{4 π ^{2}}{9}

, so

y = - \frac{8 π ^{2}}{9}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option