If $y=y(x), x \in\left(0, \frac{\pi}{2}\right)$ be the solution curve of the differential equation \[\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=\] $2 \mathrm{e}^{-4 \mathrm{x}}(2 \sin 2 \mathrm{x}+\cos 2 \mathrm{x}), \quad$ with $\quad \mathrm{y}\left(\frac{\pi}{4}\right)=\mathrm{e}^{-\pi}$, then $\mathrm{y}\left(\frac{\pi}{6}\right)$ is equal to :

(A) $\frac{2}{\sqrt{3}} \mathrm{e}^{-2 \pi / 3}$

(B) $\frac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}$

(C) $\frac{1}{\sqrt{3}} e^{-2 \pi / 3}$

(D) $\frac{1}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}$

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Differential Equations

y = y (x), x \in (0, \frac{π}{2})

be the solution curve of the differential equation

(sin^{2} 2 x) \frac{d y}{d x} + (8 sin^{2} 2 x + 2 sin 4 x) y =

2 e^{- 4 x} (2 sin 2 x + cos 2 x),

with

y (\frac{π}{4}) = e^{- π}

, then

y (\frac{π}{6})

is equal to :

Answer: (A)

Rewrite the differential equation in standard linear form:

\frac{d y}{d x} + (8 + 4 cot 2 x) y = \frac{2 e ^{- 4 x}}{s i n ^{2} 2 x} (2 sin 2 x + cos 2 x)

. The integrating factor is

e^{\int (8 + 4 c o t 2 x) d x} = e^{8 x + 2 l n (s i n 2 x)} = e^{8 x} sin^{2} 2 x

. Multiplying through and integrating:

y \cdot e^{8 x} sin^{2} 2 x = \int 2 e^{4 x} (2 sin 2 x + cos 2 x) d x = e^{4 x} sin 2 x + C

. Using

y (\frac{π}{4}) = e^{- π}

e^{- π} \cdot e^{2 π} \cdot 1 = e^{π} + C

, giving

C = 0

. Thus

y = \frac{e ^{- 4 x}}{s i n 2 x}

. At

x = \frac{π}{6}

y (\frac{π}{6}) = \frac{e ^{- 2 π /3}}{s i n ( π /3 )} = \frac{e ^{- 2 π /3}}{3 /2} = \frac{2}{3} e^{- 2 π /3}

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option