If y=y(x), x \in(0, \pi \/ 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(\pi / 4)= {\e}^{-\pi} , then y(\pi / 6) is equal to: [28-Jul-2022-Shift-1]

Correct Answer is : 2√3e−2π∕3

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

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

\frac{1}\sqrt{3} e^{-2 \pi \/ 3}

\frac{1}\sqrt{3} e^{2 \pi \/ 3}

JEE Mains 28-July-2022 Shift 1 Solved Paper

Section: Mathematics

Question : 10 of 90

Marks: +1, -0

If y=y(x),x∈(0,π∕2) be the solution curve of the differential equation
(sin22x)‌

+(8sin22x+2‌sin‌4‌x)y−2e−4x(2‌sin‌2‌x+cos‌2‌x),
with y(

)=e−π , then y(

) is equal to:

[28-Jul-2022-Shift-1]

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