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

Question

If $y=y(x), x \in(0, \pi / 2)$ be the solution curve of the differential equation

$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$,

with $y(\pi / 4)=\mathrm{e}^{-\pi}$, then $y(\pi / 6)$ is equal to :

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

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

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

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

Solution

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