MathematicsMedium38×since 2002Q3629If y=y(x),x∈(0,π/2)y=y(x), x \in(0, \pi / 2)y=y(x),x∈(0,π/2) be the solution curve of the differential equation (sin22x)dydx+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x)\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)(sin22x)dxdy+(8sin22x+2sin4x)y=2e−4x(2sin2x+cos2x), with y(π/4)=e−πy(\pi / 4)=\mathrm{e}^{-\pi}y(π/4)=e−π, then y(π/6)y(\pi / 6)y(π/6) is equal to :A23e−2π/3\frac{2}{\sqrt{3}} e^{-2 \pi / 3}32e−2π/3B23e2π/3\frac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}32e2π/3C13e−2π/3\frac{1}{\sqrt{3}} e^{-2 \pi / 3}31e−2π/3D13e2π/3\frac{1}{\sqrt{3}} e^{2 \pi / 3}31e2π/3Check answerSkip