If y = y(x),x in left(0, frac{pi}{2}right) is | vedclass

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(A) The given differential equation is $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 4 \sin 2x \cos 2x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$.Dividing by

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

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(A) The given differential equation is $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 4 \sin 2x \cos 2x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$.Dividing by $\sin^2 2x$,we get $\frac{dy}{dx} + (8 + 4 \cot 2x) y = \frac{2 e^{-4x} (2 \sin 2x + \cos 2x)}{\sin^2 2x}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = 8 + 4 \cot 2x$.The integrating factor $I.F. = e^{\int (8 + 4 \cot 2x) dx} = e^{8x + 2 \ln(\sin 2x)} = e^{8x} \sin^2 2x$.The solution is $y \cdot (e^{8x} \sin^2 2x) = \int (e^{8x} \sin^2 2x) \cdot \frac{2 e^{-4x} (2 \sin 2x + \cos 2x)}{\sin^2 2x} dx + C$.$y e^{8x} \sin^2 2x = \int 2 e^{4x} (2 \sin 2x + \cos 2x) dx + C$.Using $\int e^{ax} (a f(x) + f'(x)) dx = e^{ax} f(x) + C$,let $f(x) = \sin 2x$,then $f'(x) = 2 \cos 2x$. The integral becomes $\int e^{4x} (4 \sin 2x + 2 \cos 2x) dx = e^{4x} \sin 2x + C$.So,$y e^{8x} \sin^2 2x = e^{4x} \sin 2x + C$.Given $y(\frac{\pi}{4}) = e^{-\pi}$,we have $e^{-\pi} e^{2\pi} (1)^2 = e^{\pi} (1) + C \Rightarrow e^{\pi} = e^{\pi} + C \Rightarrow C = 0$.Thus,$y = \frac{e^{-4x}}{\sin 2x}$.For $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{e^{-4(\pi/6)}}{\sin(2 \cdot \pi/6)} = \frac{e^{-2\pi/3}}{\sin(\pi/3)} = \frac{e^{-2\pi/3}}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} e^{-2\pi/3}$.

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

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