If y = y(x) , x ∈ (0, (π)/(2)) be the solution curve of the...

Given, $\left(\sin^{2}2x\right)\dfrac{dy}{dx} + \left(8\sin^{2}2x + 2\sin 4x\right)y =$

$2{e}^{- 4x}\left(2\sin 2x + \cos 2x\right)$

Now rewriting the given differential equation we get,

$⇒ \dfrac{dy}{dx} + \left(8 + 4\cot 2x\right)y = \dfrac{2{e}^{- 4x}}{\sin^{2}2x}\left(2\sin x + \cos 2x\right)$

Which is a linear differential equation.

Now calculating $IF = {e}^{∫ \left(8 + 4\cot 2x\right)dx} = {e}^{8x + 2\ln \left(\sin 2x\right)}$

$= {e}^{8x} \cdot \sin^{2}2x$

Now solution of differential equation is given by,

$y \times IF = ∫ \dfrac{2{e}^{- 4x}\left(2\sin 2x + \cos 2x\right)}{\sin^{2}2x} \times IFdx$

$⇒ y\left({e}^{8x} \cdot \sin^{2}2x\right) = ∫ 2{e}^{4x}\left(2\sin 2x + \cos 2x\right)dx$

$⇒ y\left({e}^{8x} \cdot \sin^{2}2x\right) = {e}^{4x} \cdot \sin 2x + C$

Now given $y\left(\dfrac{\pi}{4}\right) = {e}^{- \pi} ⇒ C = 0$

So, the equation of curve is given by $y = \dfrac{{e}^{- 4x}}{\sin 2x}$

So, the value of $y\left(\dfrac{\pi}{6}\right) = \dfrac{{e}^{- 4\dfrac{\pi}{6}}}{\sin \left(2 \cdot \dfrac{\pi}{6}\right)} = \dfrac{2}{\sqrt{3}}{e}^{- \dfrac{2\pi}{3}}$