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

We are given $$\sin^2(2x)\frac{dy}{dx} + (8\sin^2(2x) + 2\sin(4x))y = 2e^{-4x}(2\sin(2x) + \cos(2x)).$$

Dividing both sides by $$\sin^2(2x)$$ yields

$$\frac{dy}{dx} + \Bigl(8 + \frac{2\sin(4x)}{\sin^2(2x)}\Bigr)y = \frac{2e^{-4x}(2\sin(2x) + \cos(2x))}{\sin^2(2x)}.$$

Since $$\sin(4x) = 2\sin(2x)\cos(2x),$$ we have $$\frac{2\sin(4x)}{\sin^2(2x)} = \frac{4\cos(2x)}{\sin(2x)} = 4\cot(2x),$$ and hence the equation becomes

$$\frac{dy}{dx} + (8 + 4\cot(2x))y = \frac{2e^{-4x}(2\sin(2x) + \cos(2x))}{\sin^2(2x)}.$$

The integrating factor is

$$\text{IF} = e^{\int (8 + 4\cot(2x))\, dx} = e^{8x + 2\ln|\sin(2x)|} = e^{8x}\sin^2(2x).$$

Multiplying the differential equation by this integrating factor gives

$$\frac{d}{dx}\bigl[y \,e^{8x}\sin^2(2x)\bigr] = 2e^{4x}(2\sin(2x) + \cos(2x)).$$

Noting that

$$\frac{d}{dx}[e^{4x}\sin(2x)] = 4e^{4x}\sin(2x) + 2e^{4x}\cos(2x) = 2e^{4x}(2\sin(2x) + \cos(2x)),$$

we integrate to obtain

$$y \,e^{8x}\sin^2(2x) = e^{4x}\sin(2x) + C.$$

Applying the initial condition $$y\bigl(\tfrac{\pi}{4}\bigr) = e^{-\pi}$$ and noting that $$\sin\bigl(\tfrac{\pi}{2}\bigr) = 1$$ leads to

$$e^{-\pi}\cdot e^{2\pi}\cdot 1 = e^{\pi} + C\quad\Rightarrow\quad C = 0.$$

Therefore

$$y\,e^{8x}\sin^2(2x) = e^{4x}\sin(2x),$$

which gives

$$y = \frac{e^{-4x}}{\sin(2x)}.$$

Finally,

$$y\bigl(\tfrac{\pi}{6}\bigr) = \frac{e^{-4\pi/6}}{\sin(\pi/3)} = \frac{e^{-2\pi/3}}{\tfrac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}e^{-2\pi/3}.$$

The answer is Option A.