Let $y=y(x)$ be the solution of the differential equation

$$\sec ^2 x d x+\left(e^{2 y} \tan ^2 x+\tan x\right) d y=0,0< x<\frac{\pi}{2}, y(\pi / 4)=0$$.

If $y(\pi / 6)=\alpha$, then $e^{8 \alpha}$ is equal to ____________.

<p>$$\begin{aligned} & \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\ & \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\ & \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\ & \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\ & \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\ & \left(\text { Put } \frac{1}{t}=u \frac{-1}{t^2} \frac{d t}{d y}=\frac{d u}{d y}\right) \\ & \frac{-d u}{d y}+u=-e^{2 y} \\ & \frac{d u}{d y}-u=e^{2 y} \\ & \text { I.F. }=e^{-\int d y}=e^{-y} \\ & u e^{-y}=\int e^{-y} \times e^{2 y} d y \\ & \frac{1}{\tan x} \times e^{-y}=e^y+c \\ & x=\frac{\pi}{4}, y=0, c=0 \\ & \begin{aligned} & \mathrm{x}=\frac{\pi}{6}, \quad \mathrm{y}=\alpha \\ & \sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^\alpha+0 \\ & \mathrm{e}^{2 \alpha}=\sqrt{3} \\ & \mathrm{e}^{8 \alpha}=9 \end{aligned} \end{aligned}$$</p>