Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + {{\sqrt 2 y} \over {2{{\cos }^4}x - {{\cos }^2}x}} = x{e^{{{\tan }^{ - 1}}(\sqrt 2 \cot 2x)}},\,0 < x < {\pi \over 2}$$ with $y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {32}}$. If $$y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over {18}}{e^{ - {{\tan }^{ - 1}}(\alpha )}}$$, then the value of 3$\alpha$² is equal to ___________.

$\frac{d y}{d x}+\frac{2 \sqrt{2} y}{1+\cos ^{2} 2 x}=x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$ <br/><br/> I.F. $=e^{\int \frac{2 \sqrt{2} d x}{1+\cos ^{2} 2 x}}=e^{\sqrt{2} \int \frac{2 \sec ^{2} 2 x}{2+\tan ^{2} 2 x} d x}$ $=e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}$ <br/><br/> $\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=\int x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$ <br/><br/> $\cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)} d x+c$ <br/><br/> $\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=e^{\frac{\pi}{2}} \cdot \frac{x^{2}}{2}+c$ <br/><br/> When $x=\frac{\pi}{4}, y=\frac{\pi^{2}}{32}$ gives $c=0$ <br/><br/> When $x=\frac{\pi}{3}, y=\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha}$ <br/><br/> So $\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha} \cdot e^{-\tan ^{-1}\left(-\sqrt{\left.\frac{3}{2}\right)}\right.}=e^{\pi / 2} \frac{\pi^{2}}{18}$ <br/><br/> $\Rightarrow \quad-\tan ^{-1} \alpha+\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi}{2}$ <br/><br/> $\Rightarrow \tan ^{-1}(-\alpha)=\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$ <br/><br/> $\Rightarrow \alpha=-\sqrt{\frac{2}{3}} \Rightarrow 3 \alpha^{2}=2$