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

Given $\dfrac{dy}{dx} + \dfrac{\sqrt{2}}{2\cos^{4}x - \cos 2x}y = x{e}^{\tan^{- 1}\left(\sqrt{2}\cot 2x\right)}$

This is a linear differential equation.

Here $I = ∫ \dfrac{dx}{2\cos^{4}x - \cos 2x} = ∫ \dfrac{dx}{\left(\dfrac{1}{2} + \dfrac{\cos^{2}2x}{2} + \cos 2x\right) - \cos 2x}$

$= 2\sqrt{2} ∫ \dfrac{\sec^{2}2xdx}{2 + \tan^{2}2x}$

Put $\tan 2x = t$

$I = \tan^{- 1}\dfrac{\tan 2x}{\sqrt{2}}$

$∴ IF = {e}^{\tan^{- 1}\left(\dfrac{\tan 2x}{\sqrt{2}}\right)} = {e}^{\cot^{- 1}\left(\sqrt{2}\cot 2x\right)}$

So general solution will be

$y{e}^{\cot^{- 1}\left(\sqrt{2}\cot 2x\right)} = ∫ x{e}^{\tan^{- 1}\left(\sqrt{2}\cot 2x\right)}{e}^{\cot^{- 1}\left(\sqrt{2}\cot 2x\right)}dx$

$y{e}^{\cot^{- 1}\left(\sqrt{2}\cot 2x\right)} = e\dfrac{\pi}{2}\left(\dfrac{{x}^{2}}{2}\right) + c$

$y\left(\dfrac{\pi}{4}\right) = \dfrac{{\pi}^{2}}{32} ⇒ c = 0$

i.e. $y = \dfrac{{x}^{2}}{2}{e}^{\tan^{- 1}\left(\sqrt{2}\cot 2x\right)}$

Given $y\left(\dfrac{\pi}{3}\right) = \dfrac{{\pi}^{2}}{18}{e}^{- \tan^{- 1}\alpha}$

$⇒ \dfrac{{\pi}^{2}}{18}{e}^{- \tan^{- 1}\alpha} = \dfrac{{\pi}^{2}}{18}{e}^{\tan^{- 1}\left(\sqrt{2}cot\dfrac{2\pi}{3}\right)}$$= \dfrac{{\pi}^{2}}{18}{e}^{- \tan^{- 1}\left(\sqrt{\dfrac{2}{3}}\right)}$

Hence $\alpha = \sqrt{\dfrac{2}{3}} ⇒ 3{\alpha}^{2} = 2$