Let $${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$$

$$\therefore$$ $$y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$$

$$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$$

$$\therefore$$ $$y = {\pi \over 4} - {\theta \over 2}$$

$$y = {\pi \over 4} - {{{x^3}} \over 2}$$

$$\therefore$$ $$y' = {{ - 3{x^2}} \over 2}$$

$$y'' = - 3x$$

$$\therefore$$ $${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$