Let ${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$ $\therefore$ $y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$ $= {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$ $\therefore$ $y = {\pi \over 4} - {\theta \over 2}$ $y = {\pi \over 4} - {{{x^3}} \over 2}$ $\therefore$ $y' = {{ - 3{x^2}} \over 2}$ $y'' = - 3x$ $\therefore$ ${x^2}y'' - 6y + {{3\pi } \over 2} = 0$