If y = ^ - 1 (sec x ^ 3 - tan x ^ 3 ), (π)/(2) < x ^ 3 < (3π)/(2) ,...

Given,

$y = \tan^{- 1}\left(sec{x}^{3} - tan{x}^{3}\right)$

$= \tan^{- 1}\left(\dfrac{1 - sin{x}^{3}}{cos{x}^{3}}\right)$

$= \tan^{- 1}\left(\dfrac{1 - \cos \left(\dfrac{\pi}{2} - {x}^{3}\right)}{\sin \left(\dfrac{\pi}{2} - {x}^{3}\right)}\right)$

$= \tan^{- 1}\left(\tan \left(\dfrac{\pi}{4} - \dfrac{{x}^{3}}{2}\right)\right)$

Since $\dfrac{\pi}{4} - \dfrac{{x}^{3}}{2} \in \left(- \dfrac{\pi}{2}, 0\right)$ as \left(\dfrac{\pi}{2} < {x}^{3} < \dfrac{3\pi}{2}\right)$</p><p>So,$y = \left(\dfrac{\pi}{4} - \dfrac{{x}^{3}}{2}\right)$</p><p>Now differentiating we get,</p><p>${y}^{'} = \dfrac{- 3{x}^{2}}{2}, {y}^{' '} = - 3x$</p><p>Now putting the value of$x$in term of${y}^{' '} in $4y = \pi - 2{x}^{3}$

We get,

4y = \pi - 2{x}^{2}\left(\dfrac{- {y}^{' '}}{3}\right)

12y = 3\pi + 2{x}^{2}{y}^{' '}

{x}^{2}{y}^{' '} - 6y + \dfrac{3\pi}{2} = 0