If $y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$, then

Question

$xy'' + 2y' = 0$

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

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

$xy'' - 4y' = 0$

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