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

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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y = {\tan ^{ - 1}}(\sec {x^3} - \tan {x^3}),\frac{\pi }{2} < {x^3} < \frac{{3\pi }}{2},

then

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Correct option: B

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

Option B (

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

) is correct because after simplifying the given equation and differentiating twice, it satisfies the given relation.

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