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

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$x^{2} y_{2} - 6 y + \frac{3 π}{2} = 0$

$x^{2} y_{2} - 6 y + 3 π = 0$

${xy}_{2} + 2 y_{1} = 0$

${xy}_{2} - 4 y_{1} = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} y = \tan^{- 1} (\sec x^{3} - \tan x^{3}) \Rightarrow y = \tan^{- 1} (\frac{1 - \sin x^{3}}{\cos x^{3}}) \Rightarrow y = \tan^{- 1} \tan (\frac{π}{4} - \frac{x^{3}}{2}) \\ \Rightarrow y = \frac{π}{4} - \frac{x^{3}}{2} ∵ \frac{π}{2} < x^{3} < \frac{3 π}{2} \Rightarrow y_{1} = \frac{- 3 x^{2}}{2}, y_{2} = - 3 x \end{array}$

$\begin{matrix} ∴ x^{2} y_{2} - 6 y + \frac{3 π}{2} = x^{2} (- 3 x) - 6 (\frac{π}{4} - \frac{x^{3}}{2}) + \frac{3 π}{2} \\ = - 3 x^{3} - \frac{3 π}{2} + 3 x^{3} + \frac{3 π}{2} = 0 \end{matrix}$