Question

12. If \( y=\tan ^{-1}\left(\sec x^{3}-\tan x^{3}\right) \cdot \frac{\pi}{2}<x^{3}<\frac{3 \pi}{2} \), then (1) \( x y^{\prime \prime}+2 y^{\prime}=0 \) (2) \( x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0 \) (3) \( x^{2} y^{\prime \prime}-6 y+3 \pi=0 \) (4) \( x y^{\prime \prime}-4 y^{\prime}=0 \)

Answer 1

Correct option is (2) \(x^2 y^" - 6 y + \frac {3\pi}{2} = 0\)

\(x^3 = \theta \Rightarrow\frac \theta 2 \in \left(\frac \pi 4, \frac {3\pi}{4}\right)\)

\(\therefore y = \tan^{-1} (\sec \theta - \tan \theta) \)

\(= \tan^{-1} \left( \frac {1 - \sin \theta}{\cos \theta}\right)\)

\(\therefore y = \frac \pi 4 - \frac {\theta}2\)

\( y = \frac \pi 4 - \frac {x^3}2\)

\(\therefore y' = \frac {-3x^2}2\)

\(y''= -3x\)

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