Question

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

• A. $x y^{\prime \prime}+2 y^{\prime}=0$
• B. $x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0$
• C. $x^{2} y^{\prime \prime}-6 y+3 \pi=0$
• D. $x y^{\prime \prime}-4 y^{\prime}=0$

Answer 1

Answer: (B)

$y=\tan ^{-1}\left(\sec x^{3}-\tan x^{3}\right)$
$=\tan ^{-1}\left(\frac{1-\sin x^{3}}{\cos x^{3}}\right)$
$=\tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}-x^{3}\right)}{\sin \left(\frac{\pi}{2}-x^{3}\right)}\right)$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x^{3}}{2}\right)\right)$
Since $\frac{\pi}{4}-\frac{x^{3}}{2} \in\left(-\frac{\pi}{2}, 0\right)$
$y=\left(\frac{\pi}{4}-\frac{x^{3}}{2}\right)$
$y^{\prime}=\frac{-3 x^{2}}{2}, y^{\prime \prime}=-3 x$
$4 y=\pi-2 x^{3}$
$4 y=\pi-2 x^{2}\left(\frac{-y^{\prime \prime}}{3}\right)$
$12 y=3 \pi+2 x^{2} y^{\prime \prime}$
$x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0$