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

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