Q. If $y = \sin{(\log_{e}{x),\ then\ x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx}\ }\ }$is equal to
A) $\sin{(\log_{e}{x)}}$
B) $\cos{(\log_{e}{x)}}$
C) $y^{2}$
D) $- y$
Explanation
(d)
Given, $y = \sin{\ (\log_{e}{x)}\ \ \ \ \ \ }$…(i)
$\Rightarrow \ \ \ \frac{dy}{dx} = \cos{(\log_{e}{x).\frac{1}{x}}}$ …(ii)
$\Rightarrow \ \ \frac{d^{2}y}{dx^{2}} = \frac{- x.\sin{(\log_{e}{x).\frac{1}{x} = \cos{(\log_{e}{x).1}}}}}{x^{2}}$
$= \frac{- \sin{\ (\log_{e}{x) - {cos\ }{(\log_{e}{x)}}}}}{x^{2}}$
$x^{2}\frac{d^{2}y}{dx^{2}} = \ - \sin{(\log_{e}{x) - x\frac{dy}{dx}}}$ [using Eq. (ii)]
$x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} = \ - y$ [using Eq. (i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.