Q. The solution of the equation $x^{2}\frac{d^{2}y}{dx^{2}} = \log x$ when $x = 1,\ y = 0$ and $\frac{dy}{dx} = - 1$ is
A) $\frac{1}{2}\left( \log x \right)^{2} + \log x$
B) $\frac{1}{2}\left( \log x \right)^{2} - \log x$
C) $- \frac{1}{2}\left( \log x \right)^{2} + \log x$
D) $- \frac{1}{2}\left( \log x \right)^{2} - \log x$
Explanation
(d)
Given, $\frac{d^{2}y}{dx^{2}} = \frac{\log x}{x^{2}}$
$\Rightarrow$ $\frac{dy}{dx} = \frac{- (\log{x + 1})}{x} + c$ [integrating]
Since, $\left( \frac{dy}{dx} \right)_{(1,0)} = - 1$
$\Rightarrow$ $\frac{- 1}{1} + c = - 1 \Rightarrow c = 0$
$\Rightarrow$ $\frac{dy}{dx} = - \frac{(\log{x + 1)}}{x} + 0$
$\Rightarrow$ $y = - \frac{1}{2}\left( \log x \right)^{2} - \log x + c_{1}$ [integrating]
At $x = 1,\ y = 0 \Rightarrow c_{1} = 0$
$\therefore$ $y = - \frac{1}{2}\left( \log x \right)^{2} - \log x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.