Explanation

(d)
Given equation is
$\frac{d^{2}y}{dx^{2}} = \frac{\log x}{x^{2}}$
On integrating, both sides we get
$\int_{}^{}{\frac{d^{2}y}{dx^{2}}dx =}\int_{}^{}{\frac{\log x}{x^{2}}dx}$
$\Rightarrow \ \frac{dy}{dx} = - \frac{\log x}{x} + \int_{}^{}{\frac{1}{x^{2}}dx + c}$
$\Rightarrow \ \frac{dy}{dx} = - \frac{\log x}{x} - \frac{1}{x} + c$
At $x = 1,\ y = 0$ and $\frac{dy}{dx} = - 1\ \Rightarrow c = 0$
$\therefore\ \frac{dy}{dx} = - \frac{(\log{x + 1)}}{x}$
Again on integrating, both sides we get
$\int_{}^{}{\frac{dy}{dx}dx = -}\int_{}^{}{\frac{\log{x + 1}}{x}dx + c_{1}}$
$y = - \frac{1}{2}({\log{x)}}^{2} - \log{x + c_{1}}$
At $x = 1,\ y = 0$
$\Rightarrow c_{1} = 0$
$\therefore y = - \frac{1}{2}({\log{x)}}^{2} - \log x$