A particle of mass m moves due to a conservative force with potential energy V(x) = $\frac{Cx}{x^2+a^2}$, where C and a are positive constants. The position(s) of stable equilibrium is/are given as

At equilibrium position

$\mathrm{F}=-\frac{dV}{dx}=0$

$\frac{dV}{dx}=\frac{C(a^2-x^2)}{(x^2+a^2{)}^2}=0$

$\therefore$ There are two equilibrium positions

$x_1=a$

$x_2=-a$

Consider $\frac{d^{2}V}{d{x}^2}=\frac{2Cx(x^2-3a^2)}{(x^2+a^2{)}^3}$

$\frac{d^{2}V}{d{x}^2}{|}_{x_1}<0$

$\frac{d^{2}V}{d{x}^2}{|}_{x_2}>0$

$\because$There is a maxima at $x=a$ and minima at $x=-a$
$\Rightarrow x_1$ is a position of unstable equilibrium and $x_2$ is a position of stable equilibrium.