A particle located in a one-dimensional potential field has its potential energy function as $\mathrm{U}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{{\mathrm{x}}^4}-\frac{\mathrm{b}}{{\mathrm{x}}^2}$, where a and b are positive constants. The position of equilibrium x corresponds to 

The position of equilibrium corresponds to F(x) = 0 

Since $\mathrm{F}\left(\mathrm{x}\right)=\frac{-\mathrm{dU}\left(\mathrm{x}\right)}{\mathrm{dx}}$

so $\mathrm{F}\left(\mathrm{x}\right)=-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{a}}{{\mathrm{x}}^4}-\frac{\mathrm{b}}{{\mathrm{x}}^2}\right)\text{ or }\mathrm{F}\left(\mathrm{x}\right)=\frac{4\mathrm{a}}{{\mathrm{x}}^5}-\frac{2\mathrm{b}}{{\mathrm{x}}^3}$

For equilibrium, F(x)= 0, therefore 

$\begin{array}{l}\frac{4\mathrm{a}}{{\mathrm{x}}^5}-\frac{2\mathrm{b}}{{\mathrm{x}}^3}=0\Rightarrow \mathrm{x}=\pm \sqrt{\frac{2\mathrm{a}}{\mathrm{b}}}\\ \frac{{\mathrm{d}}^2\mathrm{U}\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}=-\frac{20\mathrm{a}}{{\mathrm{x}}^6}+\frac{8\mathrm{b}}{{\mathrm{x}}^4}\end{array}$

Putting $\mathrm{x}=\pm \sqrt{\frac{2\mathrm{a}}{\mathrm{b}}}\text{ gives }\frac{{\mathrm{d}}^2\mathrm{U}\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}$ as negative

So U is maximum. Hence, it is position of unstable equilibrium.