 Solution

For a time-independent Hamiltonian operator and a Schrödinger picture observable with no explicit time dependence, the Heisenberg picture operator isDifferentiation gives ${\dot{O}}_H=iH{O}_H-i{O}_{H}H=i[H,O_H]$, soIf $O_S$ itself depends explicitly on time, one must add $e^{iHt}({\partial }_t{O}_S)e^{-iHt}$ to ${\dot{O}}_H$; the displayed commutator relation uses the usual convention of no explicit time dependence for the observable.