Solve the differential equation of $L-C$ circuit and obtain the expression of current.

(D) The differential equation of an $L-C$ circuit is given by,
$\frac{d^{2} q}{d t^{2}}+\frac{1}{LC} q=0$
where $q$ is the charge on the capacitor.
The general solution of this equation is,
$q(t) = q_{m} \cos(\omega_{0} t + \phi)$
where $q_{m}$ is the maximum charge,$\omega_{0} = \frac{1}{\sqrt{LC}}$ is the angular frequency,and $\phi$ is the phase constant.
Assuming the capacitor is fully charged at $t = 0$,$q(0) = q_{m}$.
$q_{m} = q_{m} \cos(\phi) \implies \cos(\phi) = 1 \implies \phi = 0$.
Thus,$q(t) = q_{m} \cos(\omega_{0} t)$.
The current $I$ is the rate of flow of charge,$I = -\frac{dq}{dt}$ (as the capacitor discharges).
$I = -\frac{d}{dt} [q_{m} \cos(\omega_{0} t)] = -q_{m} \omega_{0} (-\sin(\omega_{0} t)) = q_{m} \omega_{0} \sin(\omega_{0} t)$.
Defining the maximum current $I_{m} = q_{m} \omega_{0}$,we get,
$I = I_{m} \sin(\omega_{0} t)$.