Use the Electrical Circuit Differential Equation Where $R = 20$ Is the Resistance (In Ohms)

Use the electrical circuit differential equation $\frac { d ^ { 2 } q } { d t ^ { 2 } } + \left( \frac { R } { L } \right) \frac { d q } { d t } + \left( \frac { 1 } { L C } \right) q = \left( \frac { 1 } { L } \right) E ( t )$ where $R = 20$ is the resistance (in ohms) , $C = 0.02$ is the capacitance (in farads) , $L = 2$ is the inductance (in henrys) , $E ( t ) = 14 \sin 6 t$ is the electromotive force (in volts) , and $q$ is the charge on the capacitor (in coulombs) . Find the charge $q$ as a function of time for the electrical circuit described. Assume that $q ( 0 ) = 0$ and $q ^ { t} ( 0 ) = 0$ .

A) $y = \left( \frac { 105 } { 874 } + \frac { 42 } { 71 } t \right) e ^ { - 5 t } - \frac { 105 } { 874 } \cos 6 t - \frac { 77 } { 3,721 } \sin 6 t$
B) $y = \left( \frac { 420 } { 3,721 } + \frac { 42 } { 61 } t \right) e ^ { - 5 t } - \frac { 420 } { 3,721 } \cos 6 t - \frac { 77 } { 3,721 } \sin 6 t$
C) $y = \left( \frac { 105 } { 943 } + \frac { 42 } { 71 } t \right) e ^ { 5 t } - \frac { 105 } { 943 } \cos 6 t - \frac { 77 } { 3,721 } \sin 6 t$
D) $y = \left( \frac { 28 } { 255 } + \frac { 42 } { 61 } t \right) e ^ { - 5 t } - \frac { 28 } { 255 } \cos 6 t - \frac { 77 } { 3,721 } \sin 6 t$
E) $y = \left( \frac { 35 } { 306 } + \frac { 42 } { 61 } t \right) e ^ { 5 t } - \frac { 35 } { 306 } \cos 6 t - \frac { 77 } { 3,721 } \sin 6 t$