Answer :
If the current i(t) in the circuit at time t after the switch is closed in terms of q0 , l , c the differential equation will be V(t) = (q0/C) × e^(−t/RC )
The equation for V(t) would be V(t)=E(RC)(dV(t)/dt) if a battery with EMF E was present in the circuit. With respect to V(t), this differential equation is no longer homogenous. But it can be resolved by simply substituting Vb(t)=V(t)E. This substitution has the effect of removing the E term and producing an equation for Vb(t) that is the same as the equation you solved for V(t). When a battery is supplied, the capacitor's initial condition is typically that it is completely discharged at time t=0.
In these circumstances, the solution will appear as V(t)=E(1et/(RC)) Since the voltage across the capacitor is zero at time t=0 according to this solution, is 0 at time t=0 since the capacitor wasn't charged at that point, and it climbs asymptotically to E.
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