Note that for DC (constant in time) dv signals ( = 0 ) the capacitor acts as an open circuit (i=0). Also note the capacitor does dt not like voltage discontinuities since that would require that the current goes to infinity which is not physically possible. The constant of integration v(0) represents the voltage of the capacitor at time t=0.
Finally, this will result in the voltage at peak current equal to the initial voltage. At this point in time, the capacitor has reached its maximum current value. Now using the total electric charge equation, the amount of charge lost during the ramp up time can be found.
(A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.
Capacitors are naturally limited by its capability to handle/dissipate ripple current and pulse energy load. The limitation may be significantly different by each capacitor technology, dielectric type, its losses (and its characteristics), but also to a specific construction of the product type individual series.
For all practical purposes, though, we can say that the capacitor voltage will eventually reach 15 volts and that the current will eventually equal zero. Capacitors act somewhat like secondary-cell batteries when faced with a sudden change in applied voltage: they initially react by producing a high current which tapers off over time.
Capacitor acts like short circuit at t=0, the reason that capacitor have leading current in it. The inductor acts like an open circuit initially so the voltage leads in the inductor as voltage appears instantly across open terminals of inductor at t=0 and hence leads.