What will be the initial voltages in this circuit?
Assumption : the opamp is ideal

As soon as the switch is closed, the voltage V1 will have a value = 2V.
As per the law of virtual ground, the voltage at -ve end of opamp will be equal to 2 volts as well. All this is happening at t=0 and at t=0, the capacitor behaves as a short circuit. So, Vo will be equal to the voltage at -ve end of the opamp (as the capacitor is short now) which will make Vo = 2V as well.