Effect of time on a node with a capacitor
This question is flawed because at the point when the switch closes \$V_O\$ is undefined by any of the circuit components. Nothing in the circuit prior to t=0 defines \$V_O\$.


  At t&lt;0 vs is unplugged and the rest of circuit is dead. va=v 0 =0V


No this is not true.

Just think about it. Before t=0 the +Vin input to the op-amp is not at a defined potential - there is no resistor that might be "lightly" tying it to ground. Because of this you can't assume it is at ground. Even if you did assume it to be at ground the output is still undefined because there has been no steady state conditions set (prior to t=0) that could result in a known charge in the capacitor. This means the output voltage, \$V_O\$ is at an indeterminable level prior to t=0.

At t=0, you force a known voltage onto the +Vin input and this overcomes half the issues but you are still no closer to determining what \$V_O\$ is.

Let's make assumptions (EDIT)

So, given these issues, if I were to assume that before t=0 the capacitor was kept discharged the following would happen when the switch closes. At that instant 20mV is applied to +Vin and to combat this potential discrepancy between +Vin and -Vin, the output of the op-amp (perfect of course) instantly rises to +20mV. What happens next is that the output voltage continues to rise linearly with time.

To keep -Vin at the same potential as +Vin (20mV) there has to be a constant current through the 20k ohm resistor of 1uA. This current can only come from the 5uF capacitor and to understand what happens you need to start with the formula Q = CV.

Q is charge in the capacitor and current is rate of change of Q therefore: -

\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt} = 1\mu A\$

Capacitance is constant at 5uF therefore \$\dfrac{dV}{dt} = \dfrac{1\times 10^{-6}}{5\times 10^{-6}}\$ = 0.2V per second.

So the output ramps up at 0.2V per second to maintain 1 micro-amp through the 20k ohm resistor which in turn keeps +Vin and -Vin at equilibrium.