How do you determine voltage output with time graphically?
As long as intuition is concerned, 


Capacitor cannot change its voltage instantly. Hence, after switching, cap will still hold the previous value for an instant.
As cap act as open circuit at steady state, the steady state vout here is 2.5V which is by res division.
as soon as switch is closed, cap will hold that 2.5V across it. And now it has got a kick of 2V hence, output should go to 4.5V. BUT, as there are two caps, this 2V is divided by them and kick at the output will be of 1V. Hence, output will go to 2.5+1=3.5V. This 3.5V is there for very short time.
Then, as previously stated, cap is open at steady state, final voltage is going to be 2.5V. hence, voltage will come down from 3.5V to 2.5V. this will take certain time constant.
Time constant here by thevenin is R||R and C||C which is RC. Hence, the discharge from 3.5v to 2.5v will take 5 time constant to settle which is 5RC.


This is proved, with simulation as seen as attachment.