i m trying to solve this question by method given in most of the standard books i.e.

Vc(t) = Vc(\(\infty\)) - [Vc(0-) - Vc(\(\infty\)) ] e ( -t/RC)

first i found Vc(0-)by considering switch open and inductor behaving like short and capacitor open at steady state....then i find voltage acroos 4 ohm resistor i.e.

V4 = (4 * 18)/6 = 12 volt

now =Vc(0-)= [C(total)/2c ] * V4 = 4 volt

now i m finding [Vc(0-) by considering switch closed ....and at steady state inductor will be shorted and both capacitors will be opened....and now inductor path and capacitor c bot are in parallel and i have confused what to do here next .....please help me....

 

When the switch is open, the current through L is 3A. Since an inductor tries to keep the current constant, when you close the switch that current will have to be sourced by C and 2C.

I would say the current will divide according to the capacitances, so the 2C current should be 2A, but this is getting into the guessing area.