Thank you to everyone who helped out with my previous post concerning the singular circuit.

Now, what is the best way for solving this circuit, assuming I know my initial conditions for t=0+ ?

Do I incorporate the initial conditions as such? :

Assuming I want to solve the above circuit for V3 using superposition: I have to sum the contribution to V3 from the 3 DC sources, one at a time right? So, let's say I am finding the output at V3 due the initial charge on C1 (call this contribution V3-dueto-1). To solve for

V3-dueto1, do I "shut-off" all the DC sources except V1(0+) by replacing them with a short circuit? Then, still solving for V3-due-to-1, I expect to obtain a differential equation with "unknown constants". To figure out what these unknown constants are, I express my differential equation for time t=0, and equate it to the DC output at V3 when all the capacitors are shorted. Then, I express my differential equation for time=infinity and equate it to the DC output voltage at V3 when all the capacitors are open-circuited.

Can someone confirm the validity of this approah ?

Is there another way ? Here is the Laplace circuit:

This would be simpler I supposed because there wouldn't be any "unknown

constants" to solve for, correct?

Any suggestions / corrections ? Am I on the right track here ? I get ridiculously long expressions when I solve the Laplace circuit by superposition ...

Thanks, Dan