I have got to:

I need to find V_6 and V_3 math speaking, is there a simple sloution? which matrix should I build?

 

yes there is...

do the math yourself (good or bad), post your attempt and then you get feedback.

 

Hi,

This is a highly theoretical question/problem because there are no resistances in series with the capacitors and there are only DC sources. DC sources with zero resistance capacitors dont mix in real life because caps always have some resistance which limits the peak current. With nothing to limit the current the peak will be infinite, and that is what makes this is a purely theoretical problem

You might be able to do this in the normal way, whatever way you learned.

One idea would be to use superposition. Short out one Vo, then take the response of say V3, call this V3a. Then, relieve that short and short out the other Vo then take the response of V3 again, call that V3b. The total response for V3 is then V3=V3a+Vb.

For a simpler example, with two capacitors in series with a DC source of E volts and we want the 'divider' voltage after the circuit was first turned on, we would have:
Z=1(sC1)+1/(sC2)
V=(E/s)*(1/(sC2))/Z

and with C2=C1 this simplifies to:
V=E/(2s)

and the inverse transform of that is:
V=E/2

so the divider voltage is one half of the source voltage right after the source is turned on.
Note the current would be infinite, so this is purely theoretical and can never happen in real life. In real life, we would have resistances R1 and R2 in series with the two caps, and if R2=R1 and C2=C1 then we would get the same result but the current would be limited.

 

No need to make it so complicated. You can certainly build this circuit in real life and the lack of knowledge of the parasitic resistances only means that you don't know the initial currents or the time constants involved. But neither of those (assuming no damage occurs) will affect the final charges, which is what the TS is looking for.