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Question on RC Circuit Transient Response

I was going through some questions on Transient Response of circuits. Then I encountered this question! Though, I was a bit confused at first, I initially approached in an intuitive way. My first approach was as the circuit looked symmetric horizontally, I reduced the circuit to an equivalent circuit with a single capacitor of 2 Farad and a resistance of 0.5 ohm in series with the capacitor. Then I could easily find time constant = (1.5 Ohm) * (2 Farad)= 3 Second. Then, I could write the expression of Vx(t) as [1 - (2/3)exp(-t/3)] V. But at this time, I was a bit confused because can we really reduce the circuit in that manner just because it has symmetry and further the capacitors have different initial voltages. So, I proceeded to systematically analyse the circuit. I wrote differential equations and solved for Vx(t). But, surprisingly I got the same answer as before. Moreover, I got a first order differential equation for Vx(t). Is it because I could reduce the circuit into one capacitor format? Also, what is the order of the circuit? It looks like this circuit has order 2. But, then, while solving for Vx(t), why am I getting a first order equation? And what will be an elegant method of finding the expression for Vc1(t) and Vx(t)?

Edit:- I simulated this circuit in LTSpice and here is the waveforms.Simulation Waveforms Red, Blue and Green curves are for Vx, Vc2 and Vc1 respectively.

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    \$\begingroup\$ No matter what Vx does, the same voltage appears on both capacitors because they are identical circuits hence, what you did was fine and it is a 1st order system. End of story. \$\endgroup\$
    – Andy aka
    May 8, 2020 at 15:31
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    \$\begingroup\$ You are missing the fact that capacitors were charged at different initial voltages. Also, regarding order of the circuit, I am confused! Because, at t=0, Vx will be 1/3 V and then Vc1=1V and Vc2=-1V. Hence capacitor C1 will discharge and C2 will charge. After sometime, when Vx will rise to a certain extent, specifically when it'll surpass Vc1, then both capacitors will chrage towards 1V (Vx too). Hence capacitor C1 will at first discharge to a certain voltage and then it will charge towards 1V. How can you explain this for a first order circuit? \$\endgroup\$ May 8, 2020 at 16:13
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    \$\begingroup\$ You can solve this using superposition. As Andy says, the response to the u(t) source will be symmetric. Now superpose that with the transient decay of the response from the two capacitors' initial charges. \$\endgroup\$
    – The Photon
    May 8, 2020 at 16:19
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    \$\begingroup\$ Yes, I missed that but, the order of the circuit is unaffected by initial conditions. \$\endgroup\$
    – Andy aka
    May 8, 2020 at 16:19

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@ThePhoton provides the clearest path it seems. I worked it by hand with Laplace and simulated it in ATP. When applying superposition you will notice that the impact on Vx from the left side capacitor is exactly negative of the right side capacitor, they cancel. Thus, only the contribution from the step u(t) shows up in the closed form Laplace solution. Here are plots of Vx. Top plot is from simulation. Bottom plot is from MathCAD (used it to get the inverse Laplace).

plots

Here is a plot of all 3 voltages of interest together:

voltages

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