How to solve first-order RC circuit?
It often helps to first look for simplified answers among complex questions.  If we first revise the circuit arrangement (steady state before t=0) some of the answers start to become obvious.  







simulate this circuit &ndash; Schematic created using CircuitLab

In the more familiar arrangement above the voltage across the central resistor can easily be seen to be zero (due to the equal voltage divider in each outer leg), so we know at the initial steady state I2=0.  With all resistor values equal the voltages at the center of each leg will be 5v, so the currents will be 10/2 = 5A in each leg, (eg.: I1=5A, I4=5A, I3=5A).  Then with I3=5A you know that Vc=5v (steady state).

Now when the switch is opened the circuit changes to this.  







simulate this circuit

You can see that the three top resistors reduce to 2/3 ohm (0.667 ohm).

Forgetting about the cap for a moment you can also see that the next steady state will be a voltage divider of 0.667 ohm in series with 1 ohm.  Meaning that Vc will eventually become (10/1.667) x 1, or 6v.  So after the switch opens the cap voltage (Vc) changes from the initial 5v to 6v. 

Knowing both steady state Vc values and the equivalent resistances you should be able to determine Ix and Vc at any time in between.