Calculating the voltage out of this particular cap coupled circuit
In the instant after the transistor turns on, you can treat the two capacitors as voltage sources. So, in order to calculate the voltages across the circuit in the instant before the voltages start to change on the capacitors, you can simply start adding up voltages: +0.5 V at the collector of Q1 and -8.5V on the left side of R4. Note that the current through R7 is of no consequence initially; it affects the time constant of the voltage decay across C1, but not the initial conditions.

Now, R4 and R6 form a voltage divider, so you can simplify that part of the circuit by replacing them with their Th&eacute;venin equivalent: -7.73V and 909&Omega;. Now you can calculate the current flowing through that impedance in series with R2, which is 7.73V/(909&Omega; + 4000&Omega;) = 1.575 mA. Note that the voltage across C2 exactly cancels V1 at this moment in time.

This gives you the voltage drop across Rth and R2, which is 1.43V and 6.30V, respectively. This should give you -6.30V at the right side of R4 (NODE1), and +2.70V at VOUT.