Some confusions on a basic CMOS multivibrator question
Yes u r right completely.

your first sentence is right. so what to do next...


The simplest way is to find an equivalent circuit that represents the Vdd, R1 and R2 circuit. It's so simple, in equivalent circuit we have a voltage source of 9 volt (because of resistor division), a equivalent R of 0.9K (1K parallel with 10K), a cap with an initial value. So, we have 3 components: a 9v voltage source in series with a 0.9K resistor and in series with a cap. The cap's initial value is -1v, its final value is 9v and its time-constant (RC) value is (cap*0.9Kohm). So, place them in the exponential charging equation of capacitor and calculate the time when the cap's voltage reaches 5v.
Note that the sentence &quot;this behaviour should change when Vc reaches 0V&quot; is totally wrong here because your circuit (in charging mode) is a pure linear circuit!!!
Generally, we can calculate the equivalent circuit like this: calculate two parameters V_open_circuit and I_short_circuit. Voc in this circuit is 9v (Vdd * R2/(R1+R2)). Isc = Vdd/R1. so the equivalent resistor would be Voc/Isc = 0.9K  and the equivalent voltage source is Voc itself.
Let's evaluate and calculate my equivalent circuit with the real one. At (Vc = -1) the current into cap is (11v/1K + 1v/9K) which is equal to the equivalent circuit (10v/0.9K). At Vc=0 the current of cap is (Vdd/R1 + 0/R2) which is equivalent to (9v/0.9K)