In that case how can I calculate it ?
I thought about source transformations but I don't see how I can do that.

When the switch is closed, 12V is applied across the R1s and R2s because of the left voltage source. The right voltage source and the Rs are ignored.

 

The equation you wrote:
V = Is R+ ( Vo - IsR) e^ -(t/RC)

Is a special case of this:

\(V(t) = V_f + (V_0 - V_f)e^{-\frac{t}{RC}}\)

Where V_0 and V_f are initial and final voltage.
So basically you're saying the final voltage is Is*R... which implies you're finding the Thevenin open circuit voltage.
'Is' then is the short circuit current because that's the Thevenin / Norton relationship.

So you have R, we got it before. It's 2*(R1||R2) = 4 kilohms.
You need to find the short circuit current for this formula, meaning make the capacitor a short.

The voltage at node A = node B will be 6 V by symmetry.
Then the current must be the difference between the current through R1 and R2:
\(I_s = \frac{6V}{3k\Omega} - \frac{6V}{6k\Omega} = 1 mA\)

I've never used this exact method before... kind of interesting.
I would've done it another way - just find the Thevenin voltage, which really is what you're looking for by finding R and Is.

Two voltage dividers... 12V across them.
\(V_a = 12\frac{6}{9} = 8 V\\
V_b = 12\frac{3}{9} = 4 V\\
V_f = I_sR = V_a - V_b = 4 V\)

RC is 4k * 5u = 20 ms

So you get:
\(V_C = 4 + (2.12 - 4)e^{-\frac{t}{20ms}}\)

So just solve your equation by plugging in t = 1 ms...

I get 2.21 V, confirmed by simulation.
So... I guess you solved it wrong? The equation is right.

And just for reference, when you say t = 0+ it means approaching zero from the right... so if the switch closes at exactly 0, this is after the switch closed but 0 time has passed since then.

 

\(V(t) = V_f + (V_0 - V_f)e^{-\frac{t}{RC}}\)

According to this formula, we would have:

\(V(0) = V_f \)

This doesn't seem quite right.

 

According to this formula, we would have:

\(V(0) = V_f \)

This doesn't seem quite right.

Wouldn't it be ...

\(V(0)=V_{f}+(V_{0}-V_{f})e^{0}=V_{f}+V_{0}-V_{f}=V_{0}\)

 

You're quite right. That's what I get for staying up so late!

I was thinking of the formulation where there is a factor like:

\((1-e^{-\frac{t}{RC}})\)

which is zero when t is zero.

 

Ghar,
Thank you very much, your detailed solution helped me understand your faster way of resolution and find my mistake. The error was in RC, I forgot to multiply 4 by 10^3 so my answer were wrong.


Thank you mik3 for your following my problem
Thank you The_Electrician and t_n_k for your input.