RC circuit with two capacitors
If by intuitively you also mean qualitative, yes.

Considering both capacitors are ideal and initially discharged, you have a mathematical discontinuity in front of you if the input impedance coming at Vin is zero, and that should ideally be tackled with Laplace transformation.

Now considering the expected capacitors behavior, both capacitors will charge (the 1u cap faster than the 9u) and stabilize at certain in-between voltage.

If you want actual waveforms, you can solve the maths using Laplace transform (which is the correct way, but I will not be troubled to doing it), or take a wild guess as to what is the voltage after the discontinuity in each cap is and draw some exponential assymptotes - with a discontinuity peak at t=0 and exponential decay/growth to final voltage.

Edit: correct answer, thanks to user1582568

The voltage Vout after the discontinuity can be calculated by the divisor rule [C1/(C1 + C2)]*Vin = 1V. The final voltage is zero, and the resulting waveform is an exponential decay with time constant T = (C1 + C2)*R.