Time Constant of a tricky RC circuit
Just use the method of open circuit time constants. If you take output at C1, then you have a system with two poles and a zero.
The time constant for the poles are:
$$\tau_1 = R_1C_1$$
$$\tau_2 = (R_1+R_2)C_2$$
$$H^2 = \frac{R_2}{R_1+R_2}$$
$$H^0 = 1$$
Here, \$H^0\$ is the transfer function with all capacitors open circuit and  \$H^2\$ is transfer function with \$C_2\$ short circuit and \$C_1\$ open.
The transfer function becomes:
$$H(s) = \frac{H^0+s\tau_2H^2}{(1+s\tau_1)(1+s\tau_2)} = \frac{1+sR_2C_2}{(1+sR_1C_1)(1+s(R_1+R_2)C_2)}$$
To find the voltage with time use the step response as follows:
$$X(s) = V/s$$
$$V(s) = H(s)V/s$$
Take inverse Laplace to get your answer.