Rise time for second order RC filter?
Once you get past first order, the time domain equations start to get messy.  In fact, so messy that those equations aren't much use in designing circuits anymore.  In a real design situation, it is usually more useful to solve the output response numerically for whatever input signal you care about.

A few quick calculations are still useful in frequency space.  What you have is two RC low pass filters in series.  In the first pass, solve them as two separate filters.  The rolloff frequency (don't every say "cutoff", it gives the wrong impression) of a simple RC filter is F = 1 / (2&pi;RC).  When R is in Ohms, C in Farads, then F is in Hertz.

A single RC filter will attenuate by the ratio of the input frequency to the rolloff frequency for frequecies well past the rolloff, which can also be expressed as 20 dB/decade or about 6 dB/octave.  Two filters cascaded essentially multiply.  That means the attenuation is the square of the frequency relative to the rolloff frequency, or 40 dB/decade, which is about 12 dB/octave.

However, this does not take into account interaction between the two filters.  The output impedance of the first filter is R1//C1, with the input impedance of the second filter being R2+C2.  As long as the former is low enough compared to the latter, the two filters won't interact much and you can use the approximation of two independent filters cascaded.  Note that the magnitude of both impedances is frequency dependent.  Therefore at very low and very high frequencies the separate filters approximation is valid.

When it's not valid, things get more complicated.  Often the solution is to design the filters so that they don't significantly interact over the frequencies of interest.  For example, make R2 several times R1.