Differential Equation of RC Low Pass Filter
Here you can take two approaches. In the second approach you can see where your confusion lies..

Approach 1 - s-domain. This approach leads to an equation containing Laplace terms, which, additionally, requires you to take Laplace inverse to find the equation in the time domain. That's ok if you're familiar with the Laplace transform. For example, if the initial conditions are zero, then taking Laplace inverse of s^2V(s) gives V''(t), that of sV(s) gives V'(t), and of V(t) gives V(s). 

Approach 2 - time domain. This sometimes might be tricky to get around but still easy to handle for this problem. Assuming initial conditions are zero, what you need here is to first find the current through the circuit, which is simply 
$$ i(t)=\frac{V(t)}{R}+C_2\frac{dV(t)}{dt}\ \  (*)$$

Now a KVL leads to

$$V_{in}(t)=i(t)R_1+V_{C_1}+V(t) \ \ (**)$$
where that comes your confusion. You just need to put \$V_{C_1}=\frac{1}{C_2}\int i(t)dt\$ and get the expression as \$V_{in}(t)=i(t)R_1+\frac{1}{C_2}\int i(t)dt+V(t)\$. Now to get rid of the integration try to differentiate from both sides w.r.t time,

$$ \frac{V_{in}(t)}{dt}=R_1\frac{di(t)}{dt}+\frac{i(t)}{C_2}+\frac{dV(t)}{dt} \ \ (***)$$

Now simply substitute \$i(t)\$ from (*) into (***) and you'll get the desired differential equation. As an extra exercise for you, try to compare the resultant equation to the one you obtained from s-domain. You must get pretty the same equations..