From this derivation to get the differential of a band pass filter: Id like to find out how we set the upper frequency cut-off and the lower frequency cut-off.
$$i(t)=\frac{v_i(t)-v_o(t)}{R}=C\frac{dv_o(t)}{dt}$$
So we get
$$v_i(t)=v_o(t)+RC\frac{dv_o(t)}{dt}\tag{1}$$
For the high-pass filter we have
$$i(t)=\frac{v_o(t)}{R}=C\frac{d(v_i(t)-v_o(t))}{dt}$$ which gives
$$RC\frac{dv_i(t)}{dt}=v_o(t)+RC\frac{dv_o(t)}{dt}\tag{2}$$
Let's call the time constants of the low-pass and high-pass filters $\tau_L=R_LC_L$ and $\tau_H=R_HC_H$, respectively. For the band-pass filter we need a relation between $v_i(t)$ of the low-pass filter and $v_o(t)$ of the high-pass filter. If we use $v_{oL}(t)$ to denote the output of the low-pass filter, which equals the input of the high-pass filter, we get from (1) by taking the derivative
$$\frac{dv_i(t)}{dt}=\frac{dv_{oL}(t)}{dt}+\tau_L\frac{d^2v_{oL}(t)}{dt^2}\tag{3}$$
From (2) we have
$$\tau_H\frac{dv_{oL}(t)}{dt}=v_o(t)+\tau_H\frac{dv_o(t)}{dt}\tag{4}$$
and (by taking the derivative)
$$\tau_H\frac{d^2v_{oL}(t)}{dt^2}=\frac{dv_o(t)}{dt}+\tau_H\frac{d^2v_o(t)}{dt^2}\tag{5}$$
Plugging (4) and (5) into (3) we finally get for the band-pass filter
$$\tau_H\frac{dv_i(t)}{dt}=v_o(t)+(\tau_L+\tau_H)\frac{dv_o(t)}{dt}+\tau_L\tau_H\frac{d^2v_o(t)}{dt^2}$$