Differential equation of a LRC circuit as Voltage output/voltage input? So in this presentation i have got, there is a LRC circuit, and they get this differential equation, but there is no procedure explained how they have got it. This is basicially a series LRC circuit because current I3 is zero, so it is the same current, and i think we can ignore Uc(0) in formula. Later when we would laplace transform final differential equation we would get transfer function Vo/Vi. My question is how to get the final differential equation with Vo and Vi like this so i can get Vo/Vi, i searched the internet and couldn't find anything. English is not my first language so i couldn't express my self better. In images input voltage(Vi) is Uu, and output voltage(Vo) is Ui. LRC circuit Differential equations Final Differential equation as Vo/Vi

The first equation obtains the time-$t$ potential difference drop $u_u(t)-u_c(0)$ following from the top left-hand corner to the point where the path through the inductor, resistor and capacitor meets the bottom line. In particular, with current $i_1$ through an inductor the drop is $L\dot{i}_1$, and the other terms take a current $i_2$ instead. The second equation does the same for the voltage drop in a path omitting the inductor. Since $u_i-u_u=-L\dot{i}_1$, $RC\dot{u}_i-RC\dot{u}_u=-LRC\ddot{i}_1$, so$$LC\ddot{u}_i+RC\dot{u}_i+u_i-u_u-RC\dot{u}_u=LC\left(\ddot{u}_i-R\ddot{i}_1-\frac{\dot{i}_1}{C}\right).$$But this vanishes by the second equation. Thus$$LC\ddot{u}_i+RC\dot{u}_i+u_i=RC\dot{u}_u+u_u,$$as desired. This arrangement has the advantage of placing the i/u terms on opposite sides, in differential operators analogous to the typical LRC equation $C\dot{V}=LC\ddot{i}+RC\dot{i}+i$.