# Simple RC-Circuit (Neuron): get $u(t)$ as a function of time

I am trying to understand how a LIF-Neuron (please take a look) works and how I come from this:

$$I(t) = \frac{u(t)}{R} + C \frac{du}{dt}$$

by multiplying the equation by $R$ and call $\tau_m = R\,C$ the "leaky integrator":

$$\tau_m \frac{du}{dt} = -u(t) + R\,I(t)$$

to this expression if we assume that $u(t^{(1)}) = u_r = 0$:

$$u(t) = R\,I_0 \Bigg[1 - \text{exp}\Big\{- \frac{t - t^{(1)}}{\tau_m} \Big\} \Bigg]$$

It seems that I fail to do the integration part here. Could somebody help me to get over this?

The according circuit:

In the paper you reference (4.1.1.1) there is the assumption of a constant current: $$I(t)=I_0$$ With this substitution, your problem reduces to a "first-order linear" differential equation: $$\tau_m\frac{du}{dt}=-u(t)+RI_0 \ \ \ \ \mbox{where}\ \ u(t^{(1)})=0.$$ One method would be to use an integrating factor. Put the equation into standard form: $$\frac{du}{dt}+\frac{u(t)}{\tau_m}=\frac{RI_0}{\tau_m}$$ The integrating factor is $$exp\Big\{\int\frac{dt}{\tau_m}\Big\}=e^{t/\tau_m}$$Multiply each term by the integrating factor: $$e^{t/\tau_m}\frac{du}{dt}+e^{t/\tau_m}\frac{u(t)}{\tau_m}=e^{t/\tau_m}\frac{RI_0}{\tau_m}$$ Notice that $$e^{t/\tau_m}\frac{du}{dt}+e^{t/\tau_m}\frac{u(t)}{\tau_m}=\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]$$ Thus, the original equation is reduced to $$\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]=e^{t/\tau_m}\frac{RI_0}{\tau_m}$$ $$\int\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]\ dt=\int e^{t/\tau_m}\frac{RI_0}{\tau_m}\ dt$$ $$e^{t/\tau_m}\cdot u(t)=RI_0\ e^{t/\tau_m}+C$$ From here the remaining matter is to plug in the given initial condition to recover C, and some algebra to put into the required form. The technique above can be found in most beginning textbooks on ordinary differential equations.