RC-Circuit for a LIF-Neuron 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 by integrating the first equation if we assume that $u(t^{(1)}) = u_r = 0$:
$$
u(t) = R\,I_0 \left[1 - \exp\left\{- \frac{t - t^{(1)}}{\tau_m} \right\} \right]
$$
Having a constant input current $I_0 = 1.5$
It seems that I fail to do the integration part here. Could somebody help me to get over this?
The according circuit:

 A: We have
$$
u(t) + \tau_m \frac{du}{dt} = \Big(\text{something}\cdot u(t)\Big)+\Big(\text{something}\cdot u'(t)\Big). \tag 1
$$
If we can make this into
$$
\Big(v'(t)u(t)\Big)+\Big(v(t)u'(t)\Big) \tag 2
$$
then we can apply the product rule and it becomes
$$
\Big(v(t)u(t)\Big)'.
$$
Can we make $(1)$ into $(2)$ by multiplying it by something (so we'd also have to multiply the other side of the equation by that same thing)?  We would have
$$
\Big(\text{this thing}\cdot u(t)\Big) + \Big(\tau_m\cdot\text{this thing}\cdot u'(t)\Big)
$$
Thus we need
$$
\Big(\tau_m\cdot\text{this thing}\Big)' = \text{this thing}.
$$
Let's write it is
$$
\tau_m\frac{dw}{dt} = w.
$$
Then
$$
\frac{dw}w = \frac{dt}{\tau_m}
$$
$$
\log w = \frac t {\tau_m}
$$
$$
w = e^{t/\tau_m}. \tag 3
$$
So multiply both sides of the differential equation by $(3)$:
$$
e^{t/\tau_m} u(t) + \tau_m e^{t/\tau_m} u'(t) = e^{t/\tau_m} R I(t)
$$
$$
\frac 1{\tau_m} e^{t/\tau_m} u(t) + e^{t/\tau_m} u'(t) = \frac 1 {\tau_m} e^{t/\tau_m} R I(t)
$$
Apply the product rule
$$
\left( e^{t/\tau_m} u(t) \right)' = \frac 1 {\tau_m} e^{t/\tau_m} R I(t)
$$
To antidifferentiate both sides, you need to know what function $I$ is. To antidifferentiate the left side, just drop the "prime".
