How to solve the Leaky Integrate and Fire ODE?

Question

In the article on neuronal dynamics I read he solves the equation, $$ \tau_{m}\,{{\text{d}}u\over{\text{d}}t}=-[u(t)-u_{\rm rest}]+R\,I(t)\,. $$

as $$ u(t)=u_{\rm rest}+R\,I_{0}\left[1-\exp\left(-{t\over\tau_{m}}\right)\right]\,. $$

When $I(t) = I_0$, $u_{\text{rest}} = u(0)$, and which starts at $t=0$ and then solving for $u(t)$, however, I have little knowledge of ODE's and I was wondering if someone could show me the steps between these two equations.

Answer 1

Calling $v = u-u_{rest}$ we have

$$ \tau_m v'+v=R i(t) $$

now using the Laplace transform

$$ (\tau_m s + 1)V(s) = v(0)+R I(s) $$

with $v(0) = u_{rest}-u_{rest} = 0$ or

$$ V(s) = \frac{R}{\tau_m s+1}I(s) $$ now if $i(t) = I_0\phi(t)$ then $I(s) = \frac{I_0}{s}$ with $\phi(t)$ the Heavside unit step function.

So we have

$$ V(s) = \frac{I_0R}{s(\tau_m s+1)} $$

hence inverting we get

$$ v(t) = R I_0 \left(1-e^{-\frac{t}{\tau_m}}\right) = u(t)-u_{rest} $$

Answer 2

Try to bring you ODE into the form $a \cdot du/dt + bu = c$. Try to solve the homogeneous ODE using the ansatz $e^{\lambda t}$. Then use the sum of the homogeneous and the particular solution to find $u$.