I'd like to compute the following derivative (i.e., solve for $v$):
\begin{align} \frac{dv}{dt} = \frac{-v(t) + I_{rec}(t) + I_{ext}(t)}{\tau_m}. \end{align}
I know that if I had $\frac{dv}{dt}=\frac{-v(t)}{\tau_m}$, I'd have $v(t) = e^{-t/\tau_m}$. But it's not clear to me what to do here.
Any help would be much appreciated.
On
In principle you can use an integrating factor.
Multiply the differential equation across by $M(t)=e^{(t-t_0)/\tau_m}$ (the integrating factor) and it becomes a total derivative:
$${d\over dt}(e^{(t-t_0)/\tau_m}v(t))=\frac{I_{req}(t)+I_{ext}(t)}{\tau_m}e^{(t-t_0)/\tau_m}$$
So then $$v(t)=\frac{1}{\tau_m}e^{-t/\tau_m}\int_{t_0}^t e^{t'/\tau_m}\left(I_{req}(t')+I_{ext}(t')\right)\,\mathrm{d}t'+e^{-(t-t_0)/\tau_m}v(t_0)$$
Although in principle you'd need to know something about the $I$'s in order to integrate, and $t_0$ is some initial time you know about.
You have an equation of the form $$ v'(t) + av(t) = f(t), $$ for some $a \in \mathbb{R}$. The homogeneous solution is indeed $v(t) = Ce^{-t/a}$. Now you need to pick a particular solution $v_p(t)$ of this equation, and the general solution would be given by $$v(t) = Ce^{-t/a} + v_p(t).$$
Picking the particular solution can be done by guessing if the form of $f(x)$ is suggestive, or by variation of parameters.