integrate and fire

$V(t)$ is ‘membrane potential’ of a neuron
$f(V)$ is a decreasing linear fn of its current value, with $f(0)=20V/sec$

neuron fires when $V(t)=V_{fire}=-0.05V$
afterwards, returns to resting value $V_{rest}=-0.075V$
baseline frequency of firing is $200Hz$

what should slope of $f(V)$ be?

$f(V)=mV+20$
$\frac{dV}{dt}=mV+20$

$dt=\frac{1}{mV+20}dV$
$\int dt=\int \frac{1}{mV+20}dV$

$\int \frac{1}{mV+20}dV=\frac{1}{m}\ln |mV+20|$
$t+C=\frac{1}{m}\ln |mV+20|$
$mt+C=\ln |mV+20|$
$e^{mt+C}=mV+20$
$C{e}^{mt}=mV+20$
$V(t)=\frac{C{e}^{mt}-20}{m}$
$V(t)=C{e}^{mt}-\frac{20}{m}$

$V(0)=V_0=C{e}^{mt}-\frac{20}{m}$
$V_0=C-\frac{20}{m}$
$C=V_0+\frac{20}{m}$

$V(t)=(V_0+\frac{20}{m})e^{mt}-\frac{20}{m}$
assuming $V_0=V_{rest}$
$V(t)=(-0.075+\frac{20}{m})e^{mt}-\frac{20}{m}$
and with $200Hz$

$V(\frac{1}{200})=-0.05V=(-0.075+\frac{20}{m})e^{\frac{1}{200}m}-\frac{20}{m}$
$m=231.278$

so $f(V)=231.278V+20$

how does firing frequency depend on $f(V)$?

$V(t)=(-0.075+\frac{20}{m})e^{mt}-\frac{20}{m}$
$\frac{V+\frac{20}{m}}{e^{mt}}=-0.075+\frac{20}{m}$
$e^{mt}=\frac{V+\frac{20}{m}}{-0.075+\frac{20}{m}}$
$t=\frac{\ln (\frac{V+\frac{20}{m}}{-0.075+\frac{20}{m}})}{m}$

$t=\frac{\ln (\frac{V_{fire}+\frac{20}{m}}{-0.075+\frac{20}{m}})}{m}$ for firing based on some slope $m$ for $f(V)$