Derevation

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  • Here is the differential equation shown in Equation 17, rewritten slightly differently: $
   \tau \frac{dV}{dt} = -V$
  • If we want to find the solution as the system charges up to a final longterm value of $V_{\infty}$, which may be some value other than zero, we can rewrite this equation as $
   \frac{dV}{dt} = \frac{-(V-V_{\infty})}{\tau}. \quad\text{(Equation 17.1)}$ Adding a constant to the equation in this way does not affect its solution. Note how similar this is to the equation we developed for the gating variables of the Hodgkin/Huxley model (Equation 11 in unit Electrophysiology V).
  • To integrate the equation, collect similar variables on either side of the equation: $
   -\frac{dV}{V - V_{\infty}} = \frac{dt}{\tau}. \quad\text{(Equation 17.2)} $
  • Now integrate both sides, setting the limits of integration on the lefthand side from $V_{0}$ to $V_{m}(t)$, and the limits of integration on the righthand side from time 0 to time $t$. For simplicity, the constant $1/\tau$ is factored out of the righthand side, and the negative multiplier on the lefthand side of Equation 17.2 is moved over to the righthand side: $
   \int\limits_{V_0}^{V_{m}(t)}\frac{1}{V - V_{\infty}} \, dV = -\frac{1}{\tau} \int\limits_{0}^{t} \, dt. \quad\text{(Equation 17.3)} $
  • Both integrations are very straightforward. The lefthand side evaluates to $\ln (V - V_{\infty})$, and the righthand side to $t$. When these functions are evaluated at their limits, the result is $
   \ln (V_{m}(t) - V_{\infty}) - \ln (V_{0} - V_{\infty}) = -\frac{t}{\tau} + B, \quad\text{(Equation 17.4)} $ where $B$ is a constant of integration.
  • This can be simplified to $
   \ln \left(\frac{V_{m}(t) - V_{\infty}}{V_{0} - V_{\infty}}\right) = -\frac{t}{\tau} + B, \quad\text{(Equation 17.5)} $
  • We can easily determine the constant of integration by realizing that at time $t = 0$, $
   B = \ln \left(\frac{V_{0} - V_{\infty}}{V_{0} - V_{\infty}}\right) = \ln 1 = 0. \quad\text{(Equation 17.6)} $
  • Raising both sides of Equation 17.5 to the base of the natural logarithm $e$ leads to $
   \frac{V_{m}(t) - V_{\infty}}{V_{0} - V_{\infty}} = e^{-t/\tau}, \quad\text{(Equation 17.7)} $
  • Multiplying both sides by the denominator of the lefthand side, we obtain $
   V_{m}(t) - V_{\infty} = (V_{0} - V_{\infty}) e^{-t/\tau}, \quad\text{(Equation 17.8)} $
  • Solving for $V_{m}(t)$ yields $
   V_{m}(t) = V_{\infty} - (V_{\infty} - V_{0}) e^{-t/\tau}, \quad\text{(Equation 17.9)} $ which if differentiated and substituted back into Equation 17 satisfies that equation.
  • A simpler way of writing this equation is obtained if $V_{0}$ is added to and subtracted from the righthand side, and terms are collected: $
   V_{m}(t) = V_{\infty} + V_{0} - V_{0} - (V_{\infty} - V_{0}) e^{-t/\tau}, \quad\text{(Equation 17.10)} $ which equals $
   V_{m}(t) = V_{0} + (V_{\infty} - V_{0}) - (V_{\infty} - V_{0}) e^{-t/\tau}. \quad\text{(Equation 17.11)} $
  • Factoring out the term $(V_{\infty} -V_{0})$, one obtains $
   V_{m}(t) = V_{0} + (V_{\infty} - V_{0}) (1 -  e^{-t/\tau}). \quad\text{(Equation 18)} $