FirstOrderEquationDerivation

• The initial form of the equation is $

   \frac{dn}{dt} = \alpha_{n} (1 - n) - \beta_{n} n. \quad\text{(Equation 10)}$

• Let us focus on the right hand side. If we multiply the first rate constant through, we get $

   \alpha_{n} *1 - \alpha_{n}  n - \beta_{n} n. \quad\text{(Equation 10.1)}$

• Let's group together the factors multiplying the gate probability n: $

   \alpha_{n} - (\alpha_{n}  + \beta_{n}) n. \quad\text{(Equation 10.2)}$

• Let's factor out the term multiplying the gate probability n: $

   (\alpha_{n}  + \beta_{n}) (\frac{\alpha_{n}}{(\alpha_{n}  + \beta_{n})} -  n). \quad\text{(Equation 10.3)}$

• We can re-write this by turning the multiplying factor into a factor that divides the right hand side, if we recall that $x = 1/(1/x)$: $

   \frac{ (\frac{\alpha_{n}}{(\alpha_{n}  + \beta_{n})} -  n)}{\frac{1}{(\alpha_{n}  + \beta_{n})}}. \quad\text{(Equation 10.4)}$

• We can simplify the right hand side by defining $

   \frac{\alpha_{n}}{(\alpha_{n}  + \beta_{n})} = n_{\infty} \quad\text{(Equation 10.5)}$ and by defining $
   \frac{1}{(\alpha_{n}  + \beta_{n})} = \tau_{n} \quad\text{(Equation 10.6)}$ which allows us to re-write Equation 10.4 as $
   \frac{(n_{\infty} -  n)}{\tau_{n}}. \quad\text{(Equation 10.7)}$ and this allows us to re-write the original equation as $
   \frac{dn}{dt} = \frac{(n_{\infty} -  n)}{\tau_{n}}. \quad\text{(Equation 11)}$