Please solve problem 12.6, including finding alpha.

12.6 Analytical solution for simplified action potential Show that the function v(y) = (1 + ey)-1 solves Equation 12.23, if we take the parameter Q to be given by v2/s (5 -1). Hence derive the speed of the action potential (Equation 12.24). o is another constant, which you are to find.12.2 . SIMPLIFIED MECHANISM OF THE ACTION POTENTIAL [[STUDENT VERSION. JANUAiLv 17, 2003]] 463 In short, our model displays threshold behavior: Small disturbances get driven back to v : 0, but abovethreshold disturbances drive to the other \"xed point\" '02. Our program is now to repeat the steps in Section 12.1.3, starting from step (b) on page 451 [step (a) is unchanged). b'. Equation We rst substitute Equation 12.20 into the cable equation (Equation 12.7). Some algebra shows that 011:2 : 99M / B , so the cable equation becomes Unlike the linear cable equation, Equation 12.21 is not equivalent to a diffusion equation. In general it's very diicult to solve nonlinear, many-variable differential equations like this one. But we can simplify things, since our main interest is in nding whether there are any traveling-wave solutions to Equation 12.21 at all. Following the discussion leading to Equation 12.15, a wave traveling at speed i? can be represented by a function t) of one variable, via o(:r,t) 2 t (27/29)) (see Figure 4.12b on page 120). Substituting into Equation 12.21 leads to an ordinary (one-variable) differential equation: . (12.22) 19 ETa 1111.52 (A.....)2 (122 d2 , w We can tidy up the equation by dening the dimensionless quantities 2'; E 6/02, y E 19t/x\\&mn, s : 2.12/1\