Question

The Fitzhugh-Nagumo model for the electrical impulse in a neuron states that, in the absence of relaxation effects, the electrical potential in a neuron v(t) obeys the differential equation dv dt = −v[v2 − (1 + a)v + a] where a is a positive constant such that 0 < a < 1. (a) For what values of v is v unchanging (that is, dv/dt = 0)? (Enter your answers as a comma-separated list.) v

Answer 1

Answer:

v = 0, 1, a

Explanation:

dv/dt = v(v² - (1 + a)v + a)

When dv/dt = 0

=> 0 = v(v² - (1 + a)v + a)

=> v = 0 and (v² - (1 + a)v + a) = 0

Open up the bracket:

v² - v - av + a = 0

v(v - 1) - a(v - 1) = 0

(v - a) (v - 1) = 0

=> v = a and v = 1

Hence, the values of v for which dv/dt can be 0 are v = 0, 1, a.

Answer 2

In the Fitzhugh-Nagumo model the values of the v for the electric impulse model are 0,1 and a.

What is electric impulse?

The electric impulse is the very little electric signal, which is sent along a medium.

The differential equation for the electric potential in a neuron is given as,

$\dfrac{dv}{dt} = -v[v^2 - (1 + a)v + a]$

Here, a is a positive constant such that

$0 < a < 1$

At the point where,  dv/dt = 0. the value of v has to be find out.

Put this value in the above equation,

$0 = -v[v^2 - (1 + a)v + a]$

Solve the above equation as,

The  

$-v[v^2 - (1 + a)v + a]=0\\-v[v^2-v-va+a]=0\\-v[v(v-1)-a(v-1)]=0\\-v[(v-1)(v-a)]=0$

Equating all the factors equal to zero we get,

$v=0,1,a$

Thus, the values of the v for unchanging v are 0,1 and a.

Learn more about the electric impulse here;

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