How Do You Solve the Verhulst Equation for Logistic Population Growth?

[gpax42]

Hi all, it's been a over a year since I took my differential equations and linear algebra course and I'm currently enrolled in a class that assigned this problem as a sort of refresher on analytically solving differential equations. I can't seem to remember the proper approach to going about all this and I'm no longer in possession of my textbook from the previous course. Any help would be greatly appreciated

Homework Statement

Using methods for solving differential equations exactly, solve the Verhulst equation for logistic population growth

\frac{dP}{dt}= rP(1 - P/K)

where r is the growth rate and K is the carrying capacity.


The attempt at a solution

I began by altering the equation so as to create a more solvable form...

Dividing through by K on both sides gives me

\frac{d}{dt}\frac{P}{K} = r\frac{P}{K}(1 - P/K)

I then set x = P/K

\frac{dx}{dt} = rx(1-x)

at this point I know I need to integrate both sides

dx = rx(1-x)dt
following integration...
x(t) = [rx(1-x)]t + C

is this even remotely correct?

Thanks again for any help you can lend me

-gpax42

 

[LCKurtz]

Hi all, it's been a over a year since I took my differential equations and linear algebra course and I'm currently enrolled in a class that assigned this problem as a sort of refresher on analytically solving differential equations. I can't seem to remember the proper approach to going about all this and I'm no longer in possession of my textbook from the previous course. Any help would be greatly appreciated

Homework Statement

Using methods for solving differential equations exactly, solve the Verhulst equation for logistic population growth

\frac{dP}{dt}= rP(1 - P/K)

where r is the growth rate and K is the carrying capacity.


The attempt at a solution

I began by altering the equation so as to create a more solvable form...

Dividing through by K on both sides gives me

\frac{d}{dt}\frac{P}{K} = r\frac{P}{K}(1 - P/K)

I then set x = P/K

\frac{dx}{dt} = rx(1-x)

at this point I know I need to integrate both sides

dx = rx(1-x)dt
following integration...
x(t) = [rx(1-x)]t + C

is this even remotely correct?

-gpax42

No, it isn't. First, essentially changing the dependent variable from P to x accomplishes nothing except renaming it. Your x is an unknown function of t and its integral isn't xt.

You want to review separation of variables. Separate the P and t:

\frac {dP}{P(1-P)}= \frac r K dt

Then integrate both sides. You will want to use partial fractions on the left side.

 

[gpax42]

ahhh I see... when I separated variables, however, I found...

\intdP/[P(1-P/k)] = \intrdt

I then manipulated the left side to read

K/[P(K-P)] ... after partial fractions... = 1/P + 1/(K-P)

int[1/P + 1/(K-P) )dP] = int[rdt]

following through with integration leads to

ln(abs(P/(K-P))) = rt + C

e^(rt + C) = abs[P/(K-P)]

e^(-rt-C) = abs[(K-P)/P]

+/- e^(-rt) * e^(-C) = K/P - 1

if you let M = +/- e^(-C)
then
M*e^(-rt) = K/P - 1

solving further...


P(t) = K / (1+Me^(-rt))

at t = 0 ; P = P₀

P₀ = K / (1+M)

M =( K - P₀ ) / P₀