- #1
gpax42
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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
[tex]\frac{dP}{dt}[/tex]= rP(1 - [tex]P/K[/tex])
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
[tex]\frac{d}{dt}[/tex][tex]\frac{P}{K}[/tex] = r[tex]\frac{P}{K}[/tex](1 - [tex]P/K[/tex])
I then set x = [tex]P/K[/tex]
[tex]\frac{dx}{dt}[/tex] = 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
Homework Statement
Using methods for solving differential equations exactly, solve the Verhulst equation for logistic population growth
[tex]\frac{dP}{dt}[/tex]= rP(1 - [tex]P/K[/tex])
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
[tex]\frac{d}{dt}[/tex][tex]\frac{P}{K}[/tex] = r[tex]\frac{P}{K}[/tex](1 - [tex]P/K[/tex])
I then set x = [tex]P/K[/tex]
[tex]\frac{dx}{dt}[/tex] = 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