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Solving ordinary differential equations [growth equation]

Posted 10 years ago

Hello, Im having trouble trying to solve this equation(question). I used Dsolve but I think im inputting it wrong.

dn/dt=r n(1-n/k)

r is called the growth rate, and k is called the carrying capacity. n is the population, t is time. initial condition n = n0 at t = 0.

------> What is the symbolic solution? Also, If k = 100, n0 = 1 and r = 0.1, how long(t) until n = 50?

What I've tried...

DSolve[Derivative[1][n][t] == n^2*(1 - n/k)*r, n[t], t]

== ordinary differential equation of dn/dt = rn(1 - n/k)

Help?

POSTED BY: Brendan John
6 Replies

You need to make n a function of t everywhere.

POSTED BY: Frank Kampas
Posted 10 years ago

How do I do that?

POSTED BY: Brendan John

Replace "n" with "n[t]".

So n^2(1 - n/k)r should be written n[t]^2(1 - n[t]/k)r.

POSTED BY: Sean Clarke
Posted 10 years ago

Cheers, I get...

{{n[t] -> InverseFunction[Log[#1]/k^2 - Log[-k + #1]/k^2 - 1/(k #1) &][(r t)/ k + C[1]]}}

Look right?

POSTED BY: Brendan John
POSTED BY: Sean Clarke

Dear Brendan,

Please take into account that you are trying to solve a wrong equation: be aware of the n[t]^2 factor which should be n[t].

Just try this:

DSolve[D[n[t], t] == r n[t] (1 - n[t]/k), n[t], t][[1]]

with the answer of Mathematica 10.0.1

{n[t] -> (E^(r t + k C[1]) k)/( E^(r t + k C[1])-1)}

although the solution you might desire is

{n[t] -> (E^(r t + k C[1]) k)/( E^(r t + k C[1])+1)}

Best regards,

POSTED BY: Arturo Ramos
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