# logistic model

• Feb 12th 2009, 12:51 PM
razorfever
logistic model
the logistic model:

dN/dt = rN* ( 1 - (N/K)) , N(0) = No

where No is (N subscript 0)

by making the substitution N = 1/y and dN/dt = -1/y^2 dy/dt in the logistic model solve for y(t) and hence show that the solution is:

N(t) = NoK / [ No + (K - No)e^(-rt) ]

how do i do this????
• Feb 12th 2009, 12:57 PM
Jhevon
Quote:

Originally Posted by razorfever
the logistic model:

dN/dt = rN* ( 1 - (N/K)) , N(0) = No

where No is (N subscript 0)

by making the substitution N = 1/y and dN/dt = -1/y^2 dy/dt in the logistic model solve for y(t) and hence show that the solution is:

N(t) = NoK / [ No + (K - No)e^(-rt) ]

how do i do this????

what do you mean? they told you exactly what to do. use the substitution, you get

$\frac 1y~\frac {dy}{dt} = \frac ry \left( 1 - \frac 1{Ky} \right)$

now solve this diff eq. it is separable. once you solve for y, you can find N by doing a back-substitution.
• Feb 12th 2009, 02:54 PM
razorfever
i did what u said and got the solution in the form

N = 1 / [(1/k) + (Ce^(-rt))] with C being the constant

how do i get it in the form in the question with No???
• Feb 12th 2009, 04:43 PM
HallsofIvy
Quote:

Originally Posted by razorfever
i did what u said and got the solution in the form

N = 1 / [(1/k) + (Ce^(-rt))] with C being the constant

how do i get it in the form in the question with No???

You were told in the problem that N(0)= No.

$N(0)= No= 1/[(1/k)+ C]$ since $e^{0}= 1$. Solve for C in terms of No and k.
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