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Math Help - logistic model

  1. #1
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    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????
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by razorfever View Post
    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.
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  3. #3
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    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???
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  4. #4
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    Quote Originally Posted by razorfever View Post
    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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