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Population Growth

  1. Jan 22, 2006 #1
    I am quite sure the first approach is to use partial fractions but I am
    unclear how to finish this equation

    1/P*dP/dt=b+aP
     
  2. jcsd
  3. Jan 23, 2006 #2

    HallsofIvy

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    How to finish it? You haven't started it!

    [tex]\frac{1}{P}\frac{dP}{dt}= b+aP[/tex]
    is separable:
    [tex]\frac{dP}{P(b+aP)}= dt[/tex]
    Which I'm sure you knew since you ask about partial fractions.
    We can write
    [tex]\frac{1}{P(b+ aP)}= \frac{A}{P}+ \frac{B}{b+ aP}[/tex]
    for some A and B. Multiply both sides by P(b+ aP):
    [tex]1= A(b+ aP)+ BP[/tex]

    What do you get if P= 0? What do you get if P= -a/b? Put those values for A and B into the fractions and integrate.
     
  4. Jan 23, 2006 #3
    Solve P=? or t=?

    I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:

    A=1/b and B=-a/b. And I know both a and b from a linear regression.

    So substituting back I get

    ((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides

    (1/b)lnP +(-1/b)ln(b+aP)=t +C

    Now my question is here how do I solve for P= f(t).
     
  5. Jan 24, 2006 #4

    HallsofIvy

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    The rest is algebra:
    [tex]\frac{1}{b}ln P- \frac{1}{b}ln(b+aP)= ln\left(\frac{P}{b+aP}\right)^\frac{1}{b}= t+ c[/tex]
    Take exponential of both sides:
    [tex]\left(\frac{P}{b+aP}\right)^\frac{1}{b}= e^{t+ C}= C'e^t[/tex]
    (C= eC)
    Take ath power of both sides:
    [tex]\frac{P}{b+aP}=C"e^{at}[/tex]
    (C"= C'a)
    multiply both sides by b+ aP and expand:
    [tex] P= C"e^{at}(b+ aP)= C"be^{at}+ aC"e^{at}P[/tex]
    [tex] P- aC"e^{at}P= P(1- aC"e^{at})= C"be^{at}[/tex]
    [tex] P= \frac{C"be^{at}}{1- aC"e^{at}[/tex]
     
  6. Jan 24, 2006 #5

    mathwonk

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    i thionk that logistic equation is more usefully written as dP/dt

    = aP(1 - P/N).

    of course it is separable and solving it, shows that the populaion approaches N as time goes on.
     
  7. Jan 24, 2006 #6

    saltydog

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    Starting with:

    [tex]\frac{1}{b}ln(P)-\frac{1}{b}ln(b+aP)=t+c[/tex]

    multiplying by b and collecting logarithms:

    [tex]ln\left[\frac{P}{b+aP}\right]=b(t+c)[/tex]

    Taking exponentials:

    [tex]\frac{P}{b+aP}=e^{b(t+c)}[/tex]

    multiplying both sides by b+aP and collecting the P's:

    [tex]P\left[1-ae^{b(t+c)}\right]=be^{b(t+c)}[/tex]

    Isolating the P and then multiplying the top and bottom of the rational expression by [itex]e^{b(t+c)}[/itex] to make it cleaner leaves:

    [tex]P(t)=\frac{b}{e^{-b(t+c)}-a}[/tex]

    Now, how about completely characterizing the solutions in terms of a and b assuming some initial condition like P(0)=1. What would it look like whatever it was? (just a suggestion :smile: )
     
    Last edited: Jan 25, 2006
  8. Jan 25, 2007 #7
    I am working on the same problem; however I started out a bit different. Follwing the above; now all I need to do is substitue my a and b values; my initial condition is P sub 0 = 3.9
    Am I going in the right direction?
     
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