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Modeling Population Growth [dP/dt = k P - A P2 - h]

  1. May 18, 2013 #1
    Does anyone know how to solve dP/dt = k P - A P2 - h for P. I understand partial fractions are needed and I have already solved dP/dt = k P - A P2. Is anyone able to solve it, Cheers NZBRU.
     
  2. jcsd
  3. May 18, 2013 #2
    Hi.

    dt = dP/(k P - A P^2 - h) = -1/A dP/[ (P- k/A)^2-{(k/A)^2+h} ] = -1/A dp/[p^2 - {(k/A)^2+h}) ] , p= P- k/A

    = -1/A dp/[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]

    now you can integrate.
     
  4. May 18, 2013 #3
    If I typed that in correctly the line would be [-1/A [dp]/[[p - sqrt{(k/A)^2+h} ][p +sqrt {(k/A)^2+h} ]]]=dt or would it be: [-1/A [dp] [p +sqrt {(k/A)^2+h} ]/[p - sqrt{(k/A)^2+h} ]]=dt? (I have not used ASCIIMath extensively). Thank you for the fast response.
     
  5. May 18, 2013 #4
    Hi !
    integration gives t(P)
    The inverse fonction is P(t)
     

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