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Nasty differnation variables things

  1. Jun 11, 2008 #1
    A population grows in such a way that the rate of change of the population P at time t in days is proportional to P.

    a) Write down a dfferential equation relating P and t

    b) Show, by solving this equation, that the general solution of this equation may be written as [tex]P = Ak^{t}[/tex], where A and k are positive constants.


    a) is easy:

    dP/dt = kP

    b) I dont know where to start

    Can someone walk me through B please :)
     
  2. jcsd
  3. Jun 11, 2008 #2

    rock.freak667

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    [tex]\frac{dP}{dt} = kP[/tex]

    [tex]\Rightarrow \frac{1}{P}\frac{dP}{dt}=k[/tex]


    integrate both sides w.r.t. t
     
  4. Jun 11, 2008 #3
    Can you show me how. I dont know how to intergrate this equation

    Thanks :)
     
  5. Jun 11, 2008 #4
    hang on this is one of those stupid natural log ones

    i know if y = a ^ x then dy/dx = a^x ln a

    but how does that help?
     
  6. Jun 11, 2008 #5
    this is seperation of variables?
     
  7. Jun 11, 2008 #6

    rock.freak667

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    Yes

    [tex]\int \frac{1}{P}dP= \int k dt[/tex]



    are you able to do the left side?
     
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