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Separation of Variables

  1. Jan 19, 2012 #1
    [tex]\displaystyle\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)[/tex]


    [tex]\displaystyle\int\frac{KdN}{N\left(K-N\right)} = \int rdt[/tex]

    [tex]\displaystyle K\int\frac{dN}{N}-K\int\frac{dN}{K-N}=r\int dt[/tex]

    Now, I obtain:

    [tex]K\ln\left(\frac{N}{K-N}\right) = rt+c[/tex]

    [tex]\left(\frac{N}{K-N}\right)^K=C_0r^{rt}[/tex]

    The final solution is [tex]N(t) =\frac{C_0Ke^{rt}}{K+C_0(e^{rt}-1)}[/tex]

    I don't see how I can manipulate my equation to that. Is there a mistake or am I not seeing something.
     
    Last edited: Jan 19, 2012
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  3. Jan 19, 2012 #2

    Dick

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    On the left side, you seem to have changed [itex]\frac{K}{N (K-N)}[/itex] into [itex]\frac{K}{N} - \frac{K}{K-N}[/itex]. That's not right at all.
     
  4. Jan 19, 2012 #3
    It should be [tex]\frac{N}{K-N} = C_0e^{rt}[/tex]

    Correct?
     
  5. Jan 19, 2012 #4

    Dick

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    You didn't do partial fractions correctly the first time. That looks like a step in the right direction, yes.
     
  6. Jan 19, 2012 #5
    There is still something wrong...

    From that, I end up with:

    [tex]N = KC_0e^{rt}-NC_0e^{rt} \Rightarrow N(t)=\frac{KC_0e^{rt}}{1+C_0e^{rt}}[/tex]

    I have a sign issue and I am missing a K.
     
  7. Jan 19, 2012 #6

    SammyS

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    It's not true that
    [itex]\displaystyle \int\frac{KdN}{N\left(K-N\right)} =K\int\frac{dN}{N}-K\int\frac{dN}{K-N}[/itex]​
    You cannot break up an integral that way.

    [itex]\displaystyle\frac{1}{N}+\frac{1}{K-N}\ne\frac{1}{N(K-N)}\,,[/itex]

    However, [itex]\displaystyle \frac{1}{K}\left(\frac{1}{N}+\frac{1}{K-N}\right)=\frac{1}{N(K-N)}\,,[/itex] which is a result that can be obtained using partial fraction decomposition.
     
  8. Jan 20, 2012 #7

    I know, I have fixed that part.
     
  9. Jan 20, 2012 #8
    But I can't figure out how to manipulate that into

    [tex]N(t)=\frac{N_0Ke^{rt}}{K+N_0(e^{rt}-1)}[/tex]

    I forgot to mention that [tex]N(0)=N_0[/tex]
     
  10. Jan 20, 2012 #9

    Dick

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    Use the form you derived before.
    [tex]N(t)=\frac{KC_0e^{rt}}{1+C_0e^{rt}}[/tex].
    That tells you
    [tex]N_0=\frac{KC_0}{1+C_0}[/tex]
    That's the relation between the two constants in the two different forms.
     
  11. Jan 20, 2012 #10

    I have all that, but I still can't get to the final form. I tried solving for C but that didn't go anywhere fruitful.
     
  12. Jan 20, 2012 #11

    Dick

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    You seem to have exactly the right idea of what to do. Solve for C_0 in terms of N_0 and substitute that back into N(t). It works for me. If it doesn't go anywhere fruitful for you can you show why not?
     
  13. Jan 20, 2012 #12
    Just to verify, this what you get when you substitute correct?

    [tex]\displaystyle N(t)=\frac{K\left(\frac{N_0}{K-N_0}\right)e^{rt}}{1+\left(\frac{N_0}{K-N_0}\right)e^{rt}}[/tex]
     
  14. Jan 20, 2012 #13

    Dick

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    Sure. Continue.
     
  15. Jan 20, 2012 #14
    I got it. I don't know what the problem was earlier.

    Thanks.
     
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