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How to solve this 2nd differential eq?

  1. Aug 31, 2004 #1
    I was making a problem about population grow, and I wasn't able to solve this:
    x'=x*(M-x), for x(t).
    Can anyone help me?
    Thanks.
     
  2. jcsd
  3. Aug 31, 2004 #2

    ahrkron

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    I have the impresion that there is no analytical solution to that one. IIRC, its solution has a chaotic behavior depending on M (though there may be another parameter).
     
  4. Aug 31, 2004 #3

    Hurkyl

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    I think it's seperable.
     
  5. Aug 31, 2004 #4

    Tide

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    Integrate directly to find
    [tex]\frac{x}{x_0} \times \frac{M-x}{M-x_0} = e^{M t}[/tex]
    from which you can find x(t) by solving the quadratic equation.
     
  6. Sep 1, 2004 #5

    Galileo

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    scribbly scribbly...

    [tex]\frac{dx}{dt}=xM-x^2[/tex]
    [tex]\int \frac{dx}{xM-x^2}=\int dt[/tex]

    ... doesn't work, never mind...

    edit: wait,wait,wait... partial fractions:

    [tex]\frac{1}{xM-x^2}=\frac{1}{Mx}+\frac{1}{M(M-x)}[/tex]

    [tex]\int \frac{dx}{Mx}+\int \frac{dx}{M(M-x)}=\frac{1}{M}(\ln(\frac{x}{x_0})-\ln(\frac{M-x}{M-x_0}))=\frac{1}{M}\ln\left(\frac{x(M-x_0)}{x_0(M-x)}\right)[/tex]
    So:
    [tex]\frac{x}{x_0}\frac{M-x_0}{M-x}=e^{Mt}[/tex]
    Look solvable now...
     
    Last edited: Sep 1, 2004
  7. Sep 1, 2004 #6

    Tide

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    Thanks - I flipped a sign!
     
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