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Trying to get exact solution for a nonlinear DE

  1. May 30, 2012 #1
    Ok, this should be quite simple. I've been looking at this problem for quite some time now, and I'm tired.. Please help me!

    The equation to solve is [itex]r'[/itex] = [itex]r(1-r^2)[/itex]

    The obvious thing to do is to do partial fraction expansion and integrate(from r[itex]_{0}[/itex] to r):

    [itex] ∫ (\frac{1}{r} + \frac{1}{1-r^2})dr = t[/itex]

    After some trig substitution: [itex]ln[r\sqrt{\frac{r+1}{1-r}}] = t [/itex] (evaluated from r[itex]_{0}[/itex] to r). Then take the exponential on both sides.

    [itex]r\sqrt{\frac{r+1}{1-r}} = r_{0}\sqrt{\frac{r_{0}+1}{1-r_{0}}}e^{t} [/itex]

    This leads to kind of a nasty expression which can't be solved explicitly for [itex]r(t)[/itex] (at least it seems that it can't be solved explicitly).. So this is where I'm stuck.

    The solution to this ploblem is: [itex]r(t) = \frac{r_{0}e^{t}}{\sqrt{1+r_{0}(e^{2t}-1)}} [/itex]

    Again, this should be fairly simple, but I'm stuck.. Please help me!
     
  2. jcsd
  3. May 30, 2012 #2
    [itex] ∫ (\frac{1}{r} + \frac{1}{1-r^2})dr = t[/itex] is false. r is missing on the second fraction.
     
  4. May 31, 2012 #3
    Thank you! Got it now :)
     
  5. May 31, 2012 #4

    HallsofIvy

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    That's an incorrect "partial fraction expansion". [itex]1- x^2= (1- x)(1+ x)[/itex] so
    [tex]\frac{1}{r(1- r^2)}= \frac{1}{r}- \frac{1}{r- 1}- \frac{1}{r+ 1}[/tex]
    There is no "trig substitution" involved in integrating that. the integral is
    [tex]ln(r)- ln(r-1)- ln(r+1)+ C= ln\left(\frac{r}{r^2-1}\right)+ C[/tex]

    You have integrated incorrectly.
     
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