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Very simple ODE

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]y'=1-y^2[/itex]


    2. Relevant equations

    I've tried a few methods, including change of variables, separability. I'm pretty sure it's separable.


    3. The attempt at a solution

    If I use separation of variables, I end up with

    [itex]\frac{\text{dy}}{1-y^2}=\text{dx}[/itex]

    This integral evades me. There must be another trick that I am missing.
     
  2. jcsd
  3. Nov 26, 2013 #2

    pasmith

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    Partial fraction decomposition
     
  4. Nov 26, 2013 #3
    It appears that the integral of [itex]\frac{\text{dy}}{1-y^2}[/itex] is [itex]\tanh ^{-1}(y)[/itex].

    This would make the solution [itex]y=\frac{e^{2 c_1}+e^{2 x}}{e^{2 x}-e^{2 c_1}}[/itex]
     
  5. Nov 26, 2013 #4
    That's a good idea! I tried partial fraction decomposition, and I ended up with

    [itex]0.5 (c+\log (1-y)+\log (y+1))=c+x[/itex]
    [itex]\log (c (1-y) (y+1))=2 c+2 x[/itex]
    [itex]c (1-y) (y+1)=e^{2 c+2 x}[/itex]
    From here, it looks like it gets into a hairy quadratic formula with solutions of the form
    [itex]y=\pm \sqrt{-e^{2 c+2 x}-1}[/itex]

    If I just recognized the solution was a form of hyberbolic inverse tangent, it would have been much easier, though, I do not see how I would have seen that without guessing. I did not have that in my memory.
     
  6. Nov 26, 2013 #5

    haruspex

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    You don't need to put a constant of integration both sides (and certainly not the same one!).
    You have a sign error in the integration. You should end up with a ratio of the y terms, not the product.
     
  7. Nov 26, 2013 #6
    I do not see where the mixup is. If I expand the term into partial fractions, I end up with...
    [itex]\frac{1}{1-y^2}\to \frac{1}{(1-y) (y+1)}\to \frac{A}{y+1}+\frac{B}{1-y}\to \frac{A (1-y)+B
    (y+1)}{(1-y) (y+1)}\to A+B=1, B-A=0\to A=B=0.5[/itex]

    If that's the case for A and B, then I simply move on to... Aaaaand I see what I did. Integrating the [itex]\frac{0.5}{y+1}+\frac{0.5}{1-y}[/itex] results in a difference instead of a sum. However, I am even more baffled as to how to isolate y algebraically in this case, since I end up with...

    [itex]e^{2 c+2 x}=\frac{y+1}{1-y}[/itex]
     
  8. Nov 26, 2013 #7

    haruspex

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    It's quite easy - you're probably overthinking it
    Multiply out and collect up the y terms.
     
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