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Solving autonomous equation

  1. Jun 29, 2012 #1
    1. The problem statement, all variables and given/known data
    y*y'' = (y')² + y²*y'

    y(0) = -1/2 ,

    y'(0) = 1



    2. Relevant equations



    3. The attempt at a solution

    y' = p
    y'' = p'p

    p(p'p) = p² + y²p

    Dividing both sides by p

    p'p = p + y²

    And then I'm stuck
     
  2. jcsd
  3. Jun 29, 2012 #2

    HallsofIvy

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    Perhaps this is a typo but it should be y(p'p)= p2+ y2p

    You should have yp'= p+ y2. Also, the "p' " indicates differentiation with respect to y rather than the original independent variable.

    Now that is certainly non-trivial but at least is linear. Can you find the "integrating factor"?
     
  4. Jun 29, 2012 #3
    Thanks!

    It was not a typo, anyway, I solved yp' = p + y²:

    dp/dy - p/y = y

    Integrating factor = 1/y

    p'/y - p/y² = 1

    And that gives me y' = y(y+k)

    But that does not fit the initial conditon y'(0) = 1
     
  5. Jun 30, 2012 #4

    SammyS

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    It fits just fine with k = -3/2 .
     
  6. Jun 30, 2012 #5
    Edit: trying to solve it again
     
    Last edited: Jun 30, 2012
  7. Jun 30, 2012 #6
    I don't know what to do, can you tell me how does it fit?
     
  8. Jun 30, 2012 #7

    SammyS

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    So, you have y' = y(y+k).

    In specific that means that [itex]\displaystyle y'(0)=y(0)\left(y(0)+k\right)\ .[/itex]

    Plug the initial values into that & solve for k.

    [itex]\displaystyle (1)=\left(-\frac{1}{2}\right)\left(\left(-\frac{1}{2}\right)+k\right)[/itex]
     
  9. Jun 30, 2012 #8
    Oh!
    Of course, it didn't occur to me it was y(0) I had to use, thank you very much!
     
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