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Roots of a derivative

  1. Mar 29, 2010 #1
    So there's this ODE:
    yy''-(y')2 = y3
    After doing some work it gets to this point:
    (y')2=2y3
    Taking the root:
    y' = +-(2y3)1/2

    My question is, what do I do with the +-?
    Do I solve 2 different integrals? Assume that this is positive?
    Does it matter if it has starting conditions (y(x)=a, y'(x)=b)?
     
  2. jcsd
  3. Mar 29, 2010 #2

    tiny-tim

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    Hi manenbu! :smile:
    dunno how you got that. :confused:

    Start again.

    (Hint: can you see the anti-derivative of yy'' plus (y')2 ? if so, adapt it for minus :wink:)
     
  4. Mar 29, 2010 #3
    How I got that?

    By the substitution of y' = p(y), and then y''=p'y'=p'p.

    Then I solve whatever I get with bernoulli's method, and end up with this:
    (y')2=2y3
    After reduction of order.

    If I solve it assuming that the root is positive, I get to a solution which is correct (according to the answers). It can be also solved as a homogenous problem, reaching the same result.
    That's not the point (but if you really, want I can solve in both methods and scan the solution for you).

    The point is what I do after I get the +- thingy after taking the root.
     
  5. Mar 29, 2010 #4

    tiny-tim

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    ah i see now …

    ok, you have dy/y3/2 = ±dt√2 …

    just keep the ± :smile:

    (doesn't it get squared in the end anyway?)
     
  6. Mar 29, 2010 #5
    I need to find an expression for y, that is to integrate y'.
    So do I have 2 options for this?
     
  7. Mar 30, 2010 #6

    tiny-tim

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    (just got up :zzz: …)

    Show us what you get. :smile:

    (btw, haven't you missed out a constant of integration in your (y')2=2y3 ?)
     
  8. Apr 1, 2010 #7
    There was a starting condition, so no constant.
    Can't find the problem right now, but trust me on this. Whatever constant there was equals 0.
     
  9. Apr 1, 2010 #8

    tiny-tim

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    ok, but show us what you get. :smile:
     
  10. Apr 1, 2010 #9
    I get
    -2/y1/2 = +-21/2x + c

    I don't get your point, sorry.
     
  11. Apr 1, 2010 #10

    tiny-tim

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    (have a ± :wink:)
    ok, now square it to get y. :smile:
     
  12. Apr 2, 2010 #11
    I found the original problem!


    yy''-(y')2 = y3
    y(0) = 2
    y'(0) = 4

    Now, if we continue from where you stopped, squaring won't help because I'm squaring ±x + c and not just ±x.
    c does not equal zero.
     
  13. Apr 2, 2010 #12

    tiny-tim

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    erm :redface: … (±x + c)2 = (x ± c)2 :wink:
     
  14. Apr 2, 2010 #13
    oh - and then it's a ± before the constant, so I can drop it.

    Now I figured it out.
    thanks a lot :)
     
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