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How do I solve this differential equation?

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the following differential equation:


    2. Relevant equations

    2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'' = - (y')^2



    3. The attempt at a solution

    I don't know if the following is useful, but if you divide both sides by y^2, the LHS of the above becomes:

    (2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'')/y^2 = ( (2 y' y'')/y )'

    Which means the equation to solve is;

    ( (2 y' y'')/y )' = - (y')^2
     
  2. jcsd
  3. Sep 11, 2013 #2

    SteamKing

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    If you divide both sides of your original DE by y^2, it's not obvious how that turns into the equations in section 3. above.
     
  4. Sep 11, 2013 #3
    @SteamKing

    Reverse quotient rule. All I know is that what I wrote is definitely correct. I just have no idea how to solve it.
     
  5. Sep 11, 2013 #4

    Zondrina

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    Steam king is right, you didn't divide through properly at all.

    If this is your equation : ##2(y'')^2y + 2 y'''y'y - 2y''(y')^2 = -(y')^2##

    Then dividing through by ##y^2## would yield : ##\frac{2(y'')^2}{y} + \frac{2y'''y'}{y} + 2y'' {(\frac{y'}{y})}^2 = -{(\frac{y'}{y})}^2##
     
  6. Sep 11, 2013 #5
    Why are you guys so fixated on that line??

    That's not even the crux of the question I'm asking!

    lol


    I obtained this differential equation from applying the Euler-Lagrange equation to the following function:

    f(y,y') = (y')^2/y


    That's how I know my reverse quotient rule is correct.



    So how do I solve the original ODE?



    Thanks!
     
  7. Sep 11, 2013 #6

    vela

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    Because it's pointless to answer the question if you messed up in the first place, and despite your claim that you know it's right, there's an obvious algebra mistake in the third line. It's probably just a typo, but you should recognize why the helpers want to make sure you aren't wasting their time with the wrong question.
     
  8. Sep 11, 2013 #7
    6 replies and all of them useless.

    Fixated on something frivolous.


    I gave you guys the correct differential equation in the first line anyway, so there's absolutely no problem imo.
     
  9. Sep 12, 2013 #8

    vela

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    Well, half of those six replies are from you, so.... nyuk, nyuk, nyuk...

    How did you manage to get that differential equation from ##f(y,y') = \frac{y'^2}{y}##? In particular, how'd you manage to end up with a third derivative?
     
  10. Sep 12, 2013 #9
    By plugging (y')^2/y into the Euler-Lagrange Equation
     
  11. Sep 12, 2013 #10

    vela

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    You must have done it wrong.
     
  12. Sep 12, 2013 #11
    I've done it wrong because you can't solve the equation.... right....
     
  13. Sep 12, 2013 #12
    partial derivative of (y')^2/y wrt y = - (y')^2/y^2

    partial derivative of (y')^2/y wrt y' = (2 y' y'')/y

    derivative of (2 y' y'')/y wrt x = (2 y (y'')^2 + 2 y y''' y' -2 (y')^2 y'')/y^2


    finsihed. everything i've done is correct
     
  14. Sep 12, 2013 #13

    vela

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    Are you claiming you calculated
    $$\frac{d}{dt} \left(\frac{\partial f}{\partial y'}\right) = \frac{\partial f}{\partial y}?$$ Because if you are, you did it wrong.
     
    Last edited: Sep 12, 2013
  15. Sep 12, 2013 #14
    Yes that's the equation, just swop t for x.


    How have I done it wrong? I cross-checked with peers...
     
  16. Sep 12, 2013 #15

    vela

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    Your mistake is when you took the partial with respect to y'. It should be
    $$\frac{\partial f}{\partial y'} = \frac{2y'}{y}.$$ You seem to have applied the chain rule when you shouldn't have. You're not differentiating with respect to t.

    You'll end up with a differential equation that you'll easily be able to solve.
     
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