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Homework Help: Differential Equation (3)

  1. Feb 26, 2010 #1
    Solve the Differential Equation [tex]2yy\prime\prime+2xy\prime=0[/tex]

    set [tex]p=y\prime[/tex], and then it becomes case 1 in the textbook

    Can someone please help me solve this? It's a night mare
     
    Last edited: Feb 26, 2010
  2. jcsd
  3. Feb 26, 2010 #2

    ideasrule

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    Doesn't that simplify to y=-x?
     
  4. Feb 26, 2010 #3
    Did you consider the
    [tex]y\prime[/tex] that it's not the same as [tex]y[/tex]
     
  5. Feb 26, 2010 #4

    vela

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    What's your point? y' is a common factor in the equation you wrote. Did you not post the problem correctly?
    You do realize we don't know what textbook you're using and even less likely to have a copy, right?
     
  6. Feb 26, 2010 #5
    Okay, objection 1 makes no sense... but I'll give you you're right for objection 2.

    Here's what the textbook says for case 1:

    [tex]y\prime=p, y\prime\prime=p\prime\mbox {Dependent variable y missing}[/tex]
     
    Last edited: Feb 26, 2010
  7. Feb 26, 2010 #6

    vela

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    Really? You don't see how x=-y is a solution to 2yy'+2xy'=2y'(x+y)=0?
     
  8. Feb 26, 2010 #7
    How did you get that solution besides canceling things out?

    Cause, this is like a very annoying class, like what method did you use? Did you use [tex]p\equiv{y}\prime[/tex]
     
    Last edited: Feb 26, 2010
  9. Feb 26, 2010 #8

    vela

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    You get that solution by canceling things out. It's not the only solution to the original equation, however.
     
  10. Feb 26, 2010 #9
    could you possibly do it, by using the substitution I posted?
     
  11. Feb 26, 2010 #10

    vela

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    I don't see what that substitution buys you. Also, it's not your textbook's case 1 as y appears in the differential equation.
     
  12. Feb 26, 2010 #11
    ok I posted this wrong, i feel stupid

    [tex]2yy\prime\prime+2xy\prime=0[/tex]

    how do I solve this?

    give me a hint maybe?
     
  13. Feb 27, 2010 #12

    vela

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    I don't know. Maybe someone else here has a suggestion.
     
  14. Feb 27, 2010 #13
    What's the point of the 2's? I would have just canceled them out from the beginning.

    (Y*Y'')/Y' = -X

    (Y''/Y')(Y) = -X

    Set P = Y', then:

    (P'/P)Integral(P) = -X

    Ok, I'm sorry. This is going nowhere. Given me 24 hours to think about it.
     
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