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Homogeneous Differential Equations

  1. Mar 21, 2005 #1
    Hi,

    I need some help in finding whether this differential equation is homogeneous or not.

    3 (d^2 y / dx^2) + x (dy/dx)^2 = y^2

    I know that for example,

    x^2 dx + xy dy = 0 is homogeneous. But how can I deal with the equation that has (d^2 y / dx^2) and (dy/dx)^2 ?

    Thanks
     
  2. jcsd
  3. Mar 21, 2005 #2

    HallsofIvy

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    Is your question just to determine if the equation is homogeneous or not? If so, the fact that the equation is non-linear is not relevant: yes it is homogeneous because it does not have any terms which do NOT involve y or one of its derivatives.
    (That's the advantage of knowing the DEFINITION rather than just some examples.)

    Of course, the fact that it is non-linear pretty much means being homogeneous doesn't make it any easier to solve!
     
  4. Mar 21, 2005 #3

    saltydog

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    Just to format it:

    [tex]3 \frac{d^2y}{dx^2}+x(\frac{dy}{dx})^2-y^2=0[/tex]

    Hummmmm . . .
     
  5. Mar 21, 2005 #4
    Indeed, homogeneous but additionally nonlinear. Quite analytically insoluble, though.
     
  6. Mar 21, 2005 #5
    The simplist way to answer the question of homogeneity is to ask:

    Is Y(x) = 0 a solution?

    If the answer is yes, then the equation is homogeneous.
     
  7. Mar 21, 2005 #6

    Hurkyl

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    Hrm, does it really make sense to ask if a nonlinear DE is homogenous? I don't have a general definition handy, and Mathworld only defines homogeneity for linear differential equations.
     
  8. Mar 21, 2005 #7

    dextercioby

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    Probably Mathworld gives attempts to solve it,too...Is a nonlinear algebraic system either homogenous or nonhomogenous...?:bugeye:

    Daniel.
     
  9. Sep 3, 2007 #8
    Hey i need some help finding the general solution of

    ydy= (-x+ √(x^2 + y^2))dx

    by using the substitution y= vx and then the substitution u^2= 1 + v^2

    It would be great if someone could help.
     
  10. Sep 3, 2007 #9

    HallsofIvy

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    Do not, do not, do not "hijack" someone else's thread for a new question. It's very easy to start a thread of your own!

    In fact, I'm going to do that for you.
     
  11. Nov 9, 2007 #10
    your equation is not homogeneous:
    It follows from k-2m=2k-m=2k, so k=0, m=0
     
  12. Nov 10, 2007 #11

    Astronuc

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    The original thread was resolved 30 months ago, or 32 months ago this month. :biggrin:

    Halls of Ivy is correct. A new thread is appropriate for a new problem.
     
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