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What ODE are these called?

  1. Nov 13, 2011 #1
    Where n is a natural number, so we get polynomials of derivatives like

    [tex]\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^n + \left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{n-1} + \left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{n-3}... = 0[/tex]

    Has some ancient greek guy managed to give a name and techniques on how to solve this?

    Do ODEs become nearly impossible if I throw in a g(x,y) or h(x,y) in there? That is

    [tex]H(x,y)\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^n + G(x,y)\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{n-1} + \left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{n-3}... = 0[/tex]

    I imagine it would because we don't even have a "clean" formula for solving cubics.

    Is there a method if it was quadratic?

    How bad do the G(x,y) and H(x,y) messes things?
     
  2. jcsd
  3. Nov 14, 2011 #2

    HallsofIvy

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    Generally speaking, "ancient Greek guys" have given no names at all to things involving derivatives, because derivatives were not invented until around 1700. And as for more modern mathematicians, things tend to get named only if they are useful. I can see nothing useful about that formula.
     
  4. Nov 14, 2011 #3
    Yes we do. We have clean formulas for polynomials up to degree 4 (above which a "clean" formula is impossible).
     
  5. Nov 14, 2011 #4
    I've seen the formula, it's big and unuseful...

    Well that's not right, I am sure there are ODEs that have that form.
     
  6. Nov 14, 2011 #5
    Of course there are, at least you have just invented it :p
    Although that doesn't mean they are useful.

    Actually, the first one - without functions H,G...- doen't really need any special care.
    dy/dx are just numbers, so you need just to solve equation x^n + x^{n-1} + ... = 0 .
    If there are some real solutions x=s, you just pick one and the solution to the ODE is the function dy/dx=s.

    The one with functions H,G.. probably should have non-trivial solutions in some cases (on of the conditions is probably that H,G... have to be continuos), but again, solving it amounts to solving normal polynomial equation first and then solving equation of kind dy/dx=f(x,y), where f(x,y) is solution to the polynomial equation. (If I am not wrong:)
     
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