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Why a explicit ODE is explicited with y of higher grade?

  1. Mar 27, 2014 #1
    Given a implicit ODE like F(x, y(x), y'(x), y''(x)) = 0, why your explicit form is y''(x) = f(x, y(x), y'(x))? Why a ODE is explicited always with y of higher grade?
     
  2. jcsd
  3. Mar 27, 2014 #2

    Mark44

    Staff: Mentor

    What they are saying is that starting with an equation F( ... ) = 0 that involves x, y(x), y'(x), and y''(x), where y''(x) is given implicitly, a new equation can be written that gives y''(x) explicitly as a function of x and the lower-order derivatives.

    A very simple example would be y'' - 2y' + 2y = 0. Here the left side is F(x, y, y', y'').
    With y'' given explicitly, we have y'' = 2y' - 2y. Here the right side is f(x, y, y').
     
  4. Mar 27, 2014 #3
    ?

    You gave an example for my question, but not an answer...
     
  5. Mar 27, 2014 #4

    Mark44

    Staff: Mentor

    AFAIK, this is partly convention and partly convenience.
     
  6. Mar 27, 2014 #5

    micromass

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    2016 Award

    Right, it's to make things simpler. Solving a general equation of the type

    [tex]F(t,y,y^\prime,...,y^{(n)}) = 0[/tex]

    is pretty horrible and there are very little techniques known. On the other hand,

    [tex]y^{(n)} = f(t,y,...,y^{(n-1)})[/tex]

    is way more manageable and more results about these equations are known. Also, most equations in practice show up in this form.
     
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