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ODEs: Not sure why

  1. Jul 14, 2005 #1
    I was reading something about ODEs and I came across a section which discusses the generalisation of first-order isobaric equations to equations of general order n. The definition I have is that an n-th order isobaric equation is one in which every term can be made dimensionally consistent upon giving y and dy each a weight m, and x and dx each a weight of 1.

    The "weights" referred to are the same as those for first order isobaric equations so I follow that part. What I don't understand is the process used to determine the weights of each term in an isobaric ODE. For example:

    [tex]
    x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0
    [/tex]..eq1.

    The book says that from left to right, the "weights" of each term on the LHS are m+1, m+1, 2m, 2m, m + 1.

    From an earlier section, an example of a first order isobaric equation is:

    [tex]
    \frac{{dy}}{{dx}} = - \frac{1}{{2yx}}\left( {y^2 + \frac{2}{x}} \right)
    [/tex]...eq2

    Weights are assigned to y, dy, x and dx as before. However, unlike with eq1, to determine the weights of y, dy, x and dx, eq2 is rewritten as [tex]\left( {y^2 + \frac{2}{x}} \right)dx + 2yxdy = 0[/tex] and then the weights are determined.

    I would've thought that the method used(as in move everything to LHS) to determine the weights of the terms in eq1, would be applicable to eq2. After all, isn't eq2 just a 'special case' of equations like eq1? I'm just wondering why eq1 can be left as it is in determining the weights of y, dy, dx and x whereas eq2 needs to be rearranged prior to the assignment of weights. Any help would be good, thanks.

    Edit: Fixed first paragraph. The terms x and dx are assigned weights of 1 not m.
     
    Last edited: Jul 14, 2005
  2. jcsd
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