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Homework Help: On the regular singularities of a second order differential equation

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data
    The only singularities of the differential equation
    are regular singularities at x=1 of exponents [itex]\alpha[/itex] and [itex]\alpha'[/itex], and at x=-1 of exponents [itex]\beta[/itex] and [itex]\beta'[/itex], the point at infinity being an ordinary point.
    Prove that [itex]\beta=-\alpha[/itex] and [itex]\beta'=-\alpha'[/itex] and that the differential equation is
    [tex](x^2+1)^2 y''+2(x-1)(x-\alpha-\alpha') y'+4\alpha\alpha' y=0[/tex]
    where [itex]c_1,c_2[/itex] are constants.

    2. Relevant equations
    The solution to find is given
    3. The attempt at a solution
    I understand that if at x=1 is a regular singular point, it means that the factor [itex](x-1)[/itex] appears in the denominator of p(x) at the first power (at most), and at the second power in the denominator of q(x) (at most), then
    with [itex]\alpha,\beta \leq 1[/itex] and [itex]\alpha',\beta' \leq 2[/itex], and [itex]f(x),g(x)[/itex] without [itex]x=\pm 1[/itex] as a root. Assuming that the point at infinity is an ordinary point, it means that f(x) and g(x) are finite as [itex]x\rightarrow\infty[/itex] (for example f and g being constant). Then, for example, f and g could be a sine or cosine function of x, and [itex]\alpha=\beta=1[/itex] and [itex]\alpha'=\beta'=2[/itex], and it makes that the differential equation has only 2 singularities, both being regular, and with the point at infinity being an ordinary point. Therefore, i don't understand how is to be "proven" that [itex]\alpha[/itex] must be [itex]-\beta[/itex] and so on, and then proving the closed form of the differential equation, without any traces of [itex]f(x)[/itex] neither as [itex]g(x)[/itex].
  2. jcsd
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