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Singular points of an ODE

  1. Jan 15, 2008 #1
    1. The problem statement, all variables and given/known data
    For the ODE xy" + (2-x)y' + y = 0

    i want to show it has one singular point and identify its nature

    2. Relevant equations

    3. The attempt at a solution

    I have read the topic and I see that a point Xo is called and ordinary point of the equation if both p(x) and q(x) (once converted to SF) are anlytic at Xo.

    I really dont understand the method to work this out though.....

    Help would be much appreciated
  2. jcsd
  3. Jan 15, 2008 #2


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    What "method" are you talking about? Just finding p(x) and q(x) and determining whether they are analytic?

    Unfortunately, you haven't said what you mean by "SF" nor what p(x) and q(x) are here.

    I might guess that by "SF" (standard form?) You mean the equation solved for y". Here that would be y"= ((2-x)/x) y'+ (1/x)y and perhaps you mean p(x)= (2-x)/x and q(x)= 1/x.
  4. Jan 15, 2008 #3
    yes i am aware that when we divide by x this ODE is then in standard form. What i dont understand is how to show this equation has only one singular point?
  5. Jan 15, 2008 #4


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    What IS a singular point?

    You've already said that an "ordinary point" is a point where the coefficents of y' and y are not analytic. A "singular point" is a point where that is not true. For what values of x and y are (2-x)/x and 1/x not analytic?
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