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Series solutions near a singular point, 2nd order linear

  1. Oct 15, 2013 #1
    So, I'm trying to solve 2nd order linear differential equations (series solutions near a singular point).

    (lnx)y" + 0.5y' + y = 0 around the regular singular point x = 1

    I got the indicial equation,

    r(r-0.5) = 0,

    which leads to the roots....

    r1 = 0.5, r2 = 0

    The problem only asks us to find the first three nonzero terms in the series y1 = Ʃa_n * (x-1)^r+n from n = 0 to infinity. And we only need to find one solution, corresponding to the larger root.

    So I took the first and second derivatives of the y1 they gave and plugged it into the differential equation. Now at this point I usually factor out all the x terms. And since the left side has to equal zero for all x, I can divide by that x term to get a recurrence relation (that involves a_n terms). From the relation I can figure out what a_n is. However, in this case, I can't factor out all the x terms because there's a lnx.

    Anyone know how to get rid of the lnx?


    Last edited: Oct 15, 2013
  2. jcsd
  3. Oct 16, 2013 #2


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    Science Advisor

    Replace ln(x) with a power series about x= 1.
  4. Oct 16, 2013 #3
    Thank you.
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