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Second order ODE, I think 2 regular points

  1. Nov 30, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hello guys! I've never dealt with an ODE having 2 singularities at once, I tried to solve it but ran out of ideas. I must solve ##(x-2)y''+3y'+4\frac{y}{x^2}=0##.


    2. Relevant equations

    Not sure.

    3. The attempt at a solution
    I rewrote the ODE into the form ##y''+\frac{3}{x-2}y'+4\frac{y}{x^2(x-2)}=0##. I notice that the singularities at ##x=2## and ##x=0## are both regular, so that Frobenius method should find at least 1 solution around any of these singularities.
    So I first tried to expand the solution around ##x=2## first. Seeking solution(s) of the form ##\phi (x)=\sum _{n=0}^\infty a_n (x-2)^{n+c}##, I reached that [tex]\sum _{n=0}^\infty a_n(n+c)(n+c-1) (x-2)^{n+c-2}+ 3 \sum _{n=0}^\infty a_n (n+c) (x-2)^{n+c-2}+\frac{4}{x^2}\sum _{n=0}^\infty a_n (x-2)^{n+c}=0[/tex]. I stopped right there, because of the "1/x²" factor. But now that I think, maybe I can just "get rid of it" and it won't affect the solution of the ODE if I simply throw it away? Because that equation is satisfied for any x, so I guess this is enough of a reason to get rid of it?
     
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  3. Nov 30, 2012 #2

    haruspex

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    I was able to get this recurrence relation, which looks interesting in some ways:
    an = an-1 + an-2n(n+2)
    If you think it helps, I'll post an outline of how I got it.
     
  4. Nov 30, 2012 #3

    vela

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    You could multiply through by ##x^2 = [(x-2)+2]^2## to get everything in terms of x-2. Seems kinda of tedious though, but then Frobenius always is.
     
  5. Nov 30, 2012 #4

    fluidistic

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    Following where I left, I reach the indicial equation c(c-1)+3c=0, so that c=0 or -2. Since the roots differ by an integer Frobenius method will not necessarily find the 2 linearly independent solutions of the ODE, but I am assured to find one if I use c=0 (the greater root).
    Taking c=0, I reach that ##\sum _{n=2}^\infty (x-2)^{n-2} \{ a_n[n(n+2)]+\frac{4a_{n-2}}{x^2} \}=0##. The factor 1/x² still boggles me a lot.

    That would be awesome, though I don't want you to waste time writing down the latex so an explanation will do the job (I hope).
     
  6. Nov 30, 2012 #5

    fluidistic

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    Hmm for example I could do it in my last expression of my last post and divide by ##(x-2)^{n-2}## but even then, I'd have a term multiplied by x and another one independent from x... And since that must equal 0, I guess this implies that ##a_n [n(n+2)]=0## and therefore I get the trivial solution, ##a_n=0## for any n. Hmm I probably made a mistake somewhere then.
     
  7. Nov 30, 2012 #6

    haruspex

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    OK, here goes. Hope I made no mistakes.
    First substitution: y = x2z. A factor x cancelled out.
    2nd sub: turned all x into x+1, just to make it look a bit nicer. Got (x2-1)z''+(7x-1)z'+8z = 0
    Assuming a power series, got recurrence relation
    an+2(n+2)(n+1) = -an+1(n+1)+an(n+2)(n+4)
    3rd sub: bn=(-1)nn!an
    Got bn = bn-1+n(n+2)bn-2
    Resorted to generating samples in a spreadsheet. Looks something like n!(n+2) for n odd, (n+1)! for n even.
     
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