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Series solution near an ordinary point

  1. Oct 22, 2005 #1
    Im given y"-xy'-y=0 at x0=1.

    The problem asks for recurrene relation, and the first four terms in each of two linearly independant solutions, and the general term in each solution.

    Whats thrwoing me off is the x0=1. I tried doing y= SUM an(x-1)^n, but when i differenetiate and plug in, i get stuck with the xy' part.
    I also tried doing y=SUM an+1x^n, and it also doesnt work. I tried many other paths also, but none of them seem to work.....
     
  2. jcsd
  3. Oct 22, 2005 #2

    Pyrrhus

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    You should have
    [tex] \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} (x-1)^{n} - \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} - \sum_{n=0}^{\infty} n a_{n} (x-1)^{n} - \sum_{n=0}^{\infty} a_{n} (x-1)^{n} = 0 [/tex]
    where the term [itex] \sum_{n=0}^{\infty} n a_{n} (x-1)^{n} [/itex] comes from
    [tex] - (x-1)y' = \sum_{n=1}^{\infty} (n+1) a_{n+1} (x-1)^{n+1} = \sum_{n=0}^{\infty} (n) a_{n} (x-1)^{n} [/tex]
     
  4. Oct 22, 2005 #3
    i dont understand how you got the -(x-1)y'. does the term xy' change to (x-1)y' when the x in the y=SUM an(x-1)^n changes also? im sorry, but is there a way you can show you got the 2nd and 3rd summation in your solution in more detail please?
     
  5. Oct 22, 2005 #4
    oh wait nvm my last post, i got it, thank you.
     
  6. Oct 23, 2005 #5
    we know that
    y'= [tex]\sum_{n=0}^{\infty} n a_{n} (x-1)^{n-1}[/tex]
    y"= [tex]\sum_{n=0}^{\infty} n (n-1) a_{n} (x-1)^{n-2}[/tex]
    when we compute xy', doesnt it become just:
    [tex]\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}[/tex] ?
    i dont see how [tex]\sum_{n=0}^{\infty} n a_{n} (x-1)^{n}[/tex] and
    - [tex]\sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n}[/tex] were obtained
     
    Last edited: Oct 23, 2005
  7. Oct 23, 2005 #6

    HallsofIvy

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    Another way to do that problem, if it is the x0= 1 that is throwing you off, is to change variable. Let v= x- 1 so x= v+1. Then [tex]\frac{dy}{dv}= \frac{dy}{dx}[/tex] and [tex]\frac{d^2y}{dv^2}= \frac{d^2y}{dx^2} so we can use y" and y' to mean derivatives with respect v also. In terms of the v variable, we have y" (v+1)y'- y= y"- vy'- y'- y= 0 and
    v0= 0.
     
  8. Oct 23, 2005 #7

    Pyrrhus

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    I simply set x = 1 + (x-1), so you see on the series form above

    [tex] y'' - y'(1 + (x-1)) - y = 0 [/tex]

    [tex] y'' - y' - (x-1)y' - y = 0 [/tex]
     
  9. Oct 23, 2005 #8
    why was it necessary for you to set x= 1+(x-1)
     
  10. Oct 23, 2005 #9

    Pyrrhus

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    Look at this

    [tex] y'= \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} [/tex]

    right?

    so

    [tex] xy'= x \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-1)^{n} [/tex]

    You see a problem with that?
     
  11. Oct 23, 2005 #10
    the polynomial is not going to look like the other ones?
     
  12. Oct 4, 2006 #11
    i have this same problem and cannot figure out how you find a_0 and a_1 if all of the indexes are 0 and the (x-1)^n terms are the same as you show. if you solve the equation wouldn't you have the recurrence relation a_n+2 in terms of a_n+1 and a_n for n>or=0? Can you help me with next step?
     
  13. Oct 5, 2006 #12

    Pyrrhus

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    Actually there is a small mistake :redface:. Now you can solve this by letting n=0 in the first, second and fourth sums.
     
  14. Feb 20, 2010 #13
    Hi, I'm actually working on this same problem and I'm not having trouble getting the equation mentioned above, but what I can't seem to figure out is how to extract the recurrence relation from it.

    I see that the correct relation is

    (n+2)an+2-an+1-an=0

    But I dont understand how to get there from the four terms already mentioned above. I've tried writing the equation

    an+2(n+1)(n+2)-an+1(n+1)-an-1(n)-an=0

    which gives the correct answer for n=0 but not for subsequent terms. I've tried toying with ignoring the summation that begins at n=1 but i still can't seem to eliminate the (n+1) factor that isn't in the solution.
     
  15. Feb 20, 2010 #14
    Ah I've found my mistake, I failed to correctly shift indices.
     
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