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Solving diff. eq.

  1. Jul 28, 2008 #1
    \lambda f(y)= i b y \frac{\partial f(y)}{\partial y} + \frac{partial g(y)}{y} -\frac{k}{y}g
    \lambda g(y)= i b y \frac{\partial g(y)}{\partial y} - \frac{partial f(y)}{y} +\frac{k}{y}f

    I tried to get a hypergeometric eq. from these two but couldn't.
    Any hints to solve?
    Helps would be appreciated!
     
  2. jcsd
  3. Aug 7, 2008 #2
    I somehow got this second oder diff.eq.

    [tex]
    (1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0
    [\tex]

    where

    [tex]f_{yy}[\tex] is [tex]\frac{\partial^2}{\partial y^2}[\tex]

    Any ideas to solve this one?

    p.s. Latex is not working here or am I typing wrong?
     
    Last edited: Aug 7, 2008
  4. Aug 7, 2008 #3

    Defennder

    User Avatar
    Homework Helper

    You're using the wrong slash. The closing tag should use this "/" instead.
     
  5. Aug 7, 2008 #4
    wups, thanks very much.

    and another thing, I wrote wrong the above 2 eq.s, I put + instead of minus here

    [tex]\lambda g(y)= i m y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{y} -\frac{k}{y}f[/tex]

    so it should be:

    [tex]\lambda f(y)= i m y \frac{\partial f(y)}{\partial y} + \frac{\partial g(y)}{y} -\frac{k}{y}g[/tex]

    [tex]\lambda g(y)= i m y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{y} -\frac{k}{y}f[/tex]



    and then we get the above second oder diff.eq.:
    [tex](1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0[/tex]
     
    Last edited: Aug 7, 2008
  6. Aug 9, 2008 #5
    so any ideas?
     
  7. Aug 11, 2008 #6
    where
    [tex]m, \lambda, k[/tex] are constants.

    I am trying to put these two:
    [tex]
    f_1=\sum_{n=0}^{\infty}p_ny^{2n}, \ \ \ \ \ \ f_2=\sum_{n=0}^{\infty}a_ny^{2n+1}
    [/tex]
    and check if it is odd or even. At the end I am getting a recurrent eq.


    any other ideas?
     
  8. Aug 11, 2008 #7
    hmm, it gives zero solution.
    coefficients are zero in this case :(
     
  9. Aug 12, 2008 #8
  10. Aug 15, 2008 #9
    Ok, I got the solution.


    Now I need one thing. From Abramowitz's book I got this one


    [tex]
    F(a, a+\frac{1}{2}, \frac{3}{2}, z^2)=\frac{1}{2}z^{-1}(1-2a)^{-1}[(1+z)^{1-2a}-(1-z)^{1-2a}]
    [/tex]

    Now I need to find

    [tex]
    F(a, a+\frac{1}{2}, \frac{5}{2}, z^2)
    [/tex]


    [tex]
    F(a, a+\frac{1}{2}, \frac{7}{2}, z^2)
    [/tex]


    and, it would be great if I find

    [tex]
    F(a, a+\frac{1}{2}, n+ \frac{1}{2}, z^2)
    [/tex]


    are there any books, handbooks, or websites that I could find this guy?
     
  11. Aug 18, 2008 #10
    Hallooo???

    Anybody is viewing this thread at all?
     
  12. Aug 19, 2008 #11
    You're missing two partial symbols. Are they supposed to be:

    [tex]\lambda f(y)= i m y \frac{\partial f(y)}{\partial y} + \frac{\partial g(y)}{\partial y} -\frac{k}{y}g[/tex]

    [tex]\lambda g(y)= i m y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{\partial y} -\frac{k}{y}f[/tex]

    ???

    Also, if f and g only depend on y, then why the partials?
     
  13. Aug 20, 2008 #12
    Yeah you are right, there should be two partial symbols.

    No problem with partial. As I stated above, I got the solution for this diff. eq.

    [tex](1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0[/tex]

    from here
    http://eqworld.ipmnet.ru/en/solutions/ode/ode0226.pdf

    The solution, as you see, is a Hypergeometric function.

    Now I need some properties of the hypergeometric function. I posted it above:

    -----------
    From Abramowitz's book I got this one


    [tex]F(a, a+\frac{1}{2}, \frac{3}{2}, z^2)=\frac{1}{2}z^{-1}(1-2a)^{-1}[(1+z)^{1-2a}-(1-z)^{1-2a}][/tex]



    Now I need to find

    [tex]F(a, a+\frac{1}{2}, \frac{5}{2}, z^2)[/tex]

    and

    [tex]F(a, a+\frac{1}{2}, \frac{7}{2}, z^2)[/tex]


    and, it would be great if I find

    [tex]F(a, a+\frac{1}{2}, n+ \frac{1}{2}, z^2)[/tex]

    are there any books, handbooks, or websites that I could find this guy?


    Plz, help!
     
  14. Sep 2, 2008 #13
    I guess, I need to take a derivative:

    [tex]
    \frac{d}{dz}F(a, b, c, z^2)=\frac{ab}{2z\ c} F(a+1, b+1, c+1, z^2)
    [/tex]
     
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