solving diff. eq.

[VatanparvaR]

\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!

[VatanparvaR]

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?

[Defennder]

\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


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


where

f_{yy} \ \mbox{is} \frac{\partial^2}{\partial y^2}

You're using the wrong slash. The closing tag should use this "/" instead.

[VatanparvaR]

wups, thanks very much.

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

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

so it should be:

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

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



and then we get the above second oder diff.eq.:
(1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0

[VatanparvaR]

so any ideas?

[VatanparvaR]

where
m, \lambda, k are constants.

I am trying to put these two:

f_1=\sum_{n=0}^{\infty}p_ny^{2n}, \ \ \ \ \ \ f_2=\sum_{n=0}^{\infty}a_ny^{2n+1}

and check if it is odd or even. At the end I am getting a recurrent eq.


any other ideas?

[VatanparvaR]

hmm, it gives zero solution.
coefficients are zero in this case :(

[VatanparvaR]

I guess this one works:

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

[VatanparvaR]

Ok, I got the solution.


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



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}]


Now I need to find


F(a, a+\frac{1}{2}, \frac{5}{2}, z^2)




F(a, a+\frac{1}{2}, \frac{7}{2}, z^2)



and, it would be great if I find


F(a, a+\frac{1}{2}, n+ \frac{1}{2}, z^2)



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

[VatanparvaR]

Hallooo???

Anybody is viewing this thread at all?

[Matthew Rodman]

wups, thanks very much.

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

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



You're missing two partial symbols. Are they supposed to be:

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

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

???

Also, if f and g only depend on y, then why the partials?

[VatanparvaR]

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.

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

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


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}]



Now I need to find

F(a, a+\frac{1}{2}, \frac{5}{2}, z^2)

and

F(a, a+\frac{1}{2}, \frac{7}{2}, z^2)


and, it would be great if I find

F(a, a+\frac{1}{2}, n+ \frac{1}{2}, z^2)

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


Plz, help!

[VatanparvaR]

I guess, I need to take a derivative:


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