Solving diff. eq.

\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!
 
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:

Defennder

Homework Helper
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[tex]\lambda f(y)= i b y \frac{\partial f(y)}{\partial y} + \frac{\partial g(y)}{y} -\frac{k}{y}g[/tex]
[tex]\lambda g(y)= i b y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{y} +\frac{k}{y}f[/tex]
[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} \ \mbox{is} \frac{\partial^2}{\partial y^2}[/tex]
You're using the wrong slash. The closing tag should use this "/" instead.
 
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:
so any ideas?
 
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?
 
hmm, it gives zero solution.
coefficients are zero in this case :(
 
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?
 
Hallooo???

Anybody is viewing this thread at all?
 
wups, thanks very much.

[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]
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?
 
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!
 
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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