# 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!

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#### VatanparvaR

I somehow got this second oder diff.eq.

$$(1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0 [\tex] where [tex]f_{yy}[\tex] is [tex]\frac{\partial^2}{\partial y^2}[\tex] Any ideas to solve this one? p.s. Latex is not working here or am I typing wrong? Last edited: #### Defennder Homework Helper [tex]\lambda f(y)= i b y \frac{\partial f(y)}{\partial y} + \frac{\partial g(y)}{y} -\frac{k}{y}g$$
$$\lambda g(y)= i b y \frac{\partial g(y)}{\partial y} - \frac{\partial f(y)}{y} +\frac{k}{y}f$$
$$(1-m^2y^2)f_{yy}-2my(i\lambda+m)f_{y}+(\lambda^2-im\lambda-\frac{k(k+1)}{y^2})f=0$$

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

Last edited:

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

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:

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