I desperately with hypergeometric functions.

[Snoofleglax]

Homework Statement

I have three problems on my homework set that I can't figure out. I'll start with the longest one:

Show that by letting z=\zeta^{-1} and u=\zeta^{\alpha}v(\zeta) that the hypergeometric differential equation

z(1-z)\frac {d^2u}{dz^2} + <br /> \left[\gamma-(\alpha+\beta+1)z \right] \frac {du}{dz} - \alpha\beta u = 0

can be reduced to

\zeta(1-\zeta)\frac {d^2v}{d\zeta^2} + <br /> \left[\alpha -\beta +1 -(2\alpha+\gamma+2)\zeta \right] \frac {dv}{d\zeta} - \alpha(\alpha-\gamma+1)v = 0

and find the solutions of the first equation about z=\infty in terms of hypergeometric functions of z.

Homework Equations

Hypergeometric function:

_{2}F_{1}=\sum^{\infty}_{n=0}\frac{(\alpha)_{n}(\beta)_{n}}{(\gamma)_{n}}\frac{z^{n}}{n!}

The Attempt at a Solution

What I've gotten so far is finding how the differentials work out:

\frac{du}{dz} = -\zeta^{2}\frac{du}{d\zeta}

\frac{d^{2}u}{dz^{2}} = \zeta^{4}\frac{d^{2}u}{dz^{2}}+2\zeta^{3}\frac{du}{d\zeta}.

Then I've substituted into the hypergeometric equation to get:

\frac{1}{\zeta}\left(1-\frac{1}{\zeta} \right)\left(\zeta^{4}\frac{d^{2}u}{dz^{2}}+2\zeta^{3}\frac{du}{d\zeta}\right)+\left[\gamma-(\alpha+\beta+1)\frac{1}{\zeta} \right]\left(-\zeta^{2} \frac {du}{d\zeta}\right) - \alpha\beta u = 0

The part where I'm having trouble is trying to convert from u into v. It seems like it should be fairly elementary, but I can't figure it out. And after that I have no idea where to go to get the form my professor wants. The best I can come up with for converting u to v is

\frac{du}{d\zeta} = \frac{du}{dv}\frac{dv}{d\zeta}

but it doesn't seem to work out, or there's a wrinkle I'm missing somewhere. Any help is appreciated.

 

[clamtrox]

Just stick u = \zeta^{\alpha} v into the differential equation? :)