- #1

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where E, m, v, k_{y} are all constants and I believe turning it into hypergeometric form will help me solve it. Any help would be appreciated!

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- Thread starter thatboi
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I get##\begin{align}&z(1-z)\frac{d^2a}{dz^2}+(z-2)\frac{da}{dz} + \frac{1}{4m^2v^2z}(4E^2m^2+m^2v^2-4E^2m^2z+4Ek_{y}mv^2z+m^2v^2z\nonumber \\&-4E^2m^2z^2-4E^2k_{y}^2v^2z^2+4E^2m^2z^3+4E^f

- #1

- 75

- 9

where E, m, v, k_{y} are all constants and I believe turning it into hypergeometric form will help me solve it. Any help would be appreciated!

- #2

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- #3

- 75

- 9

It appears that ##x=\infty## is a singular point but it is not regular so I am not sure if we can use it here. Nevertheless it should still be possible to turn it into hypergeometric form right?

Also I am not sure what you mean by reducing and simplifying the constants here, I do not see anything that can be further cleaned up.

- #4

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##

z(1-z)\frac{d^2w}{dz^2} +[c-(a+b+1)z]\frac{dw}{dz} - abw = 0

##

which has three regular singular points in the extended ##z##-plane. These are at ##z=0,1,\infty##. For starters you should transform your equations singular points to the same places. Two out out of three are already where they need to be. The last may be done by the change of variable,

##

z = -\frac{m}{E}x

##

if I've done things correctly. After this is done and you've multiplied through by the ##z(1-z)## factor, if you get the hypergeometric equation, you're done. If not, then your ##a(x)## isn't a hypergeometric function. That said, it may still be related to one with the right transformations.

- #5

- 75

- 9

##

z(1-z)\frac{d^2w}{dz^2} +[c-(a+b+1)z]\frac{dw}{dz} - abw = 0

##

which has three regular singular points in the extended ##z##-plane. These are at ##z=0,1,\infty##. For starters you should transform your equations singular points to the same places. Two out out of three are already where they need to be. The last may be done by the change of variable,

##

z = -\frac{m}{E}x

##

if I've done things correctly. After this is done and you've multiplied through by the ##z(1-z)## factor, if you get the hypergeometric equation, you're done. If not, then your ##a(x)## isn't a hypergeometric function. That said, it may still be related to one with the right transformations.

- #6

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- #7

- 75

- 9

I get

##\begin{align}

&z(1-z)\frac{d^2a}{dz^2}+(z-2)\frac{da}{dz} + \frac{1}{4m^2v^2z}(4E^2m^2+m^2v^2-4E^2m^2z+4Ek_{y}mv^2z+m^2v^2z\nonumber \\

&-4E^2m^2z^2-4E^2k_{y}^2v^2z^2+4E^2m^2z^3+4E^2k_{y}^2v^2z^3)a(z) = 0 \nonumber

\end{align}

##

which I guess is not the form we wanted.

- #8

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