- #1

thatboi

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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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In summary: 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^

- #1

thatboi

- 133

- 18

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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- #2

Paul Colby

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- #3

thatboi

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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?Paul Colby said:

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

Paul Colby

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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

thatboi

- 133

- 18

Paul Colby said:

##

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

Paul Colby

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- #7

thatboi

- 133

- 18

I getPaul Colby said:

##\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

Paul Colby

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To convert a second order ODE to a hypergeometric function, you will need to use the method of Frobenius. This involves assuming a solution in the form of a power series and then solving for the coefficients using recurrence relations.

Converting a second order ODE to a hypergeometric function can make it easier to solve and analyze the equation. Hypergeometric functions have many useful properties and can be expressed in terms of simpler functions, making them a valuable tool in mathematical and scientific research.

No, not all second order ODEs can be converted to a hypergeometric function. The equation must have certain properties, such as being linear and having regular singular points, in order for the conversion to be possible.

While hypergeometric functions are useful for solving certain types of ODEs, they may not be the most efficient or accurate method for all equations. In some cases, other methods such as numerical approximation or special functions may be more suitable.

Yes, hypergeometric functions have many practical applications in fields such as physics, engineering, and economics. They can be used to model and solve various real-world problems, making them a valuable tool for scientists and researchers.

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