# Is there a name for $x(x-1)y''+[(1+c_1+c_2)x-c_3]y'+c_1c_2y=0$?

Is there a name for this differential equation? $x(x-1)y''+[(1+c_1+c_2)x-c_3]y'+c_1c_2y=0$ Thanks.

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What about it: wolframalpha.com/input/… –  Ilya Sep 25 '11 at 21:32
@Gortaur: Thanks, but Hmm... Is there a name there? I didn't mean name as in "2nd order ODE", but rather something like "Bessel's equation". –  Johnny Sep 25 '11 at 21:37
I understand that it wasn't the answer you expected, but at least we know that WA doesn't know it in its current formulation (I didn't know if you tried WA before asking here). –  Ilya Sep 25 '11 at 21:43
I don't know if it yields something useful, but have you tried putting it in Sturm-Liouville form? (Maybe WA or Mathematica can do this, saving you a lengthy calculation.) –  Gerben Sep 25 '11 at 21:59
@Gerben: Thanks, how can I get it in SL form using WA? –  Johnny Sep 25 '11 at 22:02

Maple classifies this as a Jacobi differential equation. The general solution is expressed in terms of hypergeometric functions: $y = a_{{1}}\ {{}_2F_1([c_{{1}},c_{{2}}],[c_{{3}}],\,x)}+a_{{2}}{x}^{1-c_{{3}}} \ {{}_2F_1([c_{{1}}+1-c_{{3}},c_{{2}}+1-c_{{3}}],[\,-c_{{3}}+2],\,x)}$ where $a_1$ and $a_2$ are arbitrary constants.

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Or, since it's a Jacobi DE after all, the solution is expressible in terms of Jacobi polynomials and Jacobi functions of the second kind. –  Guess who it is. Sep 26 '11 at 0:33
Jacobi polynomials if $c_1$ is a nonpositive integer, I think. Otherwise not polynomials. –  Robert Israel Sep 26 '11 at 0:54
I suppose in that case, we use "Jacobi functions of the first kind" (by analogy with the Legendre case). –  Guess who it is. Sep 26 '11 at 0:55
How does you feel that this is a Jacobi differential equation? According to mathworld.wolfram.com/JacobiDifferentialEquation.html, note that the $y''$ term of a Jacobi differential equation is $1-x^2$ while the $y''$ term of a Gaussian hypergeometric equation is $x(x-1)$ . –  doraemonpaul Oct 15 '12 at 14:05
See maplesoft.com/support/help/Maple/… An affine transformation of the independent variable takes you from $1-x^2$ to $x(1-x)$. –  Robert Israel Oct 15 '12 at 16:10

This is exactly a Gaussian hypergeometric equation.

http://eqworld.ipmnet.ru/en/solutions/ode/ode0222.pdf

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