# closed form solution of an ODEs

I posted this on mathstack and it is not a homework problem, I am doing a modeling problem and this ODE comes up. I don't know whether it could be considered to be a research problem yet. Please help me to close it if it is not placed at the right place.

I have a problem with finding the closed form solution of the following ODEs. I am NOT sure that a closed form solution exists. Closed form here means that the solution can be presented as integrals/ power series. Here is the ODE : I only consider $x\in (0,1)$ and $c_i$ are known non-zero real numbers.

$y''(x) + [\frac{c_1}{x^2}+\frac{c_2}{(1-x)^2} +c_3(\frac{1}{x}+\frac{1}{1-x})]y'(x)+[c_4(\frac{1}{x}+\frac{1}{1-x})+\frac{c_5}{x^2}+\frac{c_6}{(1-x)^2}]y(x)=0$

I can find the solution inform of power series which is very ugly. I would like to ask you all that whether you can give a method that can be used to find the solution in a nicer way. Thanks so much for your time. I really appreciate it.

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This equation, with arbitrary $c_j$ does not have explicit solutions, or a useful power series solution. You can only do qualitative study, or try to compute solutions numerically. – Alexandre Eremenko Feb 18 '13 at 1:54
@Steven: You are asking for the solutions of a second order linear equation with singular points at $x=0$, $x=1$, and $x=\infty$. When $c_1=0$ and $c_2$=0, these singular points are *regular singular points* (in which case, this is known as a *Fuchsian* equation), otherwise, they are not. Generally, for nearly all values of the constants $c_i$, these are not expressible in terms of elementary functions. – Robert Bryant Feb 18 '13 at 1:56
THANKS Alex and Bob !!! – Steven Feb 18 '13 at 3:47

All the major computer algebra systems implement this, and (not surprisingly) Mathematica does not find a solution in closed form in the generality you give. However, since your $c_i$ are known real numbers'', perhaps you are lucky, and a closed for solution is known in your special case. The relevant Mathematica incantataion is "DSolve", as in: