Solution of second order differential equation with singularities at 0,1, and ∞ I am trying to solve the following equation;
$$
U''+\left( \frac{1}{t}+\frac{3}{t-1}\right)U'+\left(\frac{1}{t}+C\right)\frac{U}{t(t-1)}=0
$$
where U is a function of t and C is constant.
The above equation is similar to a form of Riemann equation.
Could anyone please provide any support on how the above equation can be solved. I am trying to obtain the solution in form of hyper-geometric series.
 A: This is not a hypergeometric equation (singularity at infinity is irregular). It can be reduced to the prolate spheroid equation. (Also known as confluent Heun equation).
See the references in Wikiedia,
https://en.wikipedia.org/wiki/Prolate_spheroidal_wave_function
A: Maple's solution is
$$ U \left( t \right) ={\frac {a{{\rm e}^{\sqrt {-C}t}}{\it HeunC}
 \left( 2\,\sqrt {-C},0,-2,1,-2,t \right) }{ \left( t-1 \right) ^{2}}}
+{\frac {b{{\rm e}^{\sqrt {-C}t}}{\it HeunC} \left( 2\,\sqrt {-C},0,-2
,1,-2,t \right) }{ \left( t-1 \right) ^{2}}\int \!{\frac { \left( t-1
 \right) {{\rm e}^{-2\,\sqrt {-C}t}}}{t \left( {\it HeunC} \left( 2\,
\sqrt {-C},0,-2,1,-2,t \right)  \right) ^{2}}}\,{\rm d}t}
$$
EDIT:  Note that this was for an earlier version of the problem with the differential equation
$$ U'' + \left(\dfrac{1}{t} + \dfrac{3}{t-1}\right) U' + \left(\dfrac{1}{t} + C\right) U = 0 $$
The equation has now been changed to 
$$  U'' + \left(\dfrac{1}{t} + \dfrac{3}{t-1}\right) U' + \left(\dfrac{1}{t} + C\right) \dfrac{U}{t(t-1)} = 0 $$
which does have hypergeometric solutions.
Please: In future, if you want to change a question, especially after answers have been posted, don't delete the original form of the question;  rather, add a new paragraph with the change.  Otherwise, the casual reader might think we've posted wrong answers. 
A: I'm not sure if this is more useful than the Maple solution given earlier, but Wolfram Mathematica finds a slightly different solution in terms of associated Legendre functions:
$$U(t) = \frac{1}{1-t} \left( k_1 P_\ell^2(2t-1) + k_2 Q_\ell^2(2t-1) \right)$$
Here, $k_1$ and $k_2$ are constants of integration, and $P_\ell^m$ and $Q_\ell^m$ are the associated Legendre functions of the first and second kinds, with $\ell = \frac{1}{2} \left(\sqrt{9-4 C}-1\right)$.
A: The given equation can be converted to a hyper-geometric equation by the following substitution,
$$
U = tY
$$
This gives,
$$
t(1-t)Y''+(c-(a+b+1)t)Y'-abY=0
$$
with
$$
c=3, a+b+1 = 6, ab=C+4
$$
The final solution Y(t) can then be obtained using the hyper-geometric series function and hence U(t) = t Y(t).
I appreciate all for their effort and help.
