Hypergeometric equation in a particular case I have a question to make in relation to the solution of the hypergeometric differential equation. Let us consider the aforesaid equation,
\begin{equation}
y(1-y)h'' + [c-(1+a+b)y]h' -abh=0,
\end{equation}
where $h(y)$ and the prime stands for derivatives with respect to $y$. I would like to know what is the general solution when the following relation amongst the parameters holds: $a+b-c=0$. This case is always excluded in the textbooks and I did not find a book discussing such a delicate point. Perhaps, there is only one solution or the equation becomes another well-known ODE, I do not really know. Anyway, under the condition $c=a+b$ the new differential equation becomes
\begin{equation}
y(1-y)h'' + [(a+b)-(1+a+b)y]h' -abh=0.
\end{equation}
It looks quite similar to the hypergeometric differential equation but now it only has two parameters $a$ and $b$.
A second point that I would like to discover is the behavior of that general solution near $y=1$ and $y \rightarrow \infty$.
Any help in this matter it will be much appreciated.
 A: Hypergeometric equation and solution:
$$y(1-y)h'' + [a+b-(1+a+b)y]h' -abh=0$$
$$\Rightarrow h(y)=c_2 (-1)^{1-a-b} y^{1-a-b} \, _2F_1(1-a,1-b;2-a-b;y)+c_1 \, _2F_1(a,b;a+b;y),$$
with $c_1$ and $c_2$ arbitrary constants. So there are two independent solutions.
A few special cases ($B_y$ is the incomplete Beta function)
$$a=1:\;\;h(y)=c_1 \, _2F_1(1,b;b+1;y)+(-1)^{-b} c_2 y^{-b}$$
$$a=0:\;\;h(y)=(-1)^{-b} (b-1) c_2 B_y(1-b,0)+c_1$$
$$a=-1:\;\;h(y)=(-1)^{-b} c_2 y^{2-b} \, _2F_1(2,1-b;3-b;y)+\frac{c_1 (1-b y+b)}{b-1}$$
Limit $y\rightarrow 1$:
$$h(y)\rightarrow\log |y-1| \left(\frac{c_2 (-1)^{-a-b} \Gamma (2-a-b)}{\Gamma (1-a) \Gamma (1-b)}-\frac{c_1 \Gamma (a+b)}{\Gamma (a) \Gamma (b)}\right)$$
Limit $y\rightarrow\infty$:
$$h(y)\rightarrow (-1)^{-a} y^{-a} \Gamma (b-a)\frac{ c_1 \Gamma (1-a)^2 \Gamma (a+b)+c_2 \Gamma (b)^2 \Gamma (2-a-b)}{\Gamma (1-a)^2 \Gamma (b)^2}+(-1)^{-b} y^{-b} \Gamma (a-b)\frac{ c_2 \Gamma (a)^2 \Gamma (2-a-b)+c_1 \Gamma (1-b)^2 \Gamma (a+b)}{\Gamma (a)^2 \Gamma (1-b)^2}$$
A: Mathematica 12.0 answers
DSolve[y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0, h[y], y]


$ \left\{\left\{h(y)\to c_2 (-1)^{-a-b+1} y^{-a-b+1} \, _2F_1(1-a,1-b;-a-b+2;y)+c_1 \, _2F_1(a,b;a+b;y)\right\}\right\}$

Next,
AsymptoticDSolveValue[y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0, 

h[y], {y, Infinity, 1}]


$$ c_1 \left(-\frac{a b y^{-b-1}}{a b+(b+1) (-a-b+1)+(b+1) b}+\frac{b y^{-b-1}}{a b+(b+1) (-a-b+1)+(b+1) b}+y^{-b}\right)+c_2 \left(\frac{a y^{-a-1}}{(a+1) (-a-b+1)+a b+a (a+1)}-\frac{a b y^{-a-1}}{(a+1) (-a-b+1)+a b+a (a+1)}+y^{-a}\right)$$

In order to obtain the asymptotics as $y\to 1$, we need put the value of $h(y)$ at $y=1$, e.g.
AsymptoticDSolveValue[{y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0,  

 h[1] == 1}, h[y], {y, 1, 1}]


$ 1-a b (y-1)$

Maple 2018.2 answers
dsolve(y*(1-y)*(diff(h(y), y, y))+(a+b-(1+a+b)*y)*(diff(h(y), y))-a*b*h(y) = 0, h(y));


$ h \left( y \right) ={\it \_C1}\,{\mbox{$_2$F$_1$}(a,b;\,a+b;\,y)}+{
\it \_C2}\,{y}^{-a-b+1}{\mbox{$_2$F$_1$}(1-b,1-a;\,2-a-b;\,y)}
$

and
series(_C1*hypergeom([a, b], [a+b], y)+_C2*y^(-a-b+1)*hypergeom([1-b, 1-a], [2-a-b], y), y = 1, 1);


$$ -{\frac {{\it \_C1}\,\Gamma \left( a+b \right)  \left( \ln  \left( -1+
y \right) +i{\it csgn} \left( i \left( -1+y \right)  \right) \pi+2\,
\gamma+\Psi \left( a \right) +\Psi \left( b \right)  \right) }{\Gamma
 \left( a \right) \Gamma \left( b \right) }}-{\frac {{\it \_C2}\,
\Gamma \left( 2-a-b \right)  \left( \ln  \left( -1+y \right) +i{\it 
csgn} \left( i \left( -1+y \right)  \right) \pi+2\,\gamma+\Psi \left( 
1-b \right) +\Psi \left( 1-a \right)  \right) }{\Gamma \left( 1-b
 \right) \Gamma \left( 1-a \right) }}+O \left( -1+y \right) 
$$

Unfortunately,
series(_C1*hypergeom([a, b], [a+b], y)+_C2*y^(-a-b+1)*hypergeom([1-b, 1-a], [2-a-b], y), y = infinity, 1);


Error, (in asympt) unable to compute series

