Jtdylktuy

Let $a in {mathbb C}$ and $bin {mathbb C}$ and let $nge 1$ be an integer.

Consider a following family of Ordinary Differential Equations (ODEs). We have:

begin{equation}
frac{d^2 y(x)}{d x^2} - frac{n^2}{4} (a-b)^4 frac{P_n^{(2n-2)}(x)}{(x+a)^2(x+b)^{2n+2}} cdot y(x)=0
end{equation}
where $P_n^{(2n-2)}(x)$ are a polynomials of order $2n-2$ in $x$ which read:
begin{eqnarray}
P_n^{(2n-2)}(x) = left{
begin{array}{rr}
1 & mbox{if $quad n=1$}\
(a+b+2 x)^2 & mbox{if $quad n=2$}\
(a^2+ a b+b^2+3(a+b)x+3 x^2)^2 & mbox{if $quad n=3$}\
(a+b+2 x)^2(a^2+b^2+2(a+b)x+2 x^2)^2 & mbox{if $quad n=4$}\
vdots
end{array}
right.
end{eqnarray}
as a matter of fact we have:
begin{equation}
P_n^{(2n-2)}(x) =frac{((x+a)^n - (x+b)^n)^2}{(a-b)^2}
end{equation}
for $n=1,2,cdots$.
Now by using the algorithm described in my answer to How do I find a change of variables that reduces a linear 2nd order ODE to the Gaussian hypergeometric differential equation? I have found the fundamental solutions to those ODEs. They read:
begin{eqnarray}
y(x) = C_1 cdot sqrt{frac{(x+b)^{n+1}}{(x+a)^{n-1}}}W_{frac{1}{2}, {mathfrak A}_n}[left( frac{x+a}{x+b}right)^n] +
C_2 cdot sqrt{frac{(x+b)^{n+1}}{(x+a)^{n-1}}}M_{frac{1}{2}, {mathfrak A}_n}[left( frac{x+a}{x+b}right)^n]
end{eqnarray}
Here the constants read ${mathfrak A}_n = sqrt{1+n^2}/(2 n)$ for $n=1,2,cdots$ and $W$ and $M$ are the Whittaker functions https://en.wikipedia.org/wiki/Whittaker_function .
Now the following Mathematica code "proves" the result:

In[322]:= a =.; b =.; x =.;

Table[FullSimplify[(D[#, {x, 2}] - 

      n^2/4 ( (a - b)^2 ((x + a)^n - (x + b)^n)^2)/((a + x)^2 (b + 

          x)^(2 n + 2)) #) & /@ {Sqrt[(b + x)^(n + 1)]/ 

     Sqrt[((a + x)^(n - 1))]

      WhittakerW[1/2, Sqrt[1 + n^2]/(2 n), ((x + a)/(x + b))^n], 

    Sqrt[(b + x)^(n + 1)]/ Sqrt[((a + x)^(n - 1))]

      WhittakerM[1/2, Sqrt[1 + n^2]/(2 n), ((x + a)/(x + b))^n]}], {n,

   1, 6}]





Out[323]= {{0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}}

Now I would love to know what other linear second order ODEs of the form above can be mapped onto hypergeometric functions by a suitable substitution.

By applying the same algorithm to the Bessel functions we get the following answer:
begin{equation}
frac{d^2 y(x)}{d x^2} -n^2(A b-a B)^2 frac{left(-B^{2 n}(A x+a)^{2n}+A^{2 n}(B x+b)^{2 n}right)}{B^{2 n}(A x+a)^2(B x+b)^{2n+2}}cdot y(x)=0
end{equation}
is solved by
begin{eqnarray}
&&y(x)=\
&&C_1cdot sqrt{(A x+a)(B x+b)} J_{frac{sqrt{1+(A/B)^{2 n}4 n^2}}{2 n}}left[(frac{A x+a}{B x+b})^nright]+\
&&
C_2cdot sqrt{(A x+a)(B x+b)} J_{-frac{sqrt{1+(A/B)^{2 n}4 n^2}}{2 n}}left[(frac{A x+a}{B x+b})^nright]
end{eqnarray}

The result is checked by the following piece of code:

In[115]:= Table[

 FullSimplify[(D[#, {x, 2}] - ( 

       n^2 (A b - a B)^2 (-B^(2 n) (A x + a)^(2 n) + 

          A^(2 n) (B x + b)^(2 n)))/(

       B^(2 n) (a + A x)^2 (b + B x)^(2 n + 2)) #) & /@ {Sqrt[(a + 

        A x) (b + B x)]

      BesselJ[Sqrt[1 + (A/B)^(2 n) 4 n^2]/(

      2 n), ((A x + a)/(B x + b))^n], 

    Sqrt[(a + A x) (b + B x)]

      BesselJ[-(Sqrt[1 + (A/B)^(2 n) 4 n^2]/(2 n)), ((A x + a)/(

       B x + b))^n]}], {n, 1, 6}]



Out[115]= {{0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}, {0, 0}}

Let $a,b,A,B in {mathbb N}$ subject to $(a-b)^2 + (A-B)^2 > 0$.
Now let $a_1,a_2,b_1 in {mathbb R}$ and define:
begin{eqnarray}
P_0 &:=& a^2 (a_1-a_2-1) (a_1-a_2+1)+2 a b (2 a_1 a_2-a_1 b_1-a_2 b_1+b_1)+b^2 (b_1-2) b_1 \
P_1 &:=& 2 (A (a (a_1-a_2-1) (a_1-a_2+1)+b (2 a_1 a_2-a_1 b_1-a_2 b_1+b_1))+B (a (2 a_1 a_2-a_1 b_1-a_2 b_1+b_1)+b
(b_1-2) b_1)) \
P_2 &:=& A^2 (a_1-a_2-1) (a_1-a_2+1)+2 A B (2 a_1 a_2-a_1 b_1-a_2 b_1+b_1)+B^2 (b_1-2) b_1
end{eqnarray}

Then the fundamental solutions to the following ODE:
begin{eqnarray}
frac{d^2 y(x)}{d x^2} - (a B-A b)^2frac{left(P_0 + P_1 x + P_2 x^2right)}{4 (a+A x)^2 (b+B x)^2 (a-b+ (A-B)x)^2} cdot y(x) = 0
end{eqnarray}
have the following form:
begin{eqnarray}
&&y_1(x) := (a+A x)^{b_1/2} (b+B x)^{frac{1}{2} (-a_1-a_2+1)} (a+x (A-B)-b)^{frac{1}{2} (a_1+a_2-b_1+1)} , _2F_1left(a_1,a_2;b_1;frac{a+A x}{b+B x}right)\
&&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!y_2(x) := (a+A x)^{1-frac{b_1}{2}} (b+B x)^{frac{1}{2} (-a_1-a_2-1)+b_1} (a+x (A-B)-b)^{frac{1}{2} (a_1+a_2-b_1+1)} ,
_2F_1left(a_1-b_1+1,a_2-b_1+1;2-b_1;frac{a+A x}{b+B x}right)
end{eqnarray}

The following Mathematica code verifies the result:

In[1109]:= A =.; B =.; a =.; b =.; Clear[y1]; Clear[y2]; Clear[v]; x =.;

{A, B, a, b} = RandomInteger[{0, 20}, 4];

v[x_] := (-A b + a B)^2/(

   4 (a + A x)^2 (a - b + A x - B x)^2 (b + 

      B x)^2) (a^2 (-1 + a1 - a2) (1 + a1 - a2) + b^2 (-2 + b1) b1 + 

     2 a b (2 a1 a2 + b1 - a1 b1 - a2 b1) + 

     2 (A (a (-1 + a1 - a2) (1 + a1 - a2) + 

           b (2 a1 a2 + b1 - a1 b1 - a2 b1)) + 

        B (b (-2 + b1) b1 + 

           a (2 a1 a2 + b1 - a1 b1 - a2 b1))) x + (A^2 (-1 + a1 - 

           a2) (1 + a1 - a2) + B^2 (-2 + b1) b1 + 

        2 A B (2 a1 a2 + b1 - a1 b1 - a2 b1)) x^2);

y1[x_] := (a + A x)^(b1/2) (a - b + (A - B) x)^(

   1/2 (1 + a1 + a2 - b1)) (b + B x)^(1/2 (1 - a1 - a2))

    Hypergeometric2F1[a1, a2, b1, (A x + a)^1/(B x + b)^1];

y2[x_] := (a + A x)^(1 - b1/2) (a - b + (A - B) x)^(

   1/2 (1 + a1 + a2 - b1)) (b + B x)^(1/2 (-1 - a1 - a2) + b1)

    Hypergeometric2F1[1 + a1 - b1, 1 + a2 - b1, 

    2 - b1, (A x + a)^1/(B x + b)^1];

FullSimplify[(D[#, {x, 2}] - v[x] #) & /@ {y1[x], y2[x]}]





Out[1114]= {0, 0}

Update 0: The result above can be used to solve the following inverse problem.
Let $A=B=1$. Now let $a,b in {mathbb N}$ and let $P_0,P_1,P_2 in {mathbb N}$ subject to $P_0^2 + P_1^2 + P_2^2 > 0$. Then there always exist certain $a_1,a_2,b_1 in {mathbb R}$ such that the functions $y_{1,2}(x)$ above are fundamental solutions to the following ODE:
begin{eqnarray}
frac{d^2 y(x)}{d x^2} - frac{(P_0+P_1 x+P_2 x^2)}{(x+a)^2(x+b)^2} cdot y(x)=0
end{eqnarray}
Indeed if we set $A=B=1$ and then if we set $a,b in {mathbb N}$ in the top three equations defining $P_0,P_1,P_2$ above we can always solve those equations for $a_1,a_2,b_1$. Here is the Mathematica code that accomplishes that:

In[1473]:= {A, B} = {1, 1}; Clear[y1]; Clear[y2];

y1[x_] := (a + A x)^(b1/2) (a - b + (A - B) x)^(

   1/2 (1 + a1 + a2 - b1)) (b + B x)^(1/2 (1 - a1 - a2))

    Hypergeometric2F1[a1, a2, b1, (A x + a)^1/(B x + b)^1];

y2[x_] := (a + A x)^(1 - b1/2) (a - b + (A - B) x)^(

   1/2 (1 + a1 + a2 - b1)) (b + B x)^(1/2 (-1 - a1 - a2) + b1)

    Hypergeometric2F1[1 + a1 - b1, 1 + a2 - b1, 

    2 - b1, (A x + a)^1/(B x + b)^1];

{a, b} = RandomInteger[{0, 10}, 2];

{P0, P1, P2} = RandomInteger[{0, 10}, 3];

a1 =.; a2 =.; b1 =.;

subst = FullSimplify[

   Solve[{a^2 (-1 + a1 - a2) (1 + a1 - a2) + b^2 (-2 + b1) b1 + 

       2 a b (2 a1 a2 + b1 - a1 b1 - a2 b1), 

      2 (A (a (-1 + a1 - a2) (1 + a1 - a2) + 

            b (2 a1 a2 + b1 - a1 b1 - a2 b1)) + 

         B (b (-2 + b1) b1 + 

            a (2 a1 a2 + b1 - a1 b1 - a2 b1))) , (A^2 (-1 + a1 - 

           a2) (1 + a1 - a2) + B^2 (-2 + b1) b1 + 

        2 A B (2 a1 a2 + b1 - a1 b1 - a2 b1))} == 4 {P0, P1, P2}, {a1,

      a2, b1}]];

subst = Sort[{a1, a2, b1} /. subst, #1[[3]] < #2[[3]] &];

MatrixForm[subst]

aa = FullSimplify[(D[#, {x, 2}] - (

       P0 + P1 x + P2 x^2)/((x + a)^2 (x + b)^2) #) & /@ {y1[x], 

    y2[x]}]

FullSimplify[aa /. Diagonal[Thread[{a1, a2, b1} -> #] & /@ subst[[1]]]]





Out[1483]= {0, 0}

In this particular example we had ${a,b}={9,7}$,${P_0,P_1,P_2}={ 0,4,9}$ and
begin{eqnarray}
left(begin{array}{r}a_1\a_2\b_1end{array}right)=left(
begin{array}{r}frac{1}{2} left(1+sqrt{37}+3 sqrt{46}-sqrt{694}right)\frac{1}{2} left(1-sqrt{1145+12 sqrt{7981}-4 sqrt{37 left(277+3 sqrt{7981}right)}}right)\1-sqrt{694}end{array}right)
end{eqnarray}

Update 1: Now let us go back to Update 0 and let us take some particular examples where we can indeed give closed form solutions .

(A)
If we set $P_2=P_1=0$ and $P_0 neq 0$ then we get the following:
begin{eqnarray}
a_1 &=& 1\
a_2 &=& 1+frac{sqrt{(a-b)^2+4 P_0}}{a-b}\
b_1 &=& a_2
end{eqnarray}
Therefore the solutions to
begin{equation}
frac{d^2 y(x)}{d x^2} - frac{P_0}{(x+a)^2(x+b)^2} y(x)=0
end{equation}
are
begin{eqnarray}
y_1(x) &=& {(a+x)^{frac{1}{2}+frac{sqrt{(a-b)^2+4 text{P0}}}{2 (a-b)}} (b+x)^{frac{1}{2}-frac{sqrt{(a-b)^2+4 text{P0}}}{2 (a-b)}}}\
y_2(x) &=& {(a+x)^{frac{1}{2}-frac{sqrt{(a-b)^2+4 text{P0}}}{2 (a-b)}}
(b+x)^{frac{1}{2}+frac{sqrt{(a-b)^2+4 text{P0}}}{2 (a-b)}}}
end{eqnarray}
Note that :
begin{eqnarray}
lim_{brightarrow a} y_{1,2}(x) = (x+a) expleft( pm frac{sqrt{P_0}}{x+a}right)
end{eqnarray}
as it should be.

(B) Likewise let use take $P_2=P_0=0$ and $P_1 neq 0$. Then we are getting the following solution:
begin{eqnarray}
a_2 &=& 1-frac{sqrt{sqrt{16 a b P_1^2-4 P_1 (a+b) (a-b)^2+(a-b)^4}-2 P_1 (a+b)+(a-b)^2}}{sqrt{2} (a-b)}\
a_1 &=& 1-frac{P_1}{(1-a_2) (b-a)}\
b_1 &=& -1+a_1+a_2
end{eqnarray}
Therefore the solutions to
begin{equation}
frac{d^2 y(x)}{d x^2} - frac{P_1 x}{(x+a)^2(x+b)^2} y(x)=0
end{equation}
are
begin{eqnarray}
y_1(x) &=& (a+x)^{b_1/2} (b+x)^{frac{1}{2} (-a_1-a_2+1)} , _2F_1left(a_1,a_2;b_1;frac{a+x}{b+x}right)\
y_2(x) &=& (a+x)^{1-frac{b_1}{2}} (b+x)^{frac{1}{2} (-a_1-a_2-1)+b_1} , _2F_1left(a_1-b_1+1,a_2-b_1+1;2-b_1;frac{a+x}{b+x}right)
end{eqnarray}

Note that:
begin{eqnarray}
lim_{brightarrow a_+} a_2 &=& 1+ omega\
a_1 &=& 1-frac{4 a omega}{theta} \
b_1 &=& omega + 1 - frac{4 a omega}{theta}
end{eqnarray}
where $omega := frac{imath}{2} sqrt{frac{P_1}{a}}$ and $theta:=b-a$.
Therefore we have:
begin{eqnarray}
&&theta^{1+omega}cdot y_1(x) =\
&& (x+a)^{frac{omega+1}{2}-frac{2 a omega}{theta}} cdot (x+a +theta)^{frac{1}{2}(-1-omega+frac{4 a}{omega})} cdot theta^{1+omega} F_{2,1}left[
begin{array}{rr}
1+omega & 1- frac{4 a omega}{theta}\
& omega+1-frac{4 a omega}{theta}
end{array};
frac{x+a}{x+a+theta}
right]=\
&&left(1+frac{theta}{x+a}right)^{frac{2 a omega}{theta}} cdot theta^{1+omega} F_{2,1}left[
begin{array}{rr}
1+omega & 1- frac{4 a omega}{theta}\
& omega+1-frac{4 a omega}{theta}
end{array};
frac{x+a}{x+a+theta}
right] underbrace{=}_{theta rightarrow 0}\
&&e^{frac{2 a omega}{x+a}}cdot (x+a) (-4 a omega)^omega U(omega,0;-frac{4a omega}{x+a})
end{eqnarray}
See Compute a limit that involves a hypergeometric function. for explanations.

In case of the second function $(a_1,a_2,b_1) rightarrow (a_1-b_1+1,a_2-b_1+1,2-b_1)$ which is equivalent to $omega rightarrow -omega$
and therefore :
begin{eqnarray}
lim_{brightarrow a} y_{1,2}(x) = (x+a) cdot expleft(pm frac{2 a omega}{x+a}right) cdot U(pmomega,0;mpfrac{4 a omega}{x+a})
end{eqnarray}
where $U$ is the confluent hypergeometric function.

(C) Now let us assume that $P_0=P_1=0$ and $P_2neq 0$. Define $Q:=sqrt{1+4 P_2}$. Then we have:
begin{eqnarray}
&&a_2^4 (a-b)^2+\
&&-2 a_2^3 (Q+1) (a-b)^2+\
&&a_2^2 left(a^2 (4 P_2+3 Q+2)-2 a b (6 P_2+3 Q+2)+b^2 (4 P_2+3 Q+2)right)+\
&&-a_2 left(a^2 (4 P_2+Q+1)-2 a b (2 P_2 (Q+3)+Q+1)+b^2 (4 P_2+Q+1)right)+\
&&-2 a b P_2 (Q+1)=0\
&&hline\
a_1&=& frac{b (a_2 Q+a_2-4 P_2-Q-1)-a (a_2-1) (Q+1)}{(a-b) (-2 a_2+Q+1)}\
b_1&=&a_1+a_2-Q
end{eqnarray}

Therefore the solutions to
begin{equation}
frac{d^2 y(x)}{d x^2} - frac{P_2 x^2}{(x+a)^2(x+b)^2} y(x)=0
end{equation}
are
begin{eqnarray}
y_1(x) &=& (a+x)^{b_1/2} (b+x)^{frac{1}{2} (-a_1-a_2+1)} , _2F_1left(a_1,a_2;b_1;frac{a+x}{b+x}right)\
y_2(x) &=& (a+x)^{1-frac{b_1}{2}} (b+x)^{frac{1}{2} (-a_1-a_2-1)+b_1} , _2F_1left(a_1-b_1+1,a_2-b_1+1;2-b_1;frac{a+x}{b+x}right)
end{eqnarray}

Now the calculation of the limit of $b$ going to $a$ is very similar to the case before so we only present the result. We have:
begin{eqnarray}
lim_{brightarrow a} y_{1,2}(x) = left(x+a right)^{frac{1-Q}{2}}cdot expleft(mp frac{asqrt{Q^2-1}}{2(x+a)} right) cdot Uleft(frac{1}{2}(1+Qmpsqrt{Q^2-1}),1+Q;pm frac{asqrt{Q^2-1}}{x+a} right)
end{eqnarray}
where $Q:=sqrt{1+4 P_2}$.

(D) Now let us assume that $P_0$,$P_1$ and $P_2$ are arbitrary subject to $P_1 > 2 a P_2$. Then we have:
begin{eqnarray}
&&a_2^4 (a-b)^2+\
&&-2 a_2^3 (Q+1) (a-b)^2+\
&&a_2^2 left(a^2 (4 P_2+3 Q+2)+2 a (P_1-b (6 P_2+3 Q+2))+b^2 (4 P_2+3 Q+2)+2 b P_1-4 P_0right)+\
&&a_2 left(a^2 (-(4 P_2+Q+1))+2 a (b (2 P_2 (Q+3)+Q+1)-P_1
(Q+1))-b^2 (4 P_2+Q+1)-2 b P_1 (Q+1)+4 P_0 (Q+1)right)+\
&&a (Q+1) (P_1-2 b P_2)+P_1 (b Q+b+P_1)-2 P_0 (2 P_2+Q+1)=0\
&&hline\
&&a_1=frac{-a (a_2-1) (Q+1)+b (a_2 Q+a_2-4 P_2-Q-1)+2 P_1}{(a-b) (-2 a_2+Q+1)}\
&&b_1=a_1+a_2-Q
end{eqnarray}

Therefore the solutions to
begin{equation}
frac{d^2 y(x)}{d x^2} - frac{P_0+P_1 x+P_2 x^2}{(x+a)^2(x+b)^2} y(x)=0
end{equation}
are
begin{eqnarray}
y_1(x) &=& (a+x)^{b_1/2} (b+x)^{frac{1}{2} (-a_1-a_2+1)} , _2F_1left(a_1,a_2;b_1;frac{a+x}{b+x}right)\
y_2(x) &=& (a+x)^{1-frac{b_1}{2}} (b+x)^{frac{1}{2} (-a_1-a_2-1)+b_1} , _2F_1left(a_1-b_1+1,a_2-b_1+1;2-b_1;frac{a+x}{b+x}right)
end{eqnarray}
In the limit $b$ going to $a$ we have the following result:
begin{eqnarray}
lim_{brightarrow a} y_{1,2}(x) = left(x+aright)^{frac{1-Q}{2}}cdot expleft(pm frac{R}{x+a}right)cdot Uleft(frac{1}{2}(1+Qpmfrac{-P_1+2 a P_2}{R}),1+Q;mp frac{2 R}{x+a} right)
end{eqnarray}
where $Q:=sqrt{1+4 P_2}$ and $R:=sqrt{P_0-P_1 a+P_2 a^2}$.

Again by applying the same algorithm to the Gaussian hypergeometric differential equation, i.e. by rescaling the ODE in question by $x rightarrow f(x)$,$d/dx rightarrow 1/f^{'}(x) d/dx$ with $f(x):=A x^n$ and then by eliminating the term proportional to the first derivative we found the following result.

Let $a$,$b$,$c$,$A$ and $n$ be real numbers. Then the following ODE:
begin{eqnarray}
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!y^{''}(x) + left(frac{-A^2 x^{2 n} left(a^2 n^2-2 a b n^2+b^2 n^2-1right)+2 A x^n left(n^2 (a (c-2 b)+(b-1) c+1)-1right)-(c-1)^2 n^2+1}{4 x^2 left(1-A x^nright)^2}right) y(x)=0
end{eqnarray}
is solved by $y(x) = C_1 y_1(x) + C_2 y_2(x)$ where :
begin{eqnarray}
y_1(x)&=&x^{frac{1}{2} ((c-1) n+1)} left(1-A x^nright)^{frac{1}{2} (a+b-c+1)} , _2F_1left(a,b;c;A x^nright)\
y_2(x)&=&x^{frac{1}{2} (1-(c-1) n)} left(1-A x^nright)^{frac{1}{2} (a+b-c+1)} , _2F_1left(a-c+1,b-c+1;2-c;A x^nright)
end{eqnarray}

The following Mathematica code neatly verifies the result:

In[759]:= Clear[y1]; Clear[y2]; A =.; n =.; a =.; b =.; c =.;

y1[x_] = x^(1/2 ((1 + (-1 + c) n) )) (1 - A x^n)^(

   1/2 ((1 + a + b - c))) Hypergeometric2F1[a, b, c, A x^n];

y2[x_] = x^(1/2 ((1 - (-1 + c) n) )) (1 - A x^n)^(

   1/2 ((1 + a + b - c)))

    Hypergeometric2F1[a + 1 - c, b + 1 - c, 2 - c, A x^n];

FullSimplify[((

      1 - (-1 + c)^2 n^2 + 

       2 A (-1 + (1 + (-1 + b) c + a (-2 b + c)) n^2) x^n - 

       A^2 (-1 + a^2 n^2 - 2 a b n^2 + b^2 n^2) x^(2 n))/(

      4 x^2 (1 - A x^n)^2)) # + D[#, {x, 2}]] & /@ {y1[x], y2[x]}



Out[762]= {0, 0}

Let $A$,$B$,$C$, $D$ and $n$ be integers. Let $P_0$,$P_1$ and $P_2$ be another integers. Now let $a$,$b$ and $c$ be complex numbers such that such that:
begin{eqnarray}
P_0&=&B^2 left(a^2-2 a b+b^2-1right)+2 B D (2 a b-a c-b c+c)+(c-2) c D^2\
P_1&=&2 left(A left(B left(a^2-2 a b+b^2-1right)+D (2 a b-a c-b c+c)right)+C (a B (2 b-c)+c (-b B+B+(c-2)
D))right)\
P_2&=&A^2 left(a^2-2 a b+b^2-1right)+2 A C (2 a b-a c-b c+c)+(c-2) c C^2
end{eqnarray}

Consider the following ODE:
begin{eqnarray}
y^{''}(x) + frac{n}{x} y^{'}(x) + left( frac{n(n-2)}{4 x^2} - (B C - A D)^2 frac{P_0 + P_1 x+P_2 x^2}{4(B+A x)^2(B-D+(A-C)x)^2(D+C x)^2}right)y(x)=0
end{eqnarray}
then
begin{eqnarray}
&&y(x)= x^{-n/2} (A x+B)^{c/2} (C x+D)^{frac{1}{2} (-a-b+1)} (A x+B-C x-D)^{frac{1}{2} (a+b-c+1)} cdot \
&&left( C_2 left(frac{A x+B}{C x+D}right)^{1-c} , _2F_1left(a-c+1,b-c+1;2-c;frac{B+A x}{D+C x}right)+C_1 , _2F_1left(a,b;c;frac{B+A x}{D+C x}right)right)
end{eqnarray}

In[13]:= A =.; B =.; CC =.; DD =.; a =.; b =.; c =.; Clear[m]; n =.; 

x =.;

{A, B, CC, DD} = RandomSample[Range[1, 10], 4];

{P0, P1, P2} = RandomSample[Range[1, 10], 3];

subst = Solve[{(-1 + a^2 - 2 a b + b^2) B^2 + 

      2 B (2 a b + c - a c - b c) DD + (-2 + c) c DD^2, 

     2 (A ((-1 + a^2 - 2 a b + b^2) B + (2 a b + c - a c - b c) DD) + 

        CC (a B (2 b - c) + c (B - b B + (-2 + c) DD))), 

     A^2 (-1 + a^2 - 2 a b + b^2) + 

      2 A (2 a b + c - a c - b c) CC + (-2 + c) c CC^2} == {P0, P1, 

     P2}, {a, b, c}];

{a, b, c} = {a, b, c} /. subst[[1]];

m[x_] = x^(-n/2) (CC x + DD)^(1/2 (1 - a - b)) (B + A x)^(

   c/2) (B - DD + A x - CC x)^(1/2 (1 + a + b - c));

eX = (D[#, {x, 2}] + 

      n/x D[#, 

        x] + (((-2 + n) n)/(

         4 x^2) - ((B CC - A DD)^2 (P0 + P1 x + P2 x^2))/(

         4 (B + A x)^2 (B - DD + A x - CC x)^2 (DD + 

            CC x)^2)) #) & /@ {m[

      x] (C[1] Hypergeometric2F1[a, b, c, (A x + B)/(CC x + DD)] + 

       C[2] ((A x + B)/(CC x + DD))^(1 - c)

         Hypergeometric2F1[a + 1 - c, b + 1 - c, 2 - c, (A x + B)/(

         CC x + DD)])};



{n, x} = RandomReal[{1, 10}, 2, WorkingPrecision -> 50];

Simplify[eX]







Out[21]= {(0.*10^-48 + 0.*10^-49 I) C[

    1] + (0.*10^-48 + 0.*10^-48 I) C[2]}