Solutions in terms of the hypergeometric functions

Question

Is it possible to somehow express the solutions to the differential equations: $$\frac{d^2y}{dx^2} + \bigg(\frac{1}{x + 8} - \frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 4}\bigg) \frac{dy}{dx} + \bigg(\frac{1}{x^2} + \frac{3}{4x} - \frac{5}{6(x - 1)} - \frac{1}{4(x - 4)^2}\bigg) y = 0$$ and $$\frac{d^2y}{dx^2} + \bigg(\frac{1}{x + 8} + \frac{1}{3x} + \frac{1}{x - 64}\bigg) \frac{dy}{dx} + \bigg(\frac{7}{144x^2} - \frac{7}{3072x} + \frac{7}{3072(x - 64)}\bigg) y = 0$$ in terms of the hypergeometric functions?

Answer 1

Hint:

For $\dfrac{d^2y}{dx^2}+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\dfrac{dy}{dx}+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)y=0$ ,

Let $y=x^au$ ,

Then $\dfrac{dy}{dx}=x^a\dfrac{du}{dx}+ax^{a-1}u$

$\dfrac{d^2y}{dx^2}=x^a\dfrac{d^2u}{dx^2}+ax^{a-1}\dfrac{du}{dx}+ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u=x^a\dfrac{d^2u}{dx^2}+2ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u$

$\therefore x^a\dfrac{d^2u}{dx^2}+2ax^{a-1}\dfrac{du}{dx}+a(a-1)x^{a-2}u+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\left(x^a\dfrac{du}{dx}+ax^{a-1}u\right)+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)x^au=0$

$\dfrac{d^2u}{dx^2}+\dfrac{2a}{x}\dfrac{du}{dx}+\dfrac{a(a-1)}{x^2}u+\left(\dfrac{1}{x+8}+\dfrac{1}{3x}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{a}{x(x+8)}+\dfrac{a}{3x^2}+\dfrac{a}{x(x-64)}\right)u+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)u=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{6a+1}{3x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{a(3a-2)}{3x^2}+\dfrac{a}{8x}-\dfrac{a}{8(x+8)}-\dfrac{a}{64x}+\dfrac{a}{64(x-64)}\right)u+\left(\dfrac{7}{144x^2}-\dfrac{7}{3072x}+\dfrac{7}{3072(x-64)}\right)u=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{6a+1}{3x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{48a(3a-2)+7}{144x^2}+\dfrac{7(48a-1)}{3072x}-\dfrac{a}{8(x+8)}+\dfrac{48a+7}{3072(x-64)}\right)u=0$

Choose $a=\dfrac{1}{12}$ , the ODE becomes

$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{2x}+\dfrac{1}{x+8}+\dfrac{1}{x-64}\right)\dfrac{du}{dx}+\left(\dfrac{7}{1024x}-\dfrac{1}{96(x+8)}+\dfrac{11}{3072(x-64)}\right)u=0$

In fact according to http://www.wolframalpha.com/input/?i=y%27%27%2B(1%2F(x%2B8)%2B1%2F(3x)%2B1%2F(x-64))y%27%2B(7%2F(144x%5E2)-7%2F(3072x)%2B7%2F(3072(x-64)))y%3D0, it is possible to simplify to hypergeometric ODE luckily.

For $\dfrac{d^2y}{dx^2}+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\dfrac{dy}{dx}+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)y=0$ ,

Let $y=x^a(x-4)^bu$ ,

Then $\dfrac{dy}{dx}=x^a(x-4)^b\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)u$

$\dfrac{d^2y}{dx^2}=x^a(x-4)^b\dfrac{d^2u}{dx^2}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u=x^a(x-4)^b\dfrac{d^2u}{dx^2}+2x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u$

$\therefore x^a(x-4)^b\dfrac{d^2u}{dx^2}+2x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\left(x^a(x-4)^b\dfrac{du}{dx}+x^a(x-4)^b\left(\dfrac{a}{x}+\dfrac{b}{x-4}\right)u\right)+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)x^a(x-4)^bu=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{2a}{x}+\dfrac{2b}{x-4}\right)\dfrac{du}{dx}+\left(\dfrac{a(a-1)}{x^2}+\dfrac{2ab}{x(x-4)}+\dfrac{b(b-1)}{(x-4)^2}\right)u+\left(\dfrac{1}{x+8}-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x-4}\right)\dfrac{du}{dx}+\left(\dfrac{a}{x(x+8)}-\dfrac{a}{x^2}+\dfrac{a}{x(x-1)}+\dfrac{a}{x(x-4)}+\dfrac{b}{(x-4)(x+8)}-\dfrac{b}{x(x-4)}+\dfrac{b}{(x-1)(x-4)}+\dfrac{b}{(x-4)^2}\right)u+\left(\dfrac{1}{x^2}+\dfrac{3}{4x}-\dfrac{5}{6(x-1)}-\dfrac{1}{4(x-4)^2}\right)u=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{2a-1}{x}+\dfrac{1}{x-1}+\dfrac{2b+1}{x-4}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}+\left(\dfrac{a(a-2)+1}{x^2}+\dfrac{3}{4x}+\dfrac{a}{x(x-1)}-\dfrac{5}{6(x-1)}+\dfrac{2ab+a-b}{x(x-4)}+\dfrac{a}{x(x+8)}+\dfrac{b}{(x-1)(x-4)}+\dfrac{b}{(x-4)(x+8)}+\dfrac{4b^2-1}{4(x-4)^2}\right)u=0$

Choose $a=1$ and $b=-\dfrac{1}{2}$ , the ODE becomes

$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}+\left(\dfrac{3}{4x}+\dfrac{1}{x(x-1)}-\dfrac{5}{6(x-1)}+\dfrac{1}{2x(x-4)}+\dfrac{1}{x(x+8)}-\dfrac{1}{2(x-1)(x-4)}-\dfrac{1}{2(x-4)(x+8)}\right)u=0$

$\dfrac{d^2u}{dx^2}+\left(\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+8}\right)\dfrac{du}{dx}-\left(\dfrac{1}{4x}-\dfrac{1}{3(x-1)}+\dfrac{1}{12(x-4)}+\dfrac{1}{12(x+8)}\right)u=0$

Answer 2

This is an answer to the comment posted by doraemonpaul . The question is to find solutions and possibly outline differences in finding those solutions to the first ODE above. The new ODE reads: \begin{eqnarray} \frac{d^2 v(x)}{d x^2} +\underbrace{\left( \frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+8} \right)}_{a_1(x)} \frac{d v(x)}{d x} - \left( \frac{1}{4 x} - \frac{1}{3(x-1)} + \frac{1}{12(x-4)} + \frac{1}{12(x+8)}\right)v(x)=0 \end{eqnarray} In here we start from the Heun ODE and as usual we change the dependent variable $x\rightarrow f(x)$ and then the independent variable $y(x)=m(x) v(x)$ and then we choose the function $m(x)$ so that the coefficients at the first derivative match. This gives: \begin{equation} m(x)=\sqrt{x(x-1)(x+8)} (1-f(x))^{-d/2} f[x]^{-c/2} \sqrt{f^{'}(x)} (x_0-f(x))^{1/2(-a-b+c+d-1)} \end{equation} and \begin{eqnarray} &&\frac{d^2 v(x)}{d x^2} +\left( \frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+8} \right) \frac{d v(x)}{d x} + \left(\right.\\ &&\left. \frac{f'(x)^2 \left(-a^2-2 a (b-c-d)-b^2+2 b (c+d)-c^2-2 c d-d^2+1\right)}{4 (x_0-f(x))^2}\right.+\\ &&\frac{f'(x)^2 \left(c (a (x_0-1)+b (x_0-1)-2 d x_0+d+x_0-1)+x_0 (a (d-2 b)+d (b-d+1))+c^2 (-(x_0-1))+2 q\right)}{2 (x_0-1) x_0 (x_0-f(x))}+\\ &&\frac{(2-d) d f'(x)^2}{4 (f(x)-1)^2}+\\ &&\frac{f'(x)^2 \left(a (d-2 b)+d (b-c x_0+1)-d^2+2 q\right)}{2 (x_0-1) (f(x)-1)}+\\ &&\frac{f'(x)^2 \left(a (d-2 b)+d (b-c x_0+1)-d^2+2 q\right)}{2 (x_0-1) (f(x)-1)}+\\ &&\frac{f'(x)^2 \left(c (a+b+d x_0-d+1)-c^2-2 q\right)}{2 x_0 f(x)}+\\ &&\frac{1}{4} \left(2 a_1'(x)+a_1(x)^2+\underbrace{\frac{2 f^{(3)}(x) f'(x)-3 f''(x)^2}{f'(x)^2}}_{\mbox{Schwarzian derivative}}\right)\\ &&\left.\right)v(x)=0 \end{eqnarray} Now we take the good old Moebius function $f(x)=(A x+B)/(C x+D)$ (note that this anihilates the Schwarzian derivative in the last term in parentheses above) and we choose the upper case constants in the usual way i.e. via \begin{eqnarray} \left( \begin{array}{r} \frac{B}{A}\\ \frac{B-D}{A-C} \\ \frac{D}{C} \\ \frac{B-Dx_0}{A-Cx_0} \end{array}\right) = \left( \begin{array}{r} 0\\ -1 \\ -4 \\ 8\end{array}\right) \odot \pi \end{eqnarray} where $\pi$ is a permutation of length four. Having done this we decompose into partial fractions the coefficient at the zeroth derivative and then we anihilate the terms being proportional to second powers. This gives us four equations with five unknowns so we expect to get one free parameter. This is indeed the case. One of the twenty four cases reads: \begin{eqnarray} \frac{d^2 v(x)}{d x^2} +\left( \frac{1}{x} + \frac{1}{x-1} + \frac{1}{x+8} \right) \frac{d v(x)}{d x} + \left(\frac{11-24 q}{18 (x-1)}+\frac{6 q-\frac{3}{2}}{24 (x-4)}+\frac{18 q-8}{16 x}+\frac{-6 q-7}{144 (x+8)}\right) v(x)=0 \end{eqnarray} where \begin{eqnarray} v(x)=\frac{1}{m(x)} \left( C_1 \cdot Hn\left( \begin{array}{r|rr|} x_0 & a & b \\ q & c & d \end{array} f(x) \right) + C_2 \cdot [f(x)]^{1-c} Hn\left( \begin{array}{r|rr|} x_0 & a+1-c & b+1-c \\ q_1 & 2-c & d \end{array} f(x) \right) \right) \end{eqnarray} where $q_1=q-(c-1)(a+b-c-d+d x_0 +1)$ and $(a,b,c,d,q)=(1/2,1/2,1,0,q)$ and $x_0=2/3$ and $f(x)=(3 x)/(4(x-1))$.

A =.; B =.; CC =.; DD =.; x0 =.;
a =.; b =.; c =.; d =.; q =.;
A1[x_] = 1/x + 1/(x - 1) + 1/(x + 8);

perm = Permutations[{1, 2, 3, 4}];
sol = Table[{A, B, CC, DD, x0} /. 
    Solve[{B/A, 
       DD/CC, (B - DD)/(A - CC), (B - DD x0)/(A - 
          CC x0)} == {0, -1, -4, 8}[[perm[[j]]]]], {j, 1, 24}];
MatrixForm[sol]; myList = {};
Do[
 A =.; B =.; CC =.; DD =.; x0 =.;
 a =.; b =.; c =.; d =.; q =.;
 {A, B, CC, DD, x0} = First[sol[[which]]];
 f[x_] = Simplify[(A x + B)/(CC x + DD)];
 m[x_] = Sqrt[x (x - 1) (x + 8)] (1 - f[x])^(-d/2) f[x]^(-c/2) Sqrt[
    f'[x]] (x0 - f[x])^(1/2 (-a - b + d + c - 1));
 A0[x_] = 
  Apart[Together[-(-1 + a^2 + b^2 + c^2 + 2 a (b - c - d) + 2 c d + 
         d^2 - 2 b (c + d))  (Derivative[1][f][x]^2)/(
      4 (x0 - f[x])^2) + (2 q - 
        c^2 (-1 + x0) + ((1 + b - d) d + a (-2 b + d)) x0 + 
        c (-1 + d + a (-1 + x0) + b (-1 + x0) + x0 - 2 d x0))  (
      Derivative[1][f][x]^2)/(
      2 (-1 + x0) x0 (x0 - f[x])) - (-2 + d) d Derivative[1][f][x]^2/(
      4 (-1 + f[x])^2) + (-d^2 + a (-2 b + d) + 2 q + 
        d (1 + b - c x0))  (Derivative[1][f][x]^2)/(
      2 (-1 + x0) (-1 + f[x])) - (-2 + c) c  (Derivative[1][f][x]^2)/(
      4 f[x]^2) + (-c^2 - 2 q + c (1 + a + b - d + d x0))  (
      Derivative[1][f][x]^2)/(2 x0 f[x]) + 
     1/4 (A1[x]^2 + 
        2 Derivative[1][A1][x] + (-3 (f^\[Prime]\[Prime])[x]^2 + 
         2 Derivative[1][f][x] 
\!\(\*SuperscriptBox[\(f\), 
TagBox[
RowBox[{"(", "3", ")"}],
Derivative],
MultilineFunction->None]\)[x])/Derivative[1][f][x]^2)], x];
 eX = A0[x];
 subst = {a, b, c, d} /. 
   Solve[{Coefficient[eX, x, -2], Coefficient[eX, 1/(x - 1)^2], 
      Coefficient[eX, 1/(x - 4)^2], 
      Coefficient[eX, 1/(x + 8)^2]} == {0, 0, 0, 0}];
 {a, b, c, d} = subst[[1]];
 Clear[v]; Clear[y];
 myeqn = (D[
     y[x], {x, 
      2}] + (c/x + d/(x - 1) + (a + b - c - d + 1)/(x - x0)) D[y[x], 
      x] + (a b x - q)/(x (x - 1) (x - x0)) y[x]);
    subst = {x :> f[x], 
   Derivative[1][y][x] :> 1/f'[x] Derivative[1][y][x], 
   Derivative[2][y][x] :> -f''[x]/(f'[x])^3 Derivative[1][y][x] + 
     1/(f'[x])^2 Derivative[2][y][x]};
    myeqn = 
  Collect[(myeqn /. subst /. y[f[x]] :> y[x]), {y[x], y'[x], y''[x]}, 
   Simplify];
    y[x_] = m[x] v[x];
 eX = FullSimplify[
   myeqn /. Derivative[2][v][x] :> -A1[x] v'[x] - A0[x] v[x]];
 myList = Join[myList, {{f[x], x0}}];
 Print[{eX, A0[x], {a, b, c, d, q}, x0, f[x]}];
 , {which, 1, 24}]

To summarize this ODE is indeed different from the first one on top of this webpage in that in here we obtained twenty four different exactly solvable cases whereas in the other case all the twenty four cases turn out to be the same.

Answer 3

Let us focus on the second ODE since at the first glance it has three regular singular points and as such it should be possible to map it onto the hyper-geometric equation which too has three singular points. For 2nd order ODEs we always reduce it to the normal form, i.e. such that has no coefficient at the first derivative. This is done by writing $y(x)=m(x) \cdot v(x)$ where $m(x):=\exp(-1/2 \int a_1(x) dx)$and $a_1(x)$ is the coeffcient at the 1st derivative. In our case: \begin{equation} a_1(x)=\frac{1}{x+8} + \frac{1}{3 x} + \frac{1}{x-64} \end{equation} therefore $m(x)=(x+8)^{-1/2} x^{-1/6} (x-64)^{-1/2}$ and the function $v(x)$ satisfies the following ODE: \begin{equation} v^{''}(x) + \frac{48(1024+112 x+25 x^2)}{(-64+x)^2 x^2 (8+x)^2} v(x)=0 \quad (I) \end{equation}

Now what we want to do is to relate the ODE above to the appropriately transformed hypergeometric ODE. \begin{equation} x(1-x) Y^{''}(x) + (c-(a+b+1)x)Y^{'}(x)-a b Y(x)=0 \end{equation} We will start from that ODE and carry out two transformations. Let us take the Moebius function $f(x):=(A x+B)/(C x+D)$ and firstly change the abscissa $x \rightarrow f(x)$ and $d/dx \rightarrow 1/f^{'}(x) d/d x$ and then reduce the equation to the normal form by writing \begin{equation} Y(x):= (A x+B)^{-c/2} (B-D+(A-C) x)^{-(1+a+b-c)/2} (D+C x)^{(-1+a+b)/2}\cdot V(x) \end{equation} Having done this we obtain the following ODE for the function $V(x)$. We have: \begin{eqnarray} V^{''}(x)- \frac{(B C-A D)^2}{4} \cdot \frac{{\mathfrak A_0}+{\mathfrak A_1} x+{\mathfrak A_2} x^2}{(B+A x)^2 (B-D+(A-C) x)^2 (D+C x)^2} \cdot V(x)=0 \quad(II) \end{eqnarray} where: \begin{eqnarray} {\mathfrak A_0}&:=&B^2 \left(a^2-2 a b+b^2-1\right)+2 B D (2 a b-a c-b c+c)+(c-2) c D^2\\ {\mathfrak A_1}&:=&2 \left(A \left(B \left(a^2-2 a b+b^2-1\right)+D (2 a b-a c-b c+c)\right)+C (a B (2 b-c)+c (-b B+B+(c-2)D))\right)\\ {\mathfrak A}_2&:=&A^2 \left(a^2-2 a b+b^2-1\right)+2 A C (2 a b-a c-b c+c)+(c-2) c C^2 \end{eqnarray}

Note that ODE $(I)$ has exactly the same form as $(II)$. All we need to do is to adjust the parameters accordingly to map the later to the former. Firstly we find the upper case letters by matching the zeros of the denominator. Choosing \begin{eqnarray} B&=&-64 A\\ C&=&-8 A\\ D&=&-64 A \end{eqnarray} does the job. Now comes the toughest part meaning choosing the lower case parameters to match the numerators in ODEs $(I)$ and $(II)$. We have: \begin{eqnarray} {\mathfrak A_0}=-16 \left(4096 a^2+8192 a b-8192 a c+4096 b^2-8192 b c+4096 c^2-4096\right)=48 \times 1024\\ {\mathfrak A_1}=-16 \left(-128 a^2+2048 a b-896 a c-128 b^2-896 b c+1024 c^2-1152 c+128\right)=48 \times 112\\ {\mathfrak A_2}=-16 \left(a^2-34 a b+16 a c+b^2+16 b c+64 c^2-144 c-1\right)=48 \times 25 \end{eqnarray} These are just quadratic equations so they can be solved. In fact there are two sets of solutions: \begin{eqnarray} (a,b,c)&=&(\frac{1}{4},\frac{1}{4},1)\\ (a,b,c)&=&(\frac{3}{4},\frac{3}{4},1) \end{eqnarray} Now we have completed the job. We made sure that $v(x)=V(x)$. All we have to do is to bring everything together and simplify. We have: \begin{eqnarray} y(x)=C_1 x^{\frac{1}{12}} (8+x)^{-\frac{1}{4}} F_{2,1}\left[1/4,1/4,1;\frac{x-64}{-8 x-64}\right] + C_2 x^{\frac{7}{12}} (8+x)^{-\frac{3}{4}} F_{2,1}\left[3/4,3/4,1;\frac{x-64}{-8 x-64}\right] \end{eqnarray} The final step is to verify the result using a Computer Algebra System. We have:

In[295]:= 
FullSimplify[(D[#, {x, 2}] + (1/(x + 8) + 1/(3 x) + 1/(x - 64)) D[#, 
       x] + (7/(144 x^2) - 7/(3072 x) + 
        7/(3072 (x - 64))) #) & /@ { (x)^(1/(12)) ( (8 + x))^(-1/4)
     Hypergeometric2F1[1/4, 1/4, 1, (x - 64)/(-8 x - 64)],  ((x)^(
    7/(12)))/((8 + x))^(3/4)
     Hypergeometric2F1[3/4, 3/4, 1, (x - 64)/(-8 x - 64)]}]

Out[295]= {0, 0}

Answer 4

Now let us turn attention to the first ODE.

Here we start from the hypergeometric ODE \begin{eqnarray} y^{''}(x) + \left( \frac{c}{x} + \frac{a+b-c+1}{x-1} \right) y^{'}(x) + \frac{a b}{x(x-1)} y(x)=0 \end{eqnarray}

and we change the abscissa $x \rightarrow f(x)$ and after that the ordinate $y(x) = m(x) v(x)$. After some straightforward calculations we end up with the following ODE: \begin{eqnarray} v^{''}(x) + a_1(x) v^{'}(x) + a_0(x) v(x)=0 \end{eqnarray} where: \begin{eqnarray} a_0(x)&:=&\frac{m''(x)}{m(x)}+\frac{a b f'(x)^2}{(f(x)-1) f(x)} +\frac{m'(x)}{m(x)}\left(\frac{(a+b-c+1) f'(x)}{f(x)-1}+\frac{c f'(x)}{f(x)}-\frac{f''(x)}{f'(x)}\right) \quad (Ia)\\ a_1(x)&:=&\frac{(a+b-c+1) f'(x)}{f(x)-1}+\frac{c f'(x)}{f(x)}-\frac{f''(x)}{f'(x)}+\frac{2 m'(x)}{m(x)} \quad(Ib) \end{eqnarray}

Now we match the coefficients at the first derivative and solve for the function $m(x)$. Note that since $a_1(x)$ depends only on the first derivative of the function $m(x)$ we can always formally solve the resulting ODE. The solution reads: \begin{eqnarray} m(x)&=& \frac{\sqrt{(x-4) (x-1) (x+8)} f(x)^{-c/2} \sqrt{f'(x)} (1-f(x))^{\frac{1}{2} (-a-b+c-1)}}{\sqrt{x}} \quad (II) \end{eqnarray} Now all we need to do is to insert the above into the definition of $a_0(x)$ then equate the result to the coefficient at the zeroth derivative and solve for $f(x)$. This is easier said than done because the resulting differential equation is highly nonlinear.However what we can do is to assume that $f(x)$ has a particular functional form-- in this case it is a rational function and then adjust the parameters of that function such that the relevant coefficients match. This is again easier said than done because the resulting expressions quickly become unwieldy if only the degrees of the numerator and denominator get bigger than one. Therefore for the time being we only consider the old good Moebius function $f(x)=(A x+B)/(C x+D)$. If we insert this along with $(II)$ into $(Ia)$ we end up with a rational function which has the following factors in the denominator, firstly $x^2$, $(x-1)^2$, $(x-4)^2$ and $(x+8)^2$ and secondly $(A x+B)^2$, $(C x+D)^2$ and $((A-C)x+(B-D))^2$. Now there are only three ways for the relevant denominators to match. \begin{eqnarray} \left( \begin{array}{r} \frac{B}{A}\\ \frac{B-D}{A-C} \\ \frac{D}{C} \end{array}\right) = \left\{ \left( \begin{array}{r} 0\\ -1 \\ -4 \end{array}\right), \left( \begin{array}{r} 0\\ -1 \\ +8 \end{array}\right), \left( \begin{array}{r} 0\\ -4 \\ +8 \end{array}\right), \left( \begin{array}{r} -1\\ -4 \\ +8 \end{array}\right) \right\} \end{eqnarray} This leads to following solutions for the function $f(x)$. We have:

\begin{eqnarray} f(x)=\left\{ \frac{3/4 x}{-1/4 x+1}, \frac{9/8 x}{1/8 x+1}, \frac{3/8 x}{1/8 x+1}, \frac{1/2 x-1/2}{1/8 x+1} \right\} \quad (III) \end{eqnarray} Now inserting $(III)$ along with $(II)$ into $(Ia)$ and $(Ib)$ we get: \begin{eqnarray} v(x)&=&\frac{1}{m(x)} \left( C_1 F_{2,1}[a,b,c,f(x)] + C_2 f[x]^{1-c} F_{2,1}[a+1-c,b+1-c,2-c,f(x)] \right) \end{eqnarray} where \begin{eqnarray} a_1(x)&=&\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x+8}+\frac{1}{x-4} \end{eqnarray} and \begin{eqnarray} a_0(x)&=&\frac{-3 a^2+18 a b-6 a c-3 b^2-6 b c+9 c^2-12 c-8}{18 (x-1)}+\frac{-a^2-2 a b+2 a c-b^2+2 b c-c^2}{4 (x-1)^2}+\frac{4 a^2-6 a b-a c+4 b^2-b c+c-2}{24 (x-4)}+\frac{-a^2+2 a b-b^2}{4 (x-4)^2}+\frac{-12 a b+6 a c+6 b c-8 c^2+10 c+9}{16 x}+\frac{-c^2+2 c+3}{4 x^2}-\frac{5}{144 (x+8)}-\frac{1}{4 (x+8)^2}\\ a_0(x)&=&\frac{a^2+18 a b-10 a c+b^2-10 b c+9 c^2-8 c-12}{18 (x-1)}+\frac{-a^2-2 a b+2 a c-b^2+2 b c-c^2}{4 (x-1)^2}+\frac{-8 a^2+18 a b-a c-8 b^2-b c+c+3}{144 (x+8)}+\frac{-a^2+2 a b-b^2}{4 (x+8)^2}+\frac{-18 a b+9 a c+9 b c-8 c^2+7 c+9}{16 x}+\frac{-c^2+2 c+3}{4 x^2}+\frac{1}{12 (x-4)}-\frac{1}{4 (x-4)^2}\\ a_0(x)&=&\frac{a^2+6 a b-4 a c+b^2-4 b c+3 c^2-2 c+1}{24 (x-4)}+\frac{-a^2-2 a b+2 a c-b^2+2 b c-c^2}{4 (x-4)^2}+\frac{-6 a^2+18 a b-3 a c-6 b^2-3 b c+3 c+1}{144 (x+8)}+\frac{-a^2+2 a b-b^2}{4 (x+8)^2}+\frac{-6 a b+3 a c+3 b c-2 c^2+c+9}{16 x}+\frac{-c^2+2 c+3}{4 x^2}-\frac{11}{18 (x-1)}-\frac{1}{4 (x-1)^2}\\ a_0(x)&=&\frac{a^2+8 a b-5 a c+b^2-5 b c+4 c^2-3 c+1}{24 (x-4)}+\frac{-a^2-2 a b+2 a c-b^2+2 b c-c^2}{4 (x-4)^2}+\frac{-6 a^2+16 a b-2 a c-6 b^2-2 b c+2 c+1}{144 (x+8)}+\frac{-a^2+2 a b-b^2}{4 (x+8)^2}+\frac{-8 a b+4 a c+4 b c-3 c^2+2 c-11}{18 (x-1)}+\frac{-c^2+2 c-1}{4 (x-1)^2}+\frac{3}{4 x^2}+\frac{9}{16 x} \end{eqnarray} The nonbelievers can run the code below to make sure it is correct:

Clear[m]; Clear[f]; Clear[a1]; Clear[a0]; a =.; b =.; c =.; x =.;
m[x_] = Sqrt[(-4 + x) (-1 + x) (8 + x)]/
    Sqrt[x] (1 - f[x])^((-1 - a - b + c)/2) f[x]^(-c/2) Sqrt[f'[x]];
f[x_] = (3/4 x)/(-1/4 x + 1);(*(0,-1,-4)*)
f[x_] = (9/8 x)/(1/8 x + 1);(*(0,-1,+8)*)
f[x_] = (3/8 x)/(1/8 x + 1);(*(0,-4,+8)*)
f[x_] = (1/2 x - 1/2)/(1/8 x + 1);(*(-1,-4,+8)*)


a0[x_] = (
  a b Derivative[1][f][x]^2)/((-1 + f[x]) f[x]) + -((
   Derivative[1][m][
     x] (c Derivative[1][f][x]^2 - f[x] Derivative[1][f][x]^2 - 
      a f[x] Derivative[1][f][x]^2 - b f[x] Derivative[1][f][x]^2 - 
      f[x] (f^\[Prime]\[Prime])[x] + 
      f[x]^2 (f^\[Prime]\[Prime])[x]))/(
   m[x] (-1 + f[x]) f[x]  Derivative[1][f][x])) +  (
   m^\[Prime]\[Prime])[x]/ m[x];
a1[x_] = ((1 + a + b - c) Derivative[1][f][x])/(-1 + f[x]) + (
   c Derivative[1][f][x])/ f[x] + (2  Derivative[1][m][x])/ 
   m[x] - (f^\[Prime]\[Prime])[x]/ Derivative[1][f][x];

(Apart[Together[{a1[x], a0[x]}], x])
FullSimplify[(D[#, {x, 2}] + a1[x] D[#, {x, 1}] + a0[x] #) & /@ {1/
     m[x] (C[1] Hypergeometric2F1[a, b, c, f[x]] + 
      C[2] f[x]^(1 - c) Hypergeometric2F1[a + 1 - c, b + 1 - c, 2 - c,
         f[x]])}]

At the first glance it seems that it is not possible to match the coefficient $a_0(x)$ against that in our original ODE. The later simply contains not enough factors; for example it misses the very important factors $1/(x+8)$ and $1/(x+8)^2$. Therefore the lesson from this exercise is the following. In general as the complexity of our original ODE grows it will become harder and harder to match it against solutions of some known ODEs -- in this case the hypergeometric function. However we can always produce a whole family of ODEs which are close to the original one and whose solutions are known. This is what we have accomplished in here.

Answer 5

Now again we focus our attention on the first ODE. We will follow the same line of attack as in my previous answer above except that now we start from the Heun ODE rather than from the hypergeometric one. We have: \begin{eqnarray} y^{''}(x) + \left( \frac{c}{x} + \frac{d}{x-1} + \frac{a+b-c-d+1}{x-x_0} \right) y^{'}(x) + \frac{a b x - q}{x(x-1)(x-x_0)} y(x)=0 \end{eqnarray}

We change the abscissa $x→f(x)$ and after that the ordinate $y(x)=m(x)v(x)$. After some straightforward calculations we end up with the following ODE: \begin{eqnarray} v^{''}(x) + a_1(x) v^{'}(x) + a_0(x) v(x)=0 \end{eqnarray} where: \begin{eqnarray} a_0(x)&:=&\frac{m''(x)}{m(x)}+ \frac{(a b f(x)-q) f'(x)^2}{(f(x)-1) f(x) (f(x)-x_0)} + \frac{m'(x)}{m(x)}\left(\frac{(a+b-c-d+1) f'(x)}{f(x)-x_0}+d \frac{f^{'}(x)}{f(x)-1}+\frac{c f'(x)}{f(x)}-\frac{f''(x)}{f'(x)}\right) \quad (Ia)\\ a_1(x)&:=&\frac{(a+b-c-d+1) f'(x)}{f(x)-x_0}+d \frac{f^{'}(x)}{f(x)-1}+\frac{c f'(x)}{f(x)}-\frac{f''(x)}{f'(x)}+\frac{2 m'(x)}{m(x)} \quad(Ib) \end{eqnarray}

Now we match the coefficient at the first derivative meaing we solve the equation $a_1(x)= 1/(x-4) + 1/(x-1)-1/x+1/(x+8)$. This is a first order ODE with respect to $m(x)$ and it can always be solved since all terms in $a_1(x)$ are total derivatives. We have: \begin{equation} m(x)=\sqrt{\frac{(x-4) (x-1) (x+8)}{x}} (1-f(x))^{-d/2} f(x)^{-c/2} \sqrt{f'(x)} (x_0-f(x))^{\frac{1}{2} (-a-b+d+c-1)} \end{equation} We insert the above into equation $(Ia)$ and then take our old good Moebius function $f(x):=(A x+B)/(C x+D)$. Again it is not difficult to see that $a_0(x)$ is now a rational function which has the following factors in the denominator, firstly $x^2$, $(x-1)^2$, $(x-4)^2$ and $x+8)^2$ and secondly $(A x+B)^2$, $(C x+D)^2$, $((A-C)x+(B-D))^2$ and $((A-C x_0) x + (B-D x_0))^2$. Now in order to match the denominators in the coefficient in question we need to solve a following system of linear equations: \begin{eqnarray} \left( \begin{array}{r} \frac{B}{A}\\ \frac{B-D}{A-C} \\ \frac{D}{C} \\ \frac{B-Dx_0}{A-Cx_0} \end{array}\right) = \left( \begin{array}{r} 0\\ -1 \\ -4 \\ 8\end{array}\right) \end{eqnarray} which gives $(A,B,C,D,x_0)=(-3 C,0,C,-4 C,-2)$ and \begin{eqnarray} a_0(x)&=&\frac{a^2+2 c (a+b)-8 a b+3 a d+b^2+3 b d-3 d-2 c+2 q+1}{24 (x-4)}+\frac{-6 a^2+a (-16 b+14 d+15 c)-(6 b-8 d-9 c) (b-d-c)+2 d+3 c-2 q+1}{144 (x+8)}+\frac{a (8 b-4 d)+d (-4 b+d-8 c+2)-8 q-11}{18 (x-1)}+\frac{c (-3 a-3 b+9 d+c+1)+6 q+9}{16 x}-\frac{(a+b-d-c)^2}{4 (x+8)^2}-\frac{(a-b)^2}{4 (x-4)^2}-\frac{(d-1)^2}{4 (x-1)^2}-\frac{(c-3) (c+1)}{4 x^2}\\ a_1(x)&=&\frac{1}{x+8}-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-4} \end{eqnarray} All what we need to do now is to annihilate the coefficients at $1/(x+8)$, $1/(x+8)^2$, $1/(x-1)^2$ and $1/(x-4)$. Since there is five parameters $a$,$b$,$c$,$d$,$q$ and four constraints we expect to obtain one free parameter. This is indeed the case since the solution reads: \begin{eqnarray} a&=&\frac{1}{3}\left(-1+4 b + \sqrt{1-14 b+7 b^2} \right)\\ c&=&\frac{1}{3} \left(-4+7 b+ \sqrt{1-14 b+7 b^2} \right)\\ d&=&1\\ q&=& \frac{1}{6} \left(-3+25 b-16 b^2+(3-4 b) \sqrt{1-14 b+7 b^2}\right) \end{eqnarray} and \begin{eqnarray} a_0(x)&=&-\frac{7}{18} \frac{\Delta}{x^2}-\frac{4}{9} \frac{\Delta}{x} + \frac{4}{9} \frac{\Delta}{x-1} - \frac{1}{18} \frac{\Delta}{(x-4)^2}\\ a_1(x)&=&\frac{1}{x+8}-\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-4} \end{eqnarray} where \begin{equation} \Delta := 1-8 b + 4 b^2+(b-1) \sqrt{1+7(-2+b)b} \end{equation}

As usual I enclose a Mathematica code snippet that anyone can use to make sure that there no error in the equations above.

In[1]:= a =.; b =.; g =.; d =.; q =.; x0 =.; x =.; Clear[m]; \
Clear[v]; Clear[y];
a = 1/3 (-1 + 4 b + Sqrt[1 - 14 b + 7 b^2]); g = 
 1/3 (-4 + 7 b + Sqrt[1 - 14 b + 7 b^2]); d = 1; q = 
 1/6 (-3 + 25 b - 16 b^2 + (3 - 4 b) Sqrt[1 - 14 b + 7 b^2]); x0 = -2;
f[x_] = (-3 x)/(x - 4);
m[x_] = Sqrt[((-4 + x) (-1 + x) (8 + x))/x] f[x]^(-g/2) Sqrt[
   f'[x]] (1 - f[x])^(-d/2) (x0 - f[x])^(1/2 (-a - b + d + g - 1));

Clear[a0]; Clear[a1];
Delta = (1 - Sqrt[1 + 7 (-2 + b) b] + 
    b (-8 + 4 b + Sqrt[1 + 7 (-2 + b) b]));
a0[x_] = -((7 Delta)/(18 x^2)) - (4 Delta)/(9 x) + (4 Delta)/(
   9 (-1 + x)) - Delta/(18 (-4 + x)^2);
a1[x_] = 1/(-4 + x) + 1/(-1 + x) - 1/x + 1/(8 + x);

myeqn = (D[
     y[x], {x, 
      2}] + (g/x + d/(x - 1) + (a + b - g - d + 1)/(x - x0)) D[y[x], 
      x] + (a b x - q)/(x (x - 1) (x - x0)) y[x]);
subst = {x :> f[x], 
   Derivative[1][y][x] :> 1/f'[x] Derivative[1][y][x], 
   Derivative[2][y][x] :> -f''[x]/(f'[x])^3 Derivative[1][y][x] + 
     1/(f'[x])^2 Derivative[2][y][x]};
myeqn = Collect[(myeqn /. subst /. y[f[x]] :> y[x]), {y[x], y'[x], 
    y''[x]}, Simplify];
y[x_] = m[x] v[x];
FullSimplify[myeqn /. Derivative[2][v][x] :> -a1[x] v'[x] - a0[x] v[x]]


Out[13]= 0

The final conclusion of all this is that the Heun ODE can indeed be mapped onto the ODE in question(here I mean that the number of terms and their orders in each coefficient do match but the coefficients at those terms not necessarily match). As a matter of fact we even obtain a whole one parameter family of ODEs with known solutions. Unfortunately however the very particular ODE in question does not belong to this family.