Gauge transformation of differential equations I











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This is a follow-up question to Gauge transformation of differential equations. .
Let $y(x)$ be a solution to the following ODE:
begin{eqnarray}
y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
end{eqnarray}

Now define:
begin{equation}
g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
end{equation}

where
begin{equation}
r^{'}(x) + 1 - a_1(x) r(x)=0
end{equation}

Then:
begin{eqnarray}
&&g^{''}(x) + \
&&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
end{eqnarray}



In[7]:= 
Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
x0 =.;
r[x_] = Exp[Integrate[a1[x], x]] C[1] -
Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
Simplify[r'[x] + 1 - a1[x] r[x]]
g[x_] = (y[x] + r[x] y'[x])/(
r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
Collect[(g''[x] +
1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
3 Derivative[1][a0][x]^2)/a0[x]^2 + (
2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
x] :> -a1[x] y'[x] - a0[x] y[x],
Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
a0[x] y'[x]}, {y[x], y'[x]}, Simplify]

Out[9]= 0

Out[11]= 0


Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
begin{eqnarray}
a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
a_1(x)&=& frac{2}{x}\
Longrightarrow\
r(x)&=& frac{x^2}{x_0} +x
end{eqnarray}

then define:
begin{eqnarray}
{mathfrak P}_0&:=&x_0^2 x_2^2\
{mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
{mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
{mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
{mathfrak P}_4&=&-8
C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
{mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
{mathfrak P}_6&=&-8 C^2
end{eqnarray}

then we have:
begin{equation}
g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
end{equation}

Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
begin{eqnarray}
g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
end{eqnarray}



Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.










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    This is a follow-up question to Gauge transformation of differential equations. .
    Let $y(x)$ be a solution to the following ODE:
    begin{eqnarray}
    y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
    end{eqnarray}

    Now define:
    begin{equation}
    g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
    end{equation}

    where
    begin{equation}
    r^{'}(x) + 1 - a_1(x) r(x)=0
    end{equation}

    Then:
    begin{eqnarray}
    &&g^{''}(x) + \
    &&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
    a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
    end{eqnarray}



    In[7]:= 
    Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
    x0 =.;
    r[x_] = Exp[Integrate[a1[x], x]] C[1] -
    Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
    Simplify[r'[x] + 1 - a1[x] r[x]]
    g[x_] = (y[x] + r[x] y'[x])/(
    r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
    Collect[(g''[x] +
    1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
    3 Derivative[1][a0][x]^2)/a0[x]^2 + (
    2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
    2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
    x] :> -a1[x] y'[x] - a0[x] y[x],
    Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
    a0[x] y'[x]}, {y[x], y'[x]}, Simplify]

    Out[9]= 0

    Out[11]= 0


    Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
    begin{eqnarray}
    a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
    a_1(x)&=& frac{2}{x}\
    Longrightarrow\
    r(x)&=& frac{x^2}{x_0} +x
    end{eqnarray}

    then define:
    begin{eqnarray}
    {mathfrak P}_0&:=&x_0^2 x_2^2\
    {mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
    {mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
    left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
    {mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
    {mathfrak P}_4&=&-8
    C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
    {mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
    {mathfrak P}_6&=&-8 C^2
    end{eqnarray}

    then we have:
    begin{equation}
    g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
    end{equation}

    Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
    begin{eqnarray}
    g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
    end{eqnarray}



    Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.










    share|cite|improve this question
























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      up vote
      1
      down vote

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      This is a follow-up question to Gauge transformation of differential equations. .
      Let $y(x)$ be a solution to the following ODE:
      begin{eqnarray}
      y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
      end{eqnarray}

      Now define:
      begin{equation}
      g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
      end{equation}

      where
      begin{equation}
      r^{'}(x) + 1 - a_1(x) r(x)=0
      end{equation}

      Then:
      begin{eqnarray}
      &&g^{''}(x) + \
      &&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
      a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
      end{eqnarray}



      In[7]:= 
      Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
      x0 =.;
      r[x_] = Exp[Integrate[a1[x], x]] C[1] -
      Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
      Simplify[r'[x] + 1 - a1[x] r[x]]
      g[x_] = (y[x] + r[x] y'[x])/(
      r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
      Collect[(g''[x] +
      1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
      3 Derivative[1][a0][x]^2)/a0[x]^2 + (
      2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
      2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
      x] :> -a1[x] y'[x] - a0[x] y[x],
      Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
      a0[x] y'[x]}, {y[x], y'[x]}, Simplify]

      Out[9]= 0

      Out[11]= 0


      Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
      begin{eqnarray}
      a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
      a_1(x)&=& frac{2}{x}\
      Longrightarrow\
      r(x)&=& frac{x^2}{x_0} +x
      end{eqnarray}

      then define:
      begin{eqnarray}
      {mathfrak P}_0&:=&x_0^2 x_2^2\
      {mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
      {mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
      left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
      {mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
      {mathfrak P}_4&=&-8
      C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
      {mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
      {mathfrak P}_6&=&-8 C^2
      end{eqnarray}

      then we have:
      begin{equation}
      g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
      end{equation}

      Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
      begin{eqnarray}
      g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
      end{eqnarray}



      Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.










      share|cite|improve this question













      This is a follow-up question to Gauge transformation of differential equations. .
      Let $y(x)$ be a solution to the following ODE:
      begin{eqnarray}
      y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
      end{eqnarray}

      Now define:
      begin{equation}
      g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
      end{equation}

      where
      begin{equation}
      r^{'}(x) + 1 - a_1(x) r(x)=0
      end{equation}

      Then:
      begin{eqnarray}
      &&g^{''}(x) + \
      &&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
      a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
      end{eqnarray}



      In[7]:= 
      Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
      x0 =.;
      r[x_] = Exp[Integrate[a1[x], x]] C[1] -
      Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
      Simplify[r'[x] + 1 - a1[x] r[x]]
      g[x_] = (y[x] + r[x] y'[x])/(
      r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
      Collect[(g''[x] +
      1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
      3 Derivative[1][a0][x]^2)/a0[x]^2 + (
      2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
      2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
      x] :> -a1[x] y'[x] - a0[x] y[x],
      Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
      a0[x] y'[x]}, {y[x], y'[x]}, Simplify]

      Out[9]= 0

      Out[11]= 0


      Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
      begin{eqnarray}
      a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
      a_1(x)&=& frac{2}{x}\
      Longrightarrow\
      r(x)&=& frac{x^2}{x_0} +x
      end{eqnarray}

      then define:
      begin{eqnarray}
      {mathfrak P}_0&:=&x_0^2 x_2^2\
      {mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
      {mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
      left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
      {mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
      {mathfrak P}_4&=&-8
      C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
      {mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
      {mathfrak P}_6&=&-8 C^2
      end{eqnarray}

      then we have:
      begin{equation}
      g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
      end{equation}

      Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
      begin{eqnarray}
      g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
      end{eqnarray}



      Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.







      differential-equations special-functions






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      asked Nov 16 at 19:11









      Przemo

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          1 Answer
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          Here is another example which is a generalization of Example 1.3 in page 5 in https://arxiv.org/pdf/1606.01576.pdf .



          Let $a$,$b$,$c$,$a_1$,$a_2$,$a_3$,$b_2$,$b_4$ and $A$ be real parameters.
          Then let:
          begin{eqnarray}
          a_3&:=&-2 a A^2 b_2\
          b_4&:=&-A^2 b_2
          end{eqnarray}



          Now define:
          begin{eqnarray}
          p_0&:=&a_1 (a_1-2 b_2 (c-1))\
          p_1&:=&a_2 (2 a_1-2 b_2 c+b_2)\
          p_2&:=&a_2^2-2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))\
          p_3&:=&A^2 a_2 b_2 (-2 a+2 b-1)
          end{eqnarray}

          and
          begin{eqnarray}
          P_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
          P_1&:=&2 a_2 (c-2) (2 a_1-2 b_2 c+b_2)\
          P_2&:=&A^2 left(a_1^2 (-2 a-2 b+1)+2 a_1 b_2 (3 a+4 b c-7 b-3 c+6)-4 a b_2^2 (2 c-5)
          (b-c)right)+a_2^2 (2 c-5)\
          P_3&:=&2 A^2 a_2 (b_2 (5 a+4 b c-7 b-3 c+4)-2 a_1 (a+b-1))\
          P_4&:=&A^2 (2 a+2 b-3) left(2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))-a_2^2right)\
          P_5&:=&2 A^4 a_2
          b_2 (2 a-2 b+1) (a+b-2)
          end{eqnarray}

          and
          begin{eqnarray}
          Q_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
          Q_1&:=&a_2 (2 c-3) (3 a_1+b_2 (2-4 c))\
          Q_2&:=&A^2 left((2 a-1) a_1^2 (2 b-1)-2 a_1 b_2 (a (4 b (c-2)+4 c-3)-4 b c+7 b+3 c-6)-12 a b_2^2 (2
          c-3) (b-c)right)+4 a_2^2 (c-2)\
          Q_3&:=&A^2 a_2 (a_1 (a (8 b-6)-6 b+3)+2 b_2 (a (-4 b c+2 b-2 c+9)+2 (2 b-1) (2 c-3)))\
          Q_4&:=&-2 A^2 left((2 a-1) A^2 (2 b-3) b_2 (a_1 (a-b+1)+2 a b_2 (b-c))+2
          a_2^2 (a (-b)+a+b-1)right)\
          Q_5&:=&2 (1-a) A^4 a_2 (2 b-3) b_2 (2 a-2 b+1)
          end{eqnarray}

          and
          begin{equation}
          y(x):=F_{2,1}left[a,b,c,A^2 x^2right]
          end{equation}



          Then the ODE:
          begin{eqnarray}
          g^{''}(x) - frac{sumlimits_{j=0}^5 P_j x^j}{x(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g^{'}(x) + frac{sumlimits_{j=0}^5 Q_j x^j}{x^2(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g(x)=0
          end{eqnarray}

          is solved by
          begin{eqnarray}
          g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
          end{eqnarray}



          In[14]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
          =.;
          p0 =.; p1 =.; p2 =.; p3 =.;
          P0 =.; P1 =.; P2 =.; P3 =.; P4 =.; P5 =.;
          Q0 =.; Q1 =.; Q2 =.; Q3 =.; Q4 =.; Q5 =.; Clear[y];
          {a3, b4} = {-2 a A^2 b2, -A^2 b2};
          {p0, p1, p2, p3} = {a1 (a1 - 2 b2 (-1 + c)), a2 (2 a1 + b2 - 2 b2 c),
          a2^2 - 2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c)),
          A^2 a2 (-1 - 2 a + 2 b) b2};
          {P0, P1, P2, P3, P4, P5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
          2 a2 (-2 + c) (2 a1 + b2 - 2 b2 c),
          a2^2 (-5 + 2 c) +
          A^2 (a1^2 (1 - 2 a - 2 b) - 4 a b2^2 (b - c) (-5 + 2 c) +
          2 a1 b2 (6 + 3 a - 7 b - 3 c + 4 b c)),
          2 A^2 a2 (-2 a1 (-1 + a + b) + b2 (4 + 5 a - 7 b - 3 c + 4 b c)),
          A^2 (-3 + 2 a + 2 b) (-a2^2 +
          2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c))),
          2 A^4 a2 (1 + 2 a - 2 b) (-2 + a + b) b2};
          {Q0, Q1, Q2, Q3, Q4, Q5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
          a2 (3 a1 + b2 (2 - 4 c)) (-3 + 2 c),
          4 a2^2 (-2 + c) +
          A^2 ((-1 + 2 a) a1^2 (-1 + 2 b) - 12 a b2^2 (b - c) (-3 + 2 c) -
          2 a1 b2 (-6 + 7 b + 3 c - 4 b c +
          a (-3 + 4 b (-2 + c) + 4 c))),
          A^2 a2 (a1 (3 - 6 b + a (-6 + 8 b)) +
          2 b2 (2 (-1 + 2 b) (-3 + 2 c) +
          a (9 + 2 b - 2 c - 4 b c))), -2 A^2 (2 a2^2 (-1 + a + b -
          a b) + (-1 + 2 a) A^2 (-3 + 2 b) b2 (a1 (1 + a - b) +
          2 a b2 (b - c))),
          2 A^4 a2 (1 + 2 a - 2 b) (1 - a) (-3 + 2 b) b2};
          y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
          eX = (D[#, {x, 2}] - (
          P5 x^5 + P4 x^4 + P3 x^3 + P2 x^2 + P1 x^1 + P0)/(
          x (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 + p0))
          D[#, x] + (Q5 x^5 + Q4 x^4 + Q3 x^3 + Q2 x^2 + Q1 x^1 + Q0)/(
          x ^2 (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 +
          p0)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
          x] + (b4 x^4 + b2 x^2) y'[x]};

          {a, b, c, a1, a2, b2, A, x} =
          RandomReal[{0, 1}, 8, WorkingPrecision -> 50];
          Simplify[eX]

          Out[25]= {0.*10^-48}


          Update: The ODE above is a seven parameter family.
          Now, note that if in the example above we add three additional constraints and as such reduce the number of adjustable parameters to four we get another neat example:



          Firstly define:
          begin{eqnarray}
          a_1&:=& c-frac{1}{2}\
          a_2&:=& A frac{1}{sqrt{2}} sqrt{-1+4 a+8 a^2+2 c-8 a c}\
          a_3&:=&-2 a A^2\
          hline \
          b_2&:=& 1\
          b_4&:=&-A^2 \
          hline \
          b&:=&a+frac{1}{2}
          end{eqnarray}

          Then the ODE below:
          begin{eqnarray}
          &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) + frac{3-2 c+4 a A^2 x^2}{x(A x-1)(A x+1)} g^{'}(x) + frac{(-3+2 c) + sqrt{2} A sqrt{-1+4 a+8 a^2+2 c-8 a c} x+2(-1-a+2 a^2) x^2}{x^2(A x-1)(A x+1)} g(x)=0
          end{eqnarray}

          is solved by
          begin{eqnarray}
          g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
          end{eqnarray}



          In[18]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
          =.;
          {a1, a2, a3} = {(-(1/2) + c),
          A Sqrt[1/2 (-1 + 4 a + 8 a^2 + 2 c - 8 a c)], -2 a A^2};
          {b2, b4} = {1, -A^2};
          b = a + 1/2;
          y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
          eX = (D[#, {x, 2}] + (3 - 2 c + 4 a A^2 x^2)/(x (-1 + A x) (1 + A x))
          D[#, x] + ( (-3 + 2 c) +
          Sqrt[2] A Sqrt[(-1 + 4 a + 8 a^2 + 2 c - 8 a c)] x +
          2 (-1 - a + 2 a^2) A^2 x^2)/(
          x ^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
          x] + (b4 x^4 + b2 x^2) y'[x]};
          {b2, a, c, A, x} = RandomReal[{0, 1}, 5, WorkingPrecision -> 50];
          Simplify[eX]

          Out[25]= {0.*10^-49}


          Secondly define:
          begin{eqnarray}
          a_1&:=& 2c-1\
          a_2&:=& A sqrt{2} sqrt{(-1+2 a)(-1+b)}\
          a_3&:=&-2 a A^2\
          hline \
          b_2&:=& 1\
          b_4&:=&-A^2 \
          hline \
          c&:=&frac{3}{2}
          end{eqnarray}

          Then the ODE below:
          begin{eqnarray}
          &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) +
          frac{3+2 A^2(-2+a+b)x^2}{x(A x-1)(A x+1)} g^{'}(x) +
          frac{-3-sqrt{2} A sqrt{(-1+2 a)(-1+b)} x+2(-1+a)(-3+2 b) A^2 x^2}{x^2(A x-1)(A x+1)} g(x)=0
          end{eqnarray}

          is solved by
          begin{eqnarray}
          g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
          end{eqnarray}



          In[567]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; 
          x =.;
          {b2, b4} = {1, -A^2};
          {a1, a2, a3} = {2 (c - 1),
          Sqrt[2] Sqrt[-1 + 2 a] A Sqrt[-1 + b], -2 a A^2};
          c = 3/2;
          y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
          eX = (D[#, {x, 2}] + (3 + 2 A^2 (-2 + a + b) x^2)/(
          x (-1 + A x) (1 + A x))
          D[#, x] + ( -3 - Sqrt[2] A (Sqrt[-1 + 2 a] Sqrt[-1 + b]) x +
          2 (-1 + a) (-3 + 2 b) A^2 x^2)/(
          x^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
          x] + (b4 x^4 + b2 x^2) y'[x]};
          {a, b, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
          Simplify[eX]


          Out[574]= {0.*10^-47 + 0.*10^-49 I}





          share|cite|improve this answer























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            Here is another example which is a generalization of Example 1.3 in page 5 in https://arxiv.org/pdf/1606.01576.pdf .



            Let $a$,$b$,$c$,$a_1$,$a_2$,$a_3$,$b_2$,$b_4$ and $A$ be real parameters.
            Then let:
            begin{eqnarray}
            a_3&:=&-2 a A^2 b_2\
            b_4&:=&-A^2 b_2
            end{eqnarray}



            Now define:
            begin{eqnarray}
            p_0&:=&a_1 (a_1-2 b_2 (c-1))\
            p_1&:=&a_2 (2 a_1-2 b_2 c+b_2)\
            p_2&:=&a_2^2-2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))\
            p_3&:=&A^2 a_2 b_2 (-2 a+2 b-1)
            end{eqnarray}

            and
            begin{eqnarray}
            P_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
            P_1&:=&2 a_2 (c-2) (2 a_1-2 b_2 c+b_2)\
            P_2&:=&A^2 left(a_1^2 (-2 a-2 b+1)+2 a_1 b_2 (3 a+4 b c-7 b-3 c+6)-4 a b_2^2 (2 c-5)
            (b-c)right)+a_2^2 (2 c-5)\
            P_3&:=&2 A^2 a_2 (b_2 (5 a+4 b c-7 b-3 c+4)-2 a_1 (a+b-1))\
            P_4&:=&A^2 (2 a+2 b-3) left(2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))-a_2^2right)\
            P_5&:=&2 A^4 a_2
            b_2 (2 a-2 b+1) (a+b-2)
            end{eqnarray}

            and
            begin{eqnarray}
            Q_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
            Q_1&:=&a_2 (2 c-3) (3 a_1+b_2 (2-4 c))\
            Q_2&:=&A^2 left((2 a-1) a_1^2 (2 b-1)-2 a_1 b_2 (a (4 b (c-2)+4 c-3)-4 b c+7 b+3 c-6)-12 a b_2^2 (2
            c-3) (b-c)right)+4 a_2^2 (c-2)\
            Q_3&:=&A^2 a_2 (a_1 (a (8 b-6)-6 b+3)+2 b_2 (a (-4 b c+2 b-2 c+9)+2 (2 b-1) (2 c-3)))\
            Q_4&:=&-2 A^2 left((2 a-1) A^2 (2 b-3) b_2 (a_1 (a-b+1)+2 a b_2 (b-c))+2
            a_2^2 (a (-b)+a+b-1)right)\
            Q_5&:=&2 (1-a) A^4 a_2 (2 b-3) b_2 (2 a-2 b+1)
            end{eqnarray}

            and
            begin{equation}
            y(x):=F_{2,1}left[a,b,c,A^2 x^2right]
            end{equation}



            Then the ODE:
            begin{eqnarray}
            g^{''}(x) - frac{sumlimits_{j=0}^5 P_j x^j}{x(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g^{'}(x) + frac{sumlimits_{j=0}^5 Q_j x^j}{x^2(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g(x)=0
            end{eqnarray}

            is solved by
            begin{eqnarray}
            g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
            end{eqnarray}



            In[14]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
            =.;
            p0 =.; p1 =.; p2 =.; p3 =.;
            P0 =.; P1 =.; P2 =.; P3 =.; P4 =.; P5 =.;
            Q0 =.; Q1 =.; Q2 =.; Q3 =.; Q4 =.; Q5 =.; Clear[y];
            {a3, b4} = {-2 a A^2 b2, -A^2 b2};
            {p0, p1, p2, p3} = {a1 (a1 - 2 b2 (-1 + c)), a2 (2 a1 + b2 - 2 b2 c),
            a2^2 - 2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c)),
            A^2 a2 (-1 - 2 a + 2 b) b2};
            {P0, P1, P2, P3, P4, P5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
            2 a2 (-2 + c) (2 a1 + b2 - 2 b2 c),
            a2^2 (-5 + 2 c) +
            A^2 (a1^2 (1 - 2 a - 2 b) - 4 a b2^2 (b - c) (-5 + 2 c) +
            2 a1 b2 (6 + 3 a - 7 b - 3 c + 4 b c)),
            2 A^2 a2 (-2 a1 (-1 + a + b) + b2 (4 + 5 a - 7 b - 3 c + 4 b c)),
            A^2 (-3 + 2 a + 2 b) (-a2^2 +
            2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c))),
            2 A^4 a2 (1 + 2 a - 2 b) (-2 + a + b) b2};
            {Q0, Q1, Q2, Q3, Q4, Q5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
            a2 (3 a1 + b2 (2 - 4 c)) (-3 + 2 c),
            4 a2^2 (-2 + c) +
            A^2 ((-1 + 2 a) a1^2 (-1 + 2 b) - 12 a b2^2 (b - c) (-3 + 2 c) -
            2 a1 b2 (-6 + 7 b + 3 c - 4 b c +
            a (-3 + 4 b (-2 + c) + 4 c))),
            A^2 a2 (a1 (3 - 6 b + a (-6 + 8 b)) +
            2 b2 (2 (-1 + 2 b) (-3 + 2 c) +
            a (9 + 2 b - 2 c - 4 b c))), -2 A^2 (2 a2^2 (-1 + a + b -
            a b) + (-1 + 2 a) A^2 (-3 + 2 b) b2 (a1 (1 + a - b) +
            2 a b2 (b - c))),
            2 A^4 a2 (1 + 2 a - 2 b) (1 - a) (-3 + 2 b) b2};
            y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
            eX = (D[#, {x, 2}] - (
            P5 x^5 + P4 x^4 + P3 x^3 + P2 x^2 + P1 x^1 + P0)/(
            x (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 + p0))
            D[#, x] + (Q5 x^5 + Q4 x^4 + Q3 x^3 + Q2 x^2 + Q1 x^1 + Q0)/(
            x ^2 (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 +
            p0)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
            x] + (b4 x^4 + b2 x^2) y'[x]};

            {a, b, c, a1, a2, b2, A, x} =
            RandomReal[{0, 1}, 8, WorkingPrecision -> 50];
            Simplify[eX]

            Out[25]= {0.*10^-48}


            Update: The ODE above is a seven parameter family.
            Now, note that if in the example above we add three additional constraints and as such reduce the number of adjustable parameters to four we get another neat example:



            Firstly define:
            begin{eqnarray}
            a_1&:=& c-frac{1}{2}\
            a_2&:=& A frac{1}{sqrt{2}} sqrt{-1+4 a+8 a^2+2 c-8 a c}\
            a_3&:=&-2 a A^2\
            hline \
            b_2&:=& 1\
            b_4&:=&-A^2 \
            hline \
            b&:=&a+frac{1}{2}
            end{eqnarray}

            Then the ODE below:
            begin{eqnarray}
            &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) + frac{3-2 c+4 a A^2 x^2}{x(A x-1)(A x+1)} g^{'}(x) + frac{(-3+2 c) + sqrt{2} A sqrt{-1+4 a+8 a^2+2 c-8 a c} x+2(-1-a+2 a^2) x^2}{x^2(A x-1)(A x+1)} g(x)=0
            end{eqnarray}

            is solved by
            begin{eqnarray}
            g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
            end{eqnarray}



            In[18]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
            =.;
            {a1, a2, a3} = {(-(1/2) + c),
            A Sqrt[1/2 (-1 + 4 a + 8 a^2 + 2 c - 8 a c)], -2 a A^2};
            {b2, b4} = {1, -A^2};
            b = a + 1/2;
            y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
            eX = (D[#, {x, 2}] + (3 - 2 c + 4 a A^2 x^2)/(x (-1 + A x) (1 + A x))
            D[#, x] + ( (-3 + 2 c) +
            Sqrt[2] A Sqrt[(-1 + 4 a + 8 a^2 + 2 c - 8 a c)] x +
            2 (-1 - a + 2 a^2) A^2 x^2)/(
            x ^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
            x] + (b4 x^4 + b2 x^2) y'[x]};
            {b2, a, c, A, x} = RandomReal[{0, 1}, 5, WorkingPrecision -> 50];
            Simplify[eX]

            Out[25]= {0.*10^-49}


            Secondly define:
            begin{eqnarray}
            a_1&:=& 2c-1\
            a_2&:=& A sqrt{2} sqrt{(-1+2 a)(-1+b)}\
            a_3&:=&-2 a A^2\
            hline \
            b_2&:=& 1\
            b_4&:=&-A^2 \
            hline \
            c&:=&frac{3}{2}
            end{eqnarray}

            Then the ODE below:
            begin{eqnarray}
            &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) +
            frac{3+2 A^2(-2+a+b)x^2}{x(A x-1)(A x+1)} g^{'}(x) +
            frac{-3-sqrt{2} A sqrt{(-1+2 a)(-1+b)} x+2(-1+a)(-3+2 b) A^2 x^2}{x^2(A x-1)(A x+1)} g(x)=0
            end{eqnarray}

            is solved by
            begin{eqnarray}
            g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
            end{eqnarray}



            In[567]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; 
            x =.;
            {b2, b4} = {1, -A^2};
            {a1, a2, a3} = {2 (c - 1),
            Sqrt[2] Sqrt[-1 + 2 a] A Sqrt[-1 + b], -2 a A^2};
            c = 3/2;
            y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
            eX = (D[#, {x, 2}] + (3 + 2 A^2 (-2 + a + b) x^2)/(
            x (-1 + A x) (1 + A x))
            D[#, x] + ( -3 - Sqrt[2] A (Sqrt[-1 + 2 a] Sqrt[-1 + b]) x +
            2 (-1 + a) (-3 + 2 b) A^2 x^2)/(
            x^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
            x] + (b4 x^4 + b2 x^2) y'[x]};
            {a, b, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
            Simplify[eX]


            Out[574]= {0.*10^-47 + 0.*10^-49 I}





            share|cite|improve this answer



























              up vote
              0
              down vote













              Here is another example which is a generalization of Example 1.3 in page 5 in https://arxiv.org/pdf/1606.01576.pdf .



              Let $a$,$b$,$c$,$a_1$,$a_2$,$a_3$,$b_2$,$b_4$ and $A$ be real parameters.
              Then let:
              begin{eqnarray}
              a_3&:=&-2 a A^2 b_2\
              b_4&:=&-A^2 b_2
              end{eqnarray}



              Now define:
              begin{eqnarray}
              p_0&:=&a_1 (a_1-2 b_2 (c-1))\
              p_1&:=&a_2 (2 a_1-2 b_2 c+b_2)\
              p_2&:=&a_2^2-2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))\
              p_3&:=&A^2 a_2 b_2 (-2 a+2 b-1)
              end{eqnarray}

              and
              begin{eqnarray}
              P_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
              P_1&:=&2 a_2 (c-2) (2 a_1-2 b_2 c+b_2)\
              P_2&:=&A^2 left(a_1^2 (-2 a-2 b+1)+2 a_1 b_2 (3 a+4 b c-7 b-3 c+6)-4 a b_2^2 (2 c-5)
              (b-c)right)+a_2^2 (2 c-5)\
              P_3&:=&2 A^2 a_2 (b_2 (5 a+4 b c-7 b-3 c+4)-2 a_1 (a+b-1))\
              P_4&:=&A^2 (2 a+2 b-3) left(2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))-a_2^2right)\
              P_5&:=&2 A^4 a_2
              b_2 (2 a-2 b+1) (a+b-2)
              end{eqnarray}

              and
              begin{eqnarray}
              Q_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
              Q_1&:=&a_2 (2 c-3) (3 a_1+b_2 (2-4 c))\
              Q_2&:=&A^2 left((2 a-1) a_1^2 (2 b-1)-2 a_1 b_2 (a (4 b (c-2)+4 c-3)-4 b c+7 b+3 c-6)-12 a b_2^2 (2
              c-3) (b-c)right)+4 a_2^2 (c-2)\
              Q_3&:=&A^2 a_2 (a_1 (a (8 b-6)-6 b+3)+2 b_2 (a (-4 b c+2 b-2 c+9)+2 (2 b-1) (2 c-3)))\
              Q_4&:=&-2 A^2 left((2 a-1) A^2 (2 b-3) b_2 (a_1 (a-b+1)+2 a b_2 (b-c))+2
              a_2^2 (a (-b)+a+b-1)right)\
              Q_5&:=&2 (1-a) A^4 a_2 (2 b-3) b_2 (2 a-2 b+1)
              end{eqnarray}

              and
              begin{equation}
              y(x):=F_{2,1}left[a,b,c,A^2 x^2right]
              end{equation}



              Then the ODE:
              begin{eqnarray}
              g^{''}(x) - frac{sumlimits_{j=0}^5 P_j x^j}{x(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g^{'}(x) + frac{sumlimits_{j=0}^5 Q_j x^j}{x^2(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g(x)=0
              end{eqnarray}

              is solved by
              begin{eqnarray}
              g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
              end{eqnarray}



              In[14]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
              =.;
              p0 =.; p1 =.; p2 =.; p3 =.;
              P0 =.; P1 =.; P2 =.; P3 =.; P4 =.; P5 =.;
              Q0 =.; Q1 =.; Q2 =.; Q3 =.; Q4 =.; Q5 =.; Clear[y];
              {a3, b4} = {-2 a A^2 b2, -A^2 b2};
              {p0, p1, p2, p3} = {a1 (a1 - 2 b2 (-1 + c)), a2 (2 a1 + b2 - 2 b2 c),
              a2^2 - 2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c)),
              A^2 a2 (-1 - 2 a + 2 b) b2};
              {P0, P1, P2, P3, P4, P5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
              2 a2 (-2 + c) (2 a1 + b2 - 2 b2 c),
              a2^2 (-5 + 2 c) +
              A^2 (a1^2 (1 - 2 a - 2 b) - 4 a b2^2 (b - c) (-5 + 2 c) +
              2 a1 b2 (6 + 3 a - 7 b - 3 c + 4 b c)),
              2 A^2 a2 (-2 a1 (-1 + a + b) + b2 (4 + 5 a - 7 b - 3 c + 4 b c)),
              A^2 (-3 + 2 a + 2 b) (-a2^2 +
              2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c))),
              2 A^4 a2 (1 + 2 a - 2 b) (-2 + a + b) b2};
              {Q0, Q1, Q2, Q3, Q4, Q5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
              a2 (3 a1 + b2 (2 - 4 c)) (-3 + 2 c),
              4 a2^2 (-2 + c) +
              A^2 ((-1 + 2 a) a1^2 (-1 + 2 b) - 12 a b2^2 (b - c) (-3 + 2 c) -
              2 a1 b2 (-6 + 7 b + 3 c - 4 b c +
              a (-3 + 4 b (-2 + c) + 4 c))),
              A^2 a2 (a1 (3 - 6 b + a (-6 + 8 b)) +
              2 b2 (2 (-1 + 2 b) (-3 + 2 c) +
              a (9 + 2 b - 2 c - 4 b c))), -2 A^2 (2 a2^2 (-1 + a + b -
              a b) + (-1 + 2 a) A^2 (-3 + 2 b) b2 (a1 (1 + a - b) +
              2 a b2 (b - c))),
              2 A^4 a2 (1 + 2 a - 2 b) (1 - a) (-3 + 2 b) b2};
              y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
              eX = (D[#, {x, 2}] - (
              P5 x^5 + P4 x^4 + P3 x^3 + P2 x^2 + P1 x^1 + P0)/(
              x (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 + p0))
              D[#, x] + (Q5 x^5 + Q4 x^4 + Q3 x^3 + Q2 x^2 + Q1 x^1 + Q0)/(
              x ^2 (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 +
              p0)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
              x] + (b4 x^4 + b2 x^2) y'[x]};

              {a, b, c, a1, a2, b2, A, x} =
              RandomReal[{0, 1}, 8, WorkingPrecision -> 50];
              Simplify[eX]

              Out[25]= {0.*10^-48}


              Update: The ODE above is a seven parameter family.
              Now, note that if in the example above we add three additional constraints and as such reduce the number of adjustable parameters to four we get another neat example:



              Firstly define:
              begin{eqnarray}
              a_1&:=& c-frac{1}{2}\
              a_2&:=& A frac{1}{sqrt{2}} sqrt{-1+4 a+8 a^2+2 c-8 a c}\
              a_3&:=&-2 a A^2\
              hline \
              b_2&:=& 1\
              b_4&:=&-A^2 \
              hline \
              b&:=&a+frac{1}{2}
              end{eqnarray}

              Then the ODE below:
              begin{eqnarray}
              &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) + frac{3-2 c+4 a A^2 x^2}{x(A x-1)(A x+1)} g^{'}(x) + frac{(-3+2 c) + sqrt{2} A sqrt{-1+4 a+8 a^2+2 c-8 a c} x+2(-1-a+2 a^2) x^2}{x^2(A x-1)(A x+1)} g(x)=0
              end{eqnarray}

              is solved by
              begin{eqnarray}
              g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
              end{eqnarray}



              In[18]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
              =.;
              {a1, a2, a3} = {(-(1/2) + c),
              A Sqrt[1/2 (-1 + 4 a + 8 a^2 + 2 c - 8 a c)], -2 a A^2};
              {b2, b4} = {1, -A^2};
              b = a + 1/2;
              y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
              eX = (D[#, {x, 2}] + (3 - 2 c + 4 a A^2 x^2)/(x (-1 + A x) (1 + A x))
              D[#, x] + ( (-3 + 2 c) +
              Sqrt[2] A Sqrt[(-1 + 4 a + 8 a^2 + 2 c - 8 a c)] x +
              2 (-1 - a + 2 a^2) A^2 x^2)/(
              x ^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
              x] + (b4 x^4 + b2 x^2) y'[x]};
              {b2, a, c, A, x} = RandomReal[{0, 1}, 5, WorkingPrecision -> 50];
              Simplify[eX]

              Out[25]= {0.*10^-49}


              Secondly define:
              begin{eqnarray}
              a_1&:=& 2c-1\
              a_2&:=& A sqrt{2} sqrt{(-1+2 a)(-1+b)}\
              a_3&:=&-2 a A^2\
              hline \
              b_2&:=& 1\
              b_4&:=&-A^2 \
              hline \
              c&:=&frac{3}{2}
              end{eqnarray}

              Then the ODE below:
              begin{eqnarray}
              &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) +
              frac{3+2 A^2(-2+a+b)x^2}{x(A x-1)(A x+1)} g^{'}(x) +
              frac{-3-sqrt{2} A sqrt{(-1+2 a)(-1+b)} x+2(-1+a)(-3+2 b) A^2 x^2}{x^2(A x-1)(A x+1)} g(x)=0
              end{eqnarray}

              is solved by
              begin{eqnarray}
              g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
              end{eqnarray}



              In[567]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; 
              x =.;
              {b2, b4} = {1, -A^2};
              {a1, a2, a3} = {2 (c - 1),
              Sqrt[2] Sqrt[-1 + 2 a] A Sqrt[-1 + b], -2 a A^2};
              c = 3/2;
              y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
              eX = (D[#, {x, 2}] + (3 + 2 A^2 (-2 + a + b) x^2)/(
              x (-1 + A x) (1 + A x))
              D[#, x] + ( -3 - Sqrt[2] A (Sqrt[-1 + 2 a] Sqrt[-1 + b]) x +
              2 (-1 + a) (-3 + 2 b) A^2 x^2)/(
              x^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
              x] + (b4 x^4 + b2 x^2) y'[x]};
              {a, b, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
              Simplify[eX]


              Out[574]= {0.*10^-47 + 0.*10^-49 I}





              share|cite|improve this answer

























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                Here is another example which is a generalization of Example 1.3 in page 5 in https://arxiv.org/pdf/1606.01576.pdf .



                Let $a$,$b$,$c$,$a_1$,$a_2$,$a_3$,$b_2$,$b_4$ and $A$ be real parameters.
                Then let:
                begin{eqnarray}
                a_3&:=&-2 a A^2 b_2\
                b_4&:=&-A^2 b_2
                end{eqnarray}



                Now define:
                begin{eqnarray}
                p_0&:=&a_1 (a_1-2 b_2 (c-1))\
                p_1&:=&a_2 (2 a_1-2 b_2 c+b_2)\
                p_2&:=&a_2^2-2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))\
                p_3&:=&A^2 a_2 b_2 (-2 a+2 b-1)
                end{eqnarray}

                and
                begin{eqnarray}
                P_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
                P_1&:=&2 a_2 (c-2) (2 a_1-2 b_2 c+b_2)\
                P_2&:=&A^2 left(a_1^2 (-2 a-2 b+1)+2 a_1 b_2 (3 a+4 b c-7 b-3 c+6)-4 a b_2^2 (2 c-5)
                (b-c)right)+a_2^2 (2 c-5)\
                P_3&:=&2 A^2 a_2 (b_2 (5 a+4 b c-7 b-3 c+4)-2 a_1 (a+b-1))\
                P_4&:=&A^2 (2 a+2 b-3) left(2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))-a_2^2right)\
                P_5&:=&2 A^4 a_2
                b_2 (2 a-2 b+1) (a+b-2)
                end{eqnarray}

                and
                begin{eqnarray}
                Q_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
                Q_1&:=&a_2 (2 c-3) (3 a_1+b_2 (2-4 c))\
                Q_2&:=&A^2 left((2 a-1) a_1^2 (2 b-1)-2 a_1 b_2 (a (4 b (c-2)+4 c-3)-4 b c+7 b+3 c-6)-12 a b_2^2 (2
                c-3) (b-c)right)+4 a_2^2 (c-2)\
                Q_3&:=&A^2 a_2 (a_1 (a (8 b-6)-6 b+3)+2 b_2 (a (-4 b c+2 b-2 c+9)+2 (2 b-1) (2 c-3)))\
                Q_4&:=&-2 A^2 left((2 a-1) A^2 (2 b-3) b_2 (a_1 (a-b+1)+2 a b_2 (b-c))+2
                a_2^2 (a (-b)+a+b-1)right)\
                Q_5&:=&2 (1-a) A^4 a_2 (2 b-3) b_2 (2 a-2 b+1)
                end{eqnarray}

                and
                begin{equation}
                y(x):=F_{2,1}left[a,b,c,A^2 x^2right]
                end{equation}



                Then the ODE:
                begin{eqnarray}
                g^{''}(x) - frac{sumlimits_{j=0}^5 P_j x^j}{x(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g^{'}(x) + frac{sumlimits_{j=0}^5 Q_j x^j}{x^2(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[14]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
                =.;
                p0 =.; p1 =.; p2 =.; p3 =.;
                P0 =.; P1 =.; P2 =.; P3 =.; P4 =.; P5 =.;
                Q0 =.; Q1 =.; Q2 =.; Q3 =.; Q4 =.; Q5 =.; Clear[y];
                {a3, b4} = {-2 a A^2 b2, -A^2 b2};
                {p0, p1, p2, p3} = {a1 (a1 - 2 b2 (-1 + c)), a2 (2 a1 + b2 - 2 b2 c),
                a2^2 - 2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c)),
                A^2 a2 (-1 - 2 a + 2 b) b2};
                {P0, P1, P2, P3, P4, P5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
                2 a2 (-2 + c) (2 a1 + b2 - 2 b2 c),
                a2^2 (-5 + 2 c) +
                A^2 (a1^2 (1 - 2 a - 2 b) - 4 a b2^2 (b - c) (-5 + 2 c) +
                2 a1 b2 (6 + 3 a - 7 b - 3 c + 4 b c)),
                2 A^2 a2 (-2 a1 (-1 + a + b) + b2 (4 + 5 a - 7 b - 3 c + 4 b c)),
                A^2 (-3 + 2 a + 2 b) (-a2^2 +
                2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c))),
                2 A^4 a2 (1 + 2 a - 2 b) (-2 + a + b) b2};
                {Q0, Q1, Q2, Q3, Q4, Q5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
                a2 (3 a1 + b2 (2 - 4 c)) (-3 + 2 c),
                4 a2^2 (-2 + c) +
                A^2 ((-1 + 2 a) a1^2 (-1 + 2 b) - 12 a b2^2 (b - c) (-3 + 2 c) -
                2 a1 b2 (-6 + 7 b + 3 c - 4 b c +
                a (-3 + 4 b (-2 + c) + 4 c))),
                A^2 a2 (a1 (3 - 6 b + a (-6 + 8 b)) +
                2 b2 (2 (-1 + 2 b) (-3 + 2 c) +
                a (9 + 2 b - 2 c - 4 b c))), -2 A^2 (2 a2^2 (-1 + a + b -
                a b) + (-1 + 2 a) A^2 (-3 + 2 b) b2 (a1 (1 + a - b) +
                2 a b2 (b - c))),
                2 A^4 a2 (1 + 2 a - 2 b) (1 - a) (-3 + 2 b) b2};
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] - (
                P5 x^5 + P4 x^4 + P3 x^3 + P2 x^2 + P1 x^1 + P0)/(
                x (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 + p0))
                D[#, x] + (Q5 x^5 + Q4 x^4 + Q3 x^3 + Q2 x^2 + Q1 x^1 + Q0)/(
                x ^2 (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 +
                p0)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};

                {a, b, c, a1, a2, b2, A, x} =
                RandomReal[{0, 1}, 8, WorkingPrecision -> 50];
                Simplify[eX]

                Out[25]= {0.*10^-48}


                Update: The ODE above is a seven parameter family.
                Now, note that if in the example above we add three additional constraints and as such reduce the number of adjustable parameters to four we get another neat example:



                Firstly define:
                begin{eqnarray}
                a_1&:=& c-frac{1}{2}\
                a_2&:=& A frac{1}{sqrt{2}} sqrt{-1+4 a+8 a^2+2 c-8 a c}\
                a_3&:=&-2 a A^2\
                hline \
                b_2&:=& 1\
                b_4&:=&-A^2 \
                hline \
                b&:=&a+frac{1}{2}
                end{eqnarray}

                Then the ODE below:
                begin{eqnarray}
                &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) + frac{3-2 c+4 a A^2 x^2}{x(A x-1)(A x+1)} g^{'}(x) + frac{(-3+2 c) + sqrt{2} A sqrt{-1+4 a+8 a^2+2 c-8 a c} x+2(-1-a+2 a^2) x^2}{x^2(A x-1)(A x+1)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[18]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
                =.;
                {a1, a2, a3} = {(-(1/2) + c),
                A Sqrt[1/2 (-1 + 4 a + 8 a^2 + 2 c - 8 a c)], -2 a A^2};
                {b2, b4} = {1, -A^2};
                b = a + 1/2;
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] + (3 - 2 c + 4 a A^2 x^2)/(x (-1 + A x) (1 + A x))
                D[#, x] + ( (-3 + 2 c) +
                Sqrt[2] A Sqrt[(-1 + 4 a + 8 a^2 + 2 c - 8 a c)] x +
                2 (-1 - a + 2 a^2) A^2 x^2)/(
                x ^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};
                {b2, a, c, A, x} = RandomReal[{0, 1}, 5, WorkingPrecision -> 50];
                Simplify[eX]

                Out[25]= {0.*10^-49}


                Secondly define:
                begin{eqnarray}
                a_1&:=& 2c-1\
                a_2&:=& A sqrt{2} sqrt{(-1+2 a)(-1+b)}\
                a_3&:=&-2 a A^2\
                hline \
                b_2&:=& 1\
                b_4&:=&-A^2 \
                hline \
                c&:=&frac{3}{2}
                end{eqnarray}

                Then the ODE below:
                begin{eqnarray}
                &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) +
                frac{3+2 A^2(-2+a+b)x^2}{x(A x-1)(A x+1)} g^{'}(x) +
                frac{-3-sqrt{2} A sqrt{(-1+2 a)(-1+b)} x+2(-1+a)(-3+2 b) A^2 x^2}{x^2(A x-1)(A x+1)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[567]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; 
                x =.;
                {b2, b4} = {1, -A^2};
                {a1, a2, a3} = {2 (c - 1),
                Sqrt[2] Sqrt[-1 + 2 a] A Sqrt[-1 + b], -2 a A^2};
                c = 3/2;
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] + (3 + 2 A^2 (-2 + a + b) x^2)/(
                x (-1 + A x) (1 + A x))
                D[#, x] + ( -3 - Sqrt[2] A (Sqrt[-1 + 2 a] Sqrt[-1 + b]) x +
                2 (-1 + a) (-3 + 2 b) A^2 x^2)/(
                x^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};
                {a, b, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
                Simplify[eX]


                Out[574]= {0.*10^-47 + 0.*10^-49 I}





                share|cite|improve this answer














                Here is another example which is a generalization of Example 1.3 in page 5 in https://arxiv.org/pdf/1606.01576.pdf .



                Let $a$,$b$,$c$,$a_1$,$a_2$,$a_3$,$b_2$,$b_4$ and $A$ be real parameters.
                Then let:
                begin{eqnarray}
                a_3&:=&-2 a A^2 b_2\
                b_4&:=&-A^2 b_2
                end{eqnarray}



                Now define:
                begin{eqnarray}
                p_0&:=&a_1 (a_1-2 b_2 (c-1))\
                p_1&:=&a_2 (2 a_1-2 b_2 c+b_2)\
                p_2&:=&a_2^2-2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))\
                p_3&:=&A^2 a_2 b_2 (-2 a+2 b-1)
                end{eqnarray}

                and
                begin{eqnarray}
                P_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
                P_1&:=&2 a_2 (c-2) (2 a_1-2 b_2 c+b_2)\
                P_2&:=&A^2 left(a_1^2 (-2 a-2 b+1)+2 a_1 b_2 (3 a+4 b c-7 b-3 c+6)-4 a b_2^2 (2 c-5)
                (b-c)right)+a_2^2 (2 c-5)\
                P_3&:=&2 A^2 a_2 (b_2 (5 a+4 b c-7 b-3 c+4)-2 a_1 (a+b-1))\
                P_4&:=&A^2 (2 a+2 b-3) left(2 A^2 b_2 (a_1 (a-b+1)+2 a b_2 (b-c))-a_2^2right)\
                P_5&:=&2 A^4 a_2
                b_2 (2 a-2 b+1) (a+b-2)
                end{eqnarray}

                and
                begin{eqnarray}
                Q_0&:=&a_1 (2 c-3) (a_1-2 b_2 (c-1))\
                Q_1&:=&a_2 (2 c-3) (3 a_1+b_2 (2-4 c))\
                Q_2&:=&A^2 left((2 a-1) a_1^2 (2 b-1)-2 a_1 b_2 (a (4 b (c-2)+4 c-3)-4 b c+7 b+3 c-6)-12 a b_2^2 (2
                c-3) (b-c)right)+4 a_2^2 (c-2)\
                Q_3&:=&A^2 a_2 (a_1 (a (8 b-6)-6 b+3)+2 b_2 (a (-4 b c+2 b-2 c+9)+2 (2 b-1) (2 c-3)))\
                Q_4&:=&-2 A^2 left((2 a-1) A^2 (2 b-3) b_2 (a_1 (a-b+1)+2 a b_2 (b-c))+2
                a_2^2 (a (-b)+a+b-1)right)\
                Q_5&:=&2 (1-a) A^4 a_2 (2 b-3) b_2 (2 a-2 b+1)
                end{eqnarray}

                and
                begin{equation}
                y(x):=F_{2,1}left[a,b,c,A^2 x^2right]
                end{equation}



                Then the ODE:
                begin{eqnarray}
                g^{''}(x) - frac{sumlimits_{j=0}^5 P_j x^j}{x(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g^{'}(x) + frac{sumlimits_{j=0}^5 Q_j x^j}{x^2(A x-1)(A x+1) (sumlimits_{j=0}^3 p_j x^j)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[14]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
                =.;
                p0 =.; p1 =.; p2 =.; p3 =.;
                P0 =.; P1 =.; P2 =.; P3 =.; P4 =.; P5 =.;
                Q0 =.; Q1 =.; Q2 =.; Q3 =.; Q4 =.; Q5 =.; Clear[y];
                {a3, b4} = {-2 a A^2 b2, -A^2 b2};
                {p0, p1, p2, p3} = {a1 (a1 - 2 b2 (-1 + c)), a2 (2 a1 + b2 - 2 b2 c),
                a2^2 - 2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c)),
                A^2 a2 (-1 - 2 a + 2 b) b2};
                {P0, P1, P2, P3, P4, P5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
                2 a2 (-2 + c) (2 a1 + b2 - 2 b2 c),
                a2^2 (-5 + 2 c) +
                A^2 (a1^2 (1 - 2 a - 2 b) - 4 a b2^2 (b - c) (-5 + 2 c) +
                2 a1 b2 (6 + 3 a - 7 b - 3 c + 4 b c)),
                2 A^2 a2 (-2 a1 (-1 + a + b) + b2 (4 + 5 a - 7 b - 3 c + 4 b c)),
                A^2 (-3 + 2 a + 2 b) (-a2^2 +
                2 A^2 b2 (a1 (1 + a - b) + 2 a b2 (b - c))),
                2 A^4 a2 (1 + 2 a - 2 b) (-2 + a + b) b2};
                {Q0, Q1, Q2, Q3, Q4, Q5} = {a1 (a1 - 2 b2 (-1 + c)) (-3 + 2 c),
                a2 (3 a1 + b2 (2 - 4 c)) (-3 + 2 c),
                4 a2^2 (-2 + c) +
                A^2 ((-1 + 2 a) a1^2 (-1 + 2 b) - 12 a b2^2 (b - c) (-3 + 2 c) -
                2 a1 b2 (-6 + 7 b + 3 c - 4 b c +
                a (-3 + 4 b (-2 + c) + 4 c))),
                A^2 a2 (a1 (3 - 6 b + a (-6 + 8 b)) +
                2 b2 (2 (-1 + 2 b) (-3 + 2 c) +
                a (9 + 2 b - 2 c - 4 b c))), -2 A^2 (2 a2^2 (-1 + a + b -
                a b) + (-1 + 2 a) A^2 (-3 + 2 b) b2 (a1 (1 + a - b) +
                2 a b2 (b - c))),
                2 A^4 a2 (1 + 2 a - 2 b) (1 - a) (-3 + 2 b) b2};
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] - (
                P5 x^5 + P4 x^4 + P3 x^3 + P2 x^2 + P1 x^1 + P0)/(
                x (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 + p0))
                D[#, x] + (Q5 x^5 + Q4 x^4 + Q3 x^3 + Q2 x^2 + Q1 x^1 + Q0)/(
                x ^2 (-1 + A x) (1 + A x) (p3 x^3 + p2 x^2 + p1 x^1 +
                p0)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};

                {a, b, c, a1, a2, b2, A, x} =
                RandomReal[{0, 1}, 8, WorkingPrecision -> 50];
                Simplify[eX]

                Out[25]= {0.*10^-48}


                Update: The ODE above is a seven parameter family.
                Now, note that if in the example above we add three additional constraints and as such reduce the number of adjustable parameters to four we get another neat example:



                Firstly define:
                begin{eqnarray}
                a_1&:=& c-frac{1}{2}\
                a_2&:=& A frac{1}{sqrt{2}} sqrt{-1+4 a+8 a^2+2 c-8 a c}\
                a_3&:=&-2 a A^2\
                hline \
                b_2&:=& 1\
                b_4&:=&-A^2 \
                hline \
                b&:=&a+frac{1}{2}
                end{eqnarray}

                Then the ODE below:
                begin{eqnarray}
                &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) + frac{3-2 c+4 a A^2 x^2}{x(A x-1)(A x+1)} g^{'}(x) + frac{(-3+2 c) + sqrt{2} A sqrt{-1+4 a+8 a^2+2 c-8 a c} x+2(-1-a+2 a^2) x^2}{x^2(A x-1)(A x+1)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[18]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; x 
                =.;
                {a1, a2, a3} = {(-(1/2) + c),
                A Sqrt[1/2 (-1 + 4 a + 8 a^2 + 2 c - 8 a c)], -2 a A^2};
                {b2, b4} = {1, -A^2};
                b = a + 1/2;
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] + (3 - 2 c + 4 a A^2 x^2)/(x (-1 + A x) (1 + A x))
                D[#, x] + ( (-3 + 2 c) +
                Sqrt[2] A Sqrt[(-1 + 4 a + 8 a^2 + 2 c - 8 a c)] x +
                2 (-1 - a + 2 a^2) A^2 x^2)/(
                x ^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};
                {b2, a, c, A, x} = RandomReal[{0, 1}, 5, WorkingPrecision -> 50];
                Simplify[eX]

                Out[25]= {0.*10^-49}


                Secondly define:
                begin{eqnarray}
                a_1&:=& 2c-1\
                a_2&:=& A sqrt{2} sqrt{(-1+2 a)(-1+b)}\
                a_3&:=&-2 a A^2\
                hline \
                b_2&:=& 1\
                b_4&:=&-A^2 \
                hline \
                c&:=&frac{3}{2}
                end{eqnarray}

                Then the ODE below:
                begin{eqnarray}
                &&!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!g^{''}(x) +
                frac{3+2 A^2(-2+a+b)x^2}{x(A x-1)(A x+1)} g^{'}(x) +
                frac{-3-sqrt{2} A sqrt{(-1+2 a)(-1+b)} x+2(-1+a)(-3+2 b) A^2 x^2}{x^2(A x-1)(A x+1)} g(x)=0
                end{eqnarray}

                is solved by
                begin{eqnarray}
                g(x)&:=& (a_3 x^3+a_2 x^2+a_1 x) y(x) + (b_4 x^4+b_2 x^2) y^{'}(x)
                end{eqnarray}



                In[567]:= a =.; b =.; c =.; a1 =.; a2 =.; a3 =.; b2 =.; b4 =.; A =.; 
                x =.;
                {b2, b4} = {1, -A^2};
                {a1, a2, a3} = {2 (c - 1),
                Sqrt[2] Sqrt[-1 + 2 a] A Sqrt[-1 + b], -2 a A^2};
                c = 3/2;
                y[x_] = Hypergeometric2F1[a, b, c, (A x)^2];
                eX = (D[#, {x, 2}] + (3 + 2 A^2 (-2 + a + b) x^2)/(
                x (-1 + A x) (1 + A x))
                D[#, x] + ( -3 - Sqrt[2] A (Sqrt[-1 + 2 a] Sqrt[-1 + b]) x +
                2 (-1 + a) (-3 + 2 b) A^2 x^2)/(
                x^2 (-1 + A x) (1 + A x)) #) & /@ {(a3 x^3 + a2 x^2 + a1 x) y[
                x] + (b4 x^4 + b2 x^2) y'[x]};
                {a, b, A, x} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
                Simplify[eX]


                Out[574]= {0.*10^-47 + 0.*10^-49 I}






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                edited Nov 29 at 15:42

























                answered Nov 29 at 12:37









                Przemo

                4,1321928




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