Any suggestions on how to get started on solving this differential equation?

Question

The differential equation is

$$(1-x)y''(x)-y'(x)+kx^{-25/14}y(x)=0$$.

I tried a power series solution with fractional powers, like in how to obtain Frobenius series solution to ODE when one term is multiplied by x to fractional power?, but it did not work.

Answer 1

Linear ODE with rational power

$$ (1-x)\ y''(x) - \ y(x) + k\ x^{-25/14} \ y(x)=0$$

Prefactors of second derivatives go between the two derivatives by the dual transformation via integration by parts of scalar products in the Hilbert space over $(0,1)

$$ \begin{gather} \forall g,g(0,1)=0 : \int_0^1 dx &\ g(x) ( - (1-x) y''(x) ) == \int_0^1 dx ( g'(x) (1-x) y'(x) - g(x)y'(x) \end{gather} $$

We have in Mathematica code

   D[(1-x)D[ f[x],x],x]  - k\  (x^-2) (x^\[Alpha]) f[x]

    -k x^(-2+\[Alpha]) f[x]-(f^\[Prime])[x]+(1-x) (f^\[Prime]\[Prime])[x] 

By trial and error, the following group of transformations has been found

    Together /@ 
     Collect[x^(\[Alpha] + 1) D[(1 - x) D[x^\[Beta] f[x], x], x]  - 
       k  x^-2 x^(\[Alpha] + 1 + \[Beta])
         f[x] /. {\[Beta] -> -\[Alpha]} // Simplify // Expand, 
       {f[x],  f'[x], f''[x]}] /. 
        {\[Alpha] -> 3/14} 

      -(((-(51/196) + k + (9 x)/196) f[x])/ x) + 
         (-(3/7) - (4 x)/7) Derivative[1][f][x] - 
          (-x + x^2) ( f^\[Prime]\[Prime])[x]

or more readable

$$-\left(x^2-x\right) f''(x)+\left(-\frac{4 x}{7}-\frac{3}{7}\right) f'(x)-\frac{f(x) \left(k+\frac{9 x}{196}-\frac{51}{196}\right)}{x}$$

Belive it or not, there always show up relativistic factors $\sqrt{1+k^2/4} = \cosh u$:

 DSolve[ 0 == -(((-(51/196) + Sinh[u]^2/4 + (9 x)/196) f[x])/ x) +  
         (-(3/7) - (4 x)/7) Derivative[1][f][x] - 
    (-x + x^2) (  f^\[Prime]\[Prime])[x], f[x], x] // FullSimplify

$$ x^{\frac{\cosh (u)}{2}+\frac{5}{7}} \, _2F_1\left(\cosh ^2\left(\frac{u}{2}\right),\cosh ^2\left(\frac{u}{2}\right);\cosh (u)+1;x\right)$$

$$ _2F_1\left(-\sinh ^2\left(\frac{u}{2}\right), -\sinh ^2 \ \left(\frac{u}{2} \right);\ 1-\cosh (u);\ x \right)$$

There are made eliminations of constant factors and square roots are taken. So the solutions, with the powers of x added, have to be checked to satisfy the original equation (as always, because solving is not a proof, it is a process of educated guessing).