I am trying to find two independent solutions of this differential equation: $$x(1-x)y''(x)+\left[\frac d2-\left(d+\frac12\right)x\right]y'(x)-\frac{(d-1)d}{4}y(x)=0,$$ for $0<x<1$.
This is a hypergeometric equation with $a=(d-1)/2$, $b=d/2$, and $c=d/2$, so for generic values of $d$ the solutions will be the standard solutions of the hypergeometric differential equation, namely
$$y_1(x)={}_2F_1(a,b;c;x)={}_2F_1\left(\frac{d-1}2,\frac d2;\frac d2;x\right),$$ $$y_2(x)=x^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;x)=x^{(2-d)/2}{}_2F_1\left(\frac12,1,\frac{4-d}2;x\right).$$
However, I am interested in the cases where $d$ is an integer greater than $1$, so $d=2,3,4,\ldots$. For odd $d$ these solutions seem to still work just fine, but they clearly break down for even $d$ greater than or equal to $4$ because then the hypergeometric series from the second solution would be undefined.
I've looked in Abramowitz and Stegun [1] and it seems this case is referred to as the degenerate case and is treated in Bateman et al [2]. However, the two independent solutions given there for this case (given in the table starting on page 71, case 8 for $d$ odd and case 16 for $d$ even; in what follows I am talking about $d$ even as everything seems to be find for $d$ odd), $u_1={}_2F_1(a,b;c;x)$ and $u_3=(-x)^{-a}{}_2F_1(a,a+1-c,a+1-b,x^{-1})$ from page 105, seem to be the same solution according to Mathematica for this specific values of $a,b,c$ given in terms of $d$ as above. So, I am still left with only one of the two independent solutions I'm looking for.
If I plug this equation in Mathematica and ask it to solve it I get the following:
Clear[d, k];
d = 2 k;
$Assumptions = k \[Element] Integers && k > 0;
DSolve[x (1 - x) G''[x] + (d/2 - (d + 1/2) x) G'[x] - (d - 1) d G[x]/4 == 0, G[x], {x, 0, 1}]
{{G[x] -> ((-1 + x)^(1/2 - k)x^-k ((-1 + k) x^k C[1] - x C[2] Hypergeometric2F1[1 - k, 3/2 - k, 2 - k, x]))/(-1 + k)}}
so Mathematica gives again a solution that does not work for even $d$ greater than or equal to $4$ because the hypergeometric series is not defined.
Something seems very wrong in all of this, most likely something on my part which I'm missing or not understand. Any help is greatly appreciated!
[1] Abramowitz, Milton (ed.); Stegun, Irene A. (ed.), Handbook of mathematical functions with formulas, graphs and mathematical tables, Washington: U.S. Department of Commerce. xiv, 1046 pp. (1964). ZBL0171.38503.
[2] Bateman, Harry; Erdélyi, Arthur, Higher transcendental functions. The hypergeometric function. Legendre functions. Translation from the English by N.Ya. Vilenkin, Mathematische Handbücherei. Moskau: Verlag ‘Nauka’. Hauptredaktion für physikalisch-mathematische Literatur. 294 S. (1965). ZBL0146.09301.
I would like to show a solution approach for even d with ODE: $x\cdot \left(1-x\right)\cdot y''(x)+\left(\frac{d}{2}-\left(d+\frac{1}{2}\right)\cdot x\right)\cdot y'(x)-\frac{1}{4}\left(\left(d-1\right)\cdot d\right)\cdot y\left(x\right)=0$
$d=2k, k\in \mathbb{Z}\land k>0$ and $0<x<1$.
1st fundamental solution is: $$f_{1}(x)\rightarrow \frac{1}{\left(1-x\right)^{\frac{d-1}{2}}}=\frac{d-1}{2} \sum _{k=0}^{\infty} \frac{\left(\frac{d+1}{2}\right)_{k-1}}{k!}\cdot x^k$$
$(a)_{n}$ denotes the Pochhammer symbol.
2nd fundamental solution is:
$$f_{2}(x)\rightarrow \frac{P\left(\frac{d}{2}-2,x\right)\cdot \sqrt{1-x}+\tanh^{-1}\left(\sqrt{1-x}\right)\cdot x^{\left(\frac{d}{2}-1\right)}}{\left(1-x\right)^{\frac{\left(d-1\right)}{2}} \cdot x^{\left(\frac{d}{2}-1\right)}}$$
with $$P(n,x)\mapsto \sum_{i=0}^{n} p_{i}\cdot x^{i}$$
Unfortunately i can't present a general solution for $P(n,x)$.
But for a given $d$ we are able to determine the polynomial.
Example for $d=10$:
Plug in $y(x)=f_{2}(x)$ into the original ODE:
As numerator of the ODE we get: $(-3 p_2-2 p_3+4) x^3-(10 p_1+p_2-6 p_3) x^2-21 p_0 x+8 (p_1+p_2) x+25 p_0+6 p_1=0$
Now we have to determine the $p_{i}$:
We will get the equations: $25 p_0+6 p_1=0,8 (p_1+p_2)-21 p_0=0,-10 p_1-p_2+6 p_3=0,-3 p_2-2 p_3+4=0$
With the solution: $p_{0} = \frac{16}{35},p_{1} = -\frac{40}{21},p_{2} = \frac{326}{105},p_{3} = -\frac{93}{35}$
Hence we get $f_{2}(x)\mapsto \frac{\left(-\frac{93}{35}\cdot x^{3}+\frac{326}{105}\cdot x^{2}-\frac{40}{21}\cdot x+\frac{16}{35}\right)\cdot \sqrt{1-x}+\tanh^{-1}\! \left(\sqrt{1-x}\right)\cdot x^{4}}{\left(1-x\right)^{\frac{9}{2}}\cdot x^{4}}$
Now we have the general solution for $d=10$ as combination of the fundamental solutions:
$$y(x)\mapsto c_{1}\cdot f_{1} \left(x\right)+c_{2}\cdot f_{2} \left(x\right)=\frac{c_{1}}{\left(1-x\right)^{\frac{9}{2}}}+\frac{c_{2} \left(\left(-\frac{93}{35} x^{3}+\frac{326}{105} x^{2}-\frac{40}{21} x+\frac{16}{35}\right) \sqrt{1-x}+\tanh^{-1} \left(\sqrt{1-x}\right) x^{4}\right)}{\left(1-x\right)^{\frac{9}{2}} x^{4}}$$
This solution satisfies the ODE for $d=10$.
Perhaps one of the high-potentials in this forum can provide a general solution for the polynomial $P(n,x)$.