Frobenius solution to the hypergeometric equation

In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. This is usually the method we use for complicated ordinary differential equations.

The solution of the hypergeometric differential equation is very important. For instance, Legendre's differential equation can be shown to be a special case of the hypergeometric differential equation. Hence, by solving the hypergeometric differential equation, one may directly compare its solutions to get the solutions of Legendre's differential equation, after making the necessary substitutions. For more details, please check the hypergeometric differential equation.

We shall prove that this equation has three singularities, namely at x = 0, x = 1 and around x = infinity. However, as these will turn out to be regular singular points, we will be able to assume a solution on the form of a series. Since this is a second-order differential equation, we must have two linearly independent solutions.

The problem however will be that our assumed solutions may or not be independent, or worse, may not even be defined (depending on the value of the parameters of the equation). This is why we shall study the different cases for the parameters and modify our assumed solution accordingly.

Solve the hypergeometric equation around all singularities:

${\displaystyle x(1-x)y''+\left\{\gamma -(1+\alpha +\beta )x\right\}y'-\alpha \beta y=0}$

Let

${\displaystyle {\begin{aligned}P_{0}(x)&=-\alpha \beta ,\\P_{1}(x)&=\gamma -(1+\alpha +\beta )x,\\P_{2}(x)&=x(1-x)\end{aligned}}}$

Then

${\displaystyle P_{2}(0)=P_{2}(1)=0.}$

Hence, x = 0 and x = 1 are singular points. Let's start with x = 0. To see if it is regular, we study the following limits:

${\displaystyle {\begin{aligned}\lim _{x\to a}{\frac {(x-a)P_{1}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\lim _{x\to 0}{\frac {x(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\gamma \\\lim _{x\to a}{\frac {(x-a)^{2}P_{0}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)^{2}(-\alpha \beta )}{x(1-x)}}=\lim _{x\to 0}{\frac {x^{2}(-\alpha \beta )}{x(1-x)}}=0\end{aligned}}}$

Hence, both limits exist and x = 0 is a regular singular point. Therefore, we assume the solution takes the form

${\displaystyle y=\sum _{r=0}^{\infty }a_{r}x^{r+c}}$

with a₀ ≠ 0. Hence,

${\displaystyle {\begin{aligned}y'&=\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\y''&=\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}.\end{aligned}}}$

Substituting these into the hypergeometric equation, we get

${\displaystyle x\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}-x^{2}\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )x\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0}$

That is,

${\displaystyle \sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0}$

In order to simplify this equation, we need all powers to be the same, equal to r + c − 1, the smallest power. Hence, we switch the indices as follows:

${\displaystyle {\begin{aligned}&\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\&\qquad -(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}}$

Thus, isolating the first term of the sums starting from 0 we get

${\displaystyle {\begin{aligned}&a_{0}(c(c-1)+\gamma c)x^{c-1}+\sum _{r=1}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}\\&\qquad +\gamma \sum _{r=1}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}}$

Now, from the linear independence of all powers of x, that is, of the functions 1, x, x², etc., the coefficients of x^k vanish for all k. Hence, from the first term, we have

${\displaystyle a_{0}(c(c-1)+\gamma c)=0}$

which is the indicial equation. Since a₀ ≠ 0, we have

${\displaystyle c(c-1+\gamma )=0.}$

Hence,

${\displaystyle c_{1}=0,c_{2}=1-\gamma }$

Also, from the rest of the terms, we have

${\displaystyle ((r+c)(r+c-1)+\gamma (r+c))a_{r}+(-(r+c-1)(r+c-2)-(1+\alpha +\beta )(r+c-1)-\alpha \beta )a_{r-1}=0}$

Hence,

${\displaystyle {\begin{aligned}a_{r}&={\frac {(r+c-1)(r+c-2)+(1+\alpha +\beta )(r+c-1)+\alpha \beta }{(r+c)(r+c-1)+\gamma (r+c)}}a_{r-1}\\&={\frac {(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta }{(r+c)(r+c+\gamma -1)}}a_{r-1}\end{aligned}}}$

But

${\displaystyle {\begin{aligned}(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta &=(r+c-1)(r+c+\alpha -1)+(r+c-1)\beta +\alpha \beta \\&=(r+c-1)(r+c+\alpha -1)+\beta (r+c+\alpha -1)\end{aligned}}}$

Hence, we get the recurrence relation

${\displaystyle a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c+\gamma -1)}}a_{r-1},{\text{ for }}r\geq 1.}$

Let's now simplify this relation by giving a_r in terms of a₀ instead of a_r−1. From the recurrence relation (note: below, expressions of the form (u)_r refer to the Pochhammer symbol).

${\displaystyle {\begin{aligned}a_{1}&={\frac {(c+\alpha )(c+\beta )}{(c+1)(c+\gamma )}}a_{0}\\a_{2}&={\frac {(c+\alpha +1)(c+\beta +1)}{(c+2)(c+\gamma +1)}}a_{1}={\frac {(c+\alpha +1)(c+\alpha )(c+\beta )(c+\beta +1)}{(c+2)(c+1)(c+\gamma )(c+\gamma +1)}}a_{0}={\frac {(c+\alpha )_{2}(c+\beta )_{2}}{(c+1)_{2}(c+\gamma )_{2}}}a_{0}\\a_{3}&={\frac {(c+\alpha +2)(c+\beta +2)}{(c+3)(c+\gamma +2)}}a_{2}={\frac {(c+\alpha )_{2}(c+\alpha +2)(c+\beta )_{2}(c+\beta +2)}{(c+1)_{2}(c+3)(c+\gamma )_{2}(c+\gamma +2)}}a_{0}={\frac {(c+\alpha )_{3}(c+\beta )_{3}}{(c+1)_{3}(c+\gamma )_{3}}}a_{0}\end{aligned}}}$

As we can see,

${\displaystyle a_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}a_{0},{\text{ for }}r\geq 0}$

Hence, our assumed solution takes the form

${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c}.}$

We are now ready to study the solutions corresponding to the different cases for c₁ − c₂ = γ − 1 (this reduces to studying the nature of the parameter γ: whether it is an integer or not).

Then y₁ = y|_{c = 0} and y₂ = y|_{c = 1 − γ}. Since

${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c},}$

we have

${\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(\gamma )_{r}}}x^{r}=a_{0}\cdot {{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)\\y_{2}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1-\gamma +1)_{r}(1-\gamma +\gamma )_{r}}}x^{r+1-\gamma }\\&=a_{0}x^{1-\gamma }\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(2-\gamma )_{r}}}x^{r}\\&=a_{0}x^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\end{aligned}}}$

Hence, ${\displaystyle y=A'y_{1}+B'y_{2}.}$ Let A′ a₀ = a and B′ a₀ = B. Then

${\displaystyle y=A{{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)+Bx^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\,}$

Then y₁ = y|_{c = 0}. Since γ = 1, we have

${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r+c}.}$

Hence,

${\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r}}}x^{r}=a_{0}{{}_{2}F_{1}}(\alpha ,\beta ;1;x)\\y_{2}&=\left.{\frac {\partial y}{\partial c}}\right|_{c=0}.\end{aligned}}}$

To calculate this derivative, let

${\displaystyle M_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}.}$

Then

${\displaystyle \ln(M_{r})=\ln \left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\right)=\ln(c+\alpha )_{r}+\ln(c+\beta )_{r}-2\ln(c+1)_{r}}$

But

${\displaystyle \ln(c+\alpha )_{r}=\ln \left((c+\alpha )(c+\alpha +1)\cdots (c+\alpha +r-1)\right)=\sum _{k=0}^{r-1}\ln(c+\alpha +k).}$

Hence,

${\displaystyle {\begin{aligned}\ln(M_{r})&=\sum _{k=0}^{r-1}\ln(c+\alpha +k)+\sum _{k=0}^{r-1}\ln(c+\beta +k)-2\sum _{k=0}^{r-1}\ln(c+1+k)\\&=\sum _{k=0}^{r-1}\left(\ln(c+\alpha +k)+\ln(c+\beta +k)-2\ln(c+1+k)\right)\end{aligned}}}$

Differentiating both sides of the equation with respect to c, we get:

${\displaystyle {\frac {1}{M_{r}}}{\frac {\partial M_{r}}{\partial c}}=\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).}$

Hence,

${\displaystyle {\frac {\partial M_{r}}{\partial c}}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).}$

Now,

${\displaystyle y=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}=a_{0}x^{c}\sum _{r=0}^{\infty }M_{r}x^{r}.}$

Hence,

${\displaystyle {\begin{aligned}{\frac {\partial y}{\partial c}}&=a_{0}x^{c}\ln(x)\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}+a_{0}x^{c}\sum _{r=0}^{\infty }\left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\left\{\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right\}\right)x^{r}\\&=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right)x^{r}.\end{aligned}}}$

For c = 0, we get

${\displaystyle y_{2}=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}.}$

Hence, y = C′y₁ + D′y₂. Let C′a₀ = C and D′a₀ = D. Then

${\displaystyle y=C{{}_{2}F_{1}}(\alpha ,\beta ;1;x)+D\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln(x)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}}$

The value of ${\displaystyle \gamma }$ is ${\displaystyle \gamma =0,-1,-2,\cdots }$. To begin with, we shall simplify matters by concentrating a particular value of ${\displaystyle \gamma }$ and generalise the result at a later stage. We shall use the value ${\displaystyle \gamma =-2}$. The indicial equation has a root at ${\displaystyle c=0}$, and we see from the recurrence relation

${\displaystyle a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c-3)}}a_{r-1},}$

that when ${\displaystyle r=3}$ that that denominator has a factor ${\displaystyle c}$ which vanishes when ${\displaystyle c=0}$. In this case, a solution can be obtained by putting ${\displaystyle a_{0}=b_{0}c}$ where ${\displaystyle b_{0}}$ is a constant.

With this substitution, the coefficients of ${\displaystyle x^{r}}$ vanish when ${\displaystyle c=0}$ and ${\displaystyle r<3}$. The factor of ${\displaystyle c}$ in the denominator of the recurrence relation cancels with that of the numerator when ${\displaystyle r\geq 3}$. Hence, our solution takes the form

${\displaystyle y_{1}={\frac {b_{0}}{(-2)\times (-1)}}\left({\frac {(\alpha )_{3}(\beta )_{3}}{(3!0!}}x^{3}+{\frac {(\alpha )_{4}(\beta )_{4}}{4!1!}}x^{4}+{\frac {(\alpha )_{5}(\beta )_{5}}{5!2!}}x^{5}+\cdots \right)}$

${\displaystyle ={\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-3)!}}x^{r}={\frac {b_{0}}{(-2)_{2}}}{\frac {(\alpha )_{3}(\beta )_{3}}{3!}}\sum _{r=3}^{\infty }{\frac {(\alpha +3)_{r-3}(\beta +3)_{r-3}}{(1+3)_{r-3}(r-3)!}}x^{r}.}$

If we start the summation at ${\displaystyle r=0}$ rather than ${\displaystyle r=3}$ we see that

${\displaystyle y_{1}=b_{0}{\frac {(\alpha )_{3}(\beta )_{3}}{(-2)_{2}\times 3!}}x^{3}{_{2}F_{1}}(\alpha +3,\beta +3;(1+3);x).}$

The result (as we have written it) generalises easily. For ${\displaystyle \gamma =1+m}$, with ${\displaystyle m=1,2,3,\cdots }$ then

${\displaystyle y_{1}=b_{0}{\frac {(\alpha )_{m}(\beta )_{m}}{(1-m)_{m-1}\times m!}}x^{m}{_{2}F_{1}}(\alpha +m,\beta +m;(1+m);x).}$

Obviously, if ${\displaystyle \gamma =-2}$, then ${\displaystyle m=3}$. The expression for ${\displaystyle y_{1}(x)}$ we have just given looks a little inelegant since we have a multiplicative constant apart from the usual arbitrary multiplicative constant ${\displaystyle b_{0}}$. Later, we shall see that we can recast things in such a way that this extra constant never appears

The other root to the indicial equation is ${\displaystyle c=1-\gamma =3}$, but this gives us (apart from a multiplicative constant) the same result as found using ${\displaystyle c=0}$. This means we must take the partial derivative (w.r.t. ${\displaystyle c}$) of the usual trial solution in order to find a second independent solution. If we define the linear operator ${\displaystyle L}$ as

${\displaystyle L=x(1-x){\frac {d^{2}}{dx^{2}}}-(\alpha +\beta +1)x{\frac {d}{dx}}+\gamma {\frac {d}{dx}}-\alpha \beta ,}$

then since ${\displaystyle \gamma =-2}$ in our case,

${\displaystyle Lc\sum _{r=0}^{\infty }b_{r}(c)x^{r}=b_{0}c^{2}(c-3).}$

(We insist that ${\displaystyle b_{0}\neq 0}$.) Taking the partial derivative w.r.t ${\displaystyle c}$,

${\displaystyle L{\frac {\partial }{\partial c}}c\sum _{r=0}^{\infty }b_{r}(c)x^{r+c}=b_{0}(3c^{2}-6c).}$

Note that we must evaluate the partial derivative at ${\displaystyle c=0}$ (and not at the other root ${\displaystyle c=3}$). Otherwise the right hand side is non-zero in the above, and we do not have a solution of ${\displaystyle Ly(x)=0}$. The factor ${\displaystyle c}$ is not cancelled for ${\displaystyle r=0,1}$ and ${\displaystyle r=2}$. This part of the second independent solution is

${\displaystyle {\bigg [}{\frac {\partial }{\partial c}}b_{0}{\bigg (}c+c{\frac {(c+\alpha )(c+\beta )}{(c+1)(c-2)}}x+c{\frac {(c+\alpha )(c+\alpha +1)(c+\beta )(c+\beta +1)}{(c+1)(c+2)(c-2)(c-1)}}x^{2}{\bigg )}{\bigg ]}{\bigg \vert }_{c=0}.}$ ${\displaystyle =b_{0}\left(1+{\frac {\alpha \beta }{1!\times (-2)}}x+{\frac {\alpha (\alpha +1)\beta (\beta +1)}{2!\times (-2)\times (-1)}}x^{2}\right)=b_{0}\sum _{r=0}^{3-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-3)_{r}}}x^{r}.}$

Now we can turn our attention to the terms where the factor ${\displaystyle c}$ cancels. First

${\displaystyle cb_{3}={\frac {b_{0}}{(c-1)(c-2)}}{\cancel {c}}{\frac {(c+\alpha )(c+\alpha +1)(c+\alpha +2)(c+\beta )(c+\beta +1)(c+\beta +2)}{{\cancel {c}}(c+1)(c+2)(c+3)}}.}$

After this, the recurrence relations give us

${\displaystyle cb_{4}=cb_{3}(c){\frac {(c+\alpha +3)(c+\beta +3)}{(c+1)(c+4))}}.}$

${\displaystyle cb_{5}=cb_{3}(c){\frac {(c+\alpha +3)(c+\alpha +4)(c+\beta +3)(c+\beta +4))}{(c+2)(c+1)(c+5)(c+4)}}.}$

So, if ${\displaystyle r\geq 3}$ we have

${\displaystyle cb_{r}={\frac {b_{0}}{(c-1)(c-2)}}{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r-3}(c+1)_{r}}}.}$

We need the partial derivatives

${\displaystyle {\frac {\partial cb_{3}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{3}(\beta )_{3}}{0!3!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{\alpha }}+{\frac {1}{\alpha +1}}+{\frac {1}{\alpha +2}}}$ ${\displaystyle +{\frac {1}{\beta }}+{\frac {1}{\beta +1}}+{\frac {1}{\beta +2}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}{\bigg ]}.}$

Similarly, we can write

${\displaystyle {\frac {\partial cb_{4}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{4}(\beta )_{4}}{1!4!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}}$ ${\displaystyle +\sum _{k=0}^{k=3}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=3}{\frac {1}{\beta +k}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{4}}-{\frac {1}{1}}{\bigg ]},}$

and

${\displaystyle {\frac {\partial cb_{5}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{5}(\beta )_{5}}{2!5!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}}$ ${\displaystyle +\sum _{k=0}^{k=4}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=4}{\frac {1}{\beta +k}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{4}}-{\frac {1}{5}}-{\frac {1}{1}}-{\frac {1}{2}}{\bigg ]}.}$

It becomes clear that for ${\displaystyle r\geq 3}$

${\displaystyle {\frac {\partial cb_{r}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{r}(\beta )_{r}}{(r-3)!r!}}{\bigg [}H_{2}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-3}{\bigg ]}.}$

Here, ${\displaystyle H_{k}}$ is the ${\displaystyle k}$th partial sum of the harmonic series, and by definition ${\displaystyle H_{0}=0}$ and ${\displaystyle H_{1}=1}$.

Putting these together, for the case ${\displaystyle \gamma =-2}$ we have a second solution

${\displaystyle y_{2}(x)=\log x\times {\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-3)!}}x^{r}+b_{0}\sum _{r=0}^{3-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-3)_{r}}}x^{r}}$

${\displaystyle +{\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(r-3)!r!}}{\bigg [}H_{2}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-3}{{\bigg ]}x^{r}}.}$

The two independent solutions for ${\displaystyle \gamma =1-m}$ (where ${\displaystyle m}$ is a positive integer) are then

${\displaystyle y_{1}(x)={\frac {1}{(1-m)_{m-1}}}\sum _{r=m}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-m)!}}x^{r}}$

and

${\displaystyle y_{2}(x)=\log x\times y_{1}(x)+\sum _{r=0}^{m-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-m)_{r}}}x^{r}}$

${\displaystyle +{\frac {1}{(1-m)_{m-1}}}\sum _{r=m}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(r-m)!r!}}{\bigg [}H_{m-1}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-m}{\bigg ]}x^{r}.}$

The general solution is as usual ${\displaystyle y(x)=Ay_{1}(x)+By_{2}(x)}$ where ${\displaystyle A}$ and ${\displaystyle B}$ are arbitrary constants. Now, if the reader consults a ``standard solution" for this case, such as given by Abramowitz and Stegun ^[1] in §15.5.21 (which we shall write down at the end of the next section) it shall be found that the ${\displaystyle y_{2}}$ solution we have found looks somewhat different from the standard solution. In our solution for ${\displaystyle y_{2}}$, the first term in the infinite series part of ${\displaystyle y_{2}}$ is a term in ${\displaystyle x^{m}}$. The first term in the corresponding infinite series in the standard solution is a term in ${\displaystyle x^{m+1}}$. The ${\displaystyle x^{m}}$ term is missing from the standard solution. Nonetheless, the two solutions are entirely equivalent.

The reason for the apparent discrepancy between the solution given above and the standard solution in Abramowitz and Stegun ^[1] §15.5.21 is that there are an infinite number of ways in which to represent the two independent solutions of the hypergeometric ODE. In the last section, for instance, we replaced ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}c}$. Suppose though, we are given some function ${\displaystyle h(c)}$ which is continuous and finite everywhere in an arbitrarily small interval about ${\displaystyle c=0}$. Suppose we are also given

${\displaystyle h(c)\vert _{c=0}\neq 0,}$ and ${\displaystyle {\frac {dh}{dc}}{\bigg \vert }_{c=0}\neq 0.}$

Then, if instead of replacing ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}c}$ we replace ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}h(c)c}$, we still find we have a valid solution of the hypergeometric equation. Clearly, we have an infinity of possibilities for ${\displaystyle h(c)}$. There is however a ``natural choice" for ${\displaystyle h(c)}$. Suppose that ${\displaystyle cb_{N}(c)=b_{0}f(c)}$ is the first non zero term in the first ${\displaystyle y_{1}(x)}$ solution with ${\displaystyle c=0}$. If we make ${\displaystyle h(c)}$ the reciprocal of ${\displaystyle f(c)}$, then we won't have a multiplicative constant involved in ${\displaystyle y_{1}(x)}$ as we did in the previous section. From another point of view, we get the same result if we ``insist" that ${\displaystyle a_{N}}$ is independent of ${\displaystyle c}$, and find ${\displaystyle a_{0}(c)}$ by using the recurrence relations backwards.

For the first ${\displaystyle (c=0)}$ solution, the function ${\displaystyle h(c)}$ gives us (apart from multiplicative constant) the same ${\displaystyle y_{1}(x)}$ as we would have obtained using ${\displaystyle h(c)=1}$. Suppose that using ${\displaystyle h(c)=1}$ gives rise to two independent solutions ${\displaystyle y_{1}(x)}$ and ${\displaystyle y_{2}(x)}$. In the following we shall denote the solutions arrived at given some ${\displaystyle h(c)\neq 1}$ as ${\displaystyle {\tilde {y}}_{1}(x)}$ and ${\displaystyle {\tilde {y}}_{2}(x)}$.

The second solution requires us to take the partial derivative w.r.t ${\displaystyle c}$, and substituting the usual trial solution gives us

${\displaystyle L{\frac {\partial }{\partial c}}\sum _{r=0}^{\infty }ch(c)b_{r}x^{r+c}=b_{0}\left({\frac {dh}{dc}}c^{2}(c-1)+2ch(c)(c-1)+h(c)c^{2}\right).}$

The operator ${\displaystyle L}$ is the same linear operator discussed in the previous section. That is to say, the hypergeometric ODE is represented as ${\displaystyle Ly(x)=0}$.

Evaluating the left hand side at ${\displaystyle c=0}$ will give us a second independent solution. Note that this second solution ${\displaystyle {{\tilde {y}}_{2}}}$ is in fact a linear combination of ${\displaystyle y_{1}(x)}$ and ${\displaystyle y_{2}(x)}$.

Any two independent linear combinations (${\displaystyle {\tilde {y}}_{1}}$ and ${\displaystyle {\tilde {y}}_{2}}$) of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are independent solutions of ${\displaystyle Ly=0}$.

The general solution can be written as a linear combination of ${\displaystyle {\tilde {y}}_{1}}$ and ${\displaystyle {\tilde {y}}_{2}}$ just as well as linear combinations of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$.

We shall review the special case where ${\displaystyle \gamma =1-3=-2}$ that was considered in the last section. If we ``insist" ${\displaystyle a_{3}(c)=const.}$, then the recurrence relations yield

${\displaystyle a_{2}=a_{3}{\frac {c(3+c)}{(2+\alpha +c)(2+\beta +c)}},}$ ${\displaystyle a_{1}=a_{3}{\frac {c(2+c)(3+c)(c-1)}{(1+\alpha +c)(2+\alpha +c)(1+\beta +c)(2+\beta +c)}},}$

and

${\displaystyle a_{0}=a_{3}{\frac {c(1+c)(2+c)(3+c)(c-1)(c-2)}{(\alpha +c)_{3}(\beta +c)_{3}}}=b_{0}ch(c).}$

These three coefficients are all zero at ${\displaystyle c=0}$ as expected. We have three terms involved in ${\displaystyle y_{2}(x)}$ by taking the partial derivative w.r.t ${\displaystyle c}$, we denote the sum of the three terms involving these coefficients as ${\displaystyle S_{3}}$ where

${\displaystyle S_{3}=\left[{\frac {\partial }{\partial c}}\left(a_{0}(c)x^{c}+a_{1}(c)x^{c+1}+a_{2}(c)x^{c+2}\right)\right]_{c=0},}$ ${\displaystyle =a_{3}\left[{\frac {3\times 2\times 1(-2)\times (-1)}{(\alpha )_{3}(\beta )_{3}}}x^{3-3}+{\frac {3\times 2\times (-1)}{(\alpha +1)(\alpha +2)(\beta +1)(\beta +2)}}x^{3-2}+{\frac {3}{(\alpha +2)(\beta +2)_{1}}}x^{3-1}\right].}$