In the book "Theorie der gewöhnlichen Differentialgleichungen" by Bieberbach, page 240, there is a solution to the hypergeometric differential equation $z(z1)w^{\prime \prime}+(2z1)w^{\prime}+\frac{w}{4}=0$. This equation corresponds to the hypergeometric function $F(\frac{1}{2},\frac{1}{2},1;z)$. In general, a basis for the solution of the hypergeometric differential equation $$z(z1)w^{\prime \prime}+(z(1+\alpha+\beta)\gamma )w^{\prime}+\alpha \beta w=0$$ is given by $F(\alpha, \beta, \gamma;z)$ and $z^{1\gamma}F(\alpha\gamma+1, \beta\gamma+1,2\gamma; z)$. However, when $\alpha = \beta=\frac{1}{2}$ and $\gamma=1$, these two solutions coincide. In order to remedy this, the author considers for the first basis element the function $w_1=F(\frac{1}{2},\frac{1}{2},1;z)$ and for the second $w_2=F_1(\frac{1}{2},\frac{1}{2},1;z)+F(\frac{1}{2},\frac{1}{2},1;z) \log z$, where $F_1(\frac{1}{2},\frac{1}{2},1;z)=(\frac{\partial F(\alpha, \beta, \gamma;z)}{\partial \alpha}+\frac{\partial F(\alpha, \beta, \gamma;z)}{\partial \beta}+2\frac{\partial F(\alpha, \beta, \gamma;z)}{\partial \gamma})_{\alpha=\frac{1}{2},\beta=\frac{1}{2}, \gamma=1}$. Of course one can check that this is really a solution, but my question is from where, exactly, comes the operator $\frac{\partial}{\partial \alpha}+\frac{\partial}{\partial \beta}+2\frac{\partial}{\partial \gamma}$? How one should get this idea?
1 Answer
Cannot be submitted as a comment. Your question should be adjusted in view of the output of the Maple 2017.3 command
dsolve(z*(z1)*((D@@2)(w))(z)+(2*z1)*(D(w))(z)+2*w(z) = 0);
$$w \left( z \right) ={\it \_C1}\, {\mbox{$_2$F$_1$}(1/2i/2\sqrt {7},1/2i/2\sqrt {7};\,i\sqrt {7}+1;\,{z}^{1})} {z}^{1/2+i/2\sqrt {7}}+{\it \_C2}\, {\mbox{$_2$F$_1$}(1/2+i/2\sqrt {7},1/2+i/2\sqrt {7};\,i\sqrt {7}+1;\,{z}^{1})} {z}^{1/2i/2\sqrt {7}} $$ as well as the output of the Mathematica 11.2.0.0 command
DSolve[z*(z  1)*w''[z] + (2 z  1)*w'[z] + 2*w[z] == 0, w[z], z]
$$\left\{\left\{w(z)\to c_1 P_{\frac{1}{2} i \left(i+\sqrt{7}\right)}(2 z1)+c_2 Q_{\frac{1}{2} i \left(i+\sqrt{7}\right)}(2 z1)\right\}\right\} $$

$\begingroup$ @ user64494 : Sorry, but I do not understand what you mean! Could you be more precise? What are $C_1$ and $C_2$? $\endgroup$– CyrusCommented Nov 30, 2017 at 15:15

$\begingroup$ @Cyrus: The outputs are not of the form $ F(\frac{1}{2},\frac{1}{2},1;z)$ stated by you. As usualy, $C\,(c)$ stand for constants. $\endgroup$ Commented Nov 30, 2017 at 15:17

$\begingroup$ @ user64494: Yes! you are right, I edited the question, the last coefficeitn should be $\frac{1}{4}$ not $2$! $\endgroup$– CyrusCommented Nov 30, 2017 at 15:21

$\begingroup$ @Cyrus: In the case $+\frac 1 4 w(z)$ Maple outputs $$w \left( z \right) ={\it \_C1}\,{\it EllipticK} \left( \sqrt {z} \right) +{\it \_C2}\,{\it EllipticCK} \left( \sqrt {z} \right) $$. The executed code on demand. $\endgroup$ Commented Nov 30, 2017 at 15:23

$\begingroup$ @Cyrus: Also Mathematica performs $$ \left\{\left\{w(z)\to \frac{2 c_1 K\left(\frac{1}{2} (22 z)\right)}{\pi }+c_2 Q_{\frac{1}{2}}(2 z1)\right\}\right\}$$ in this case. $\endgroup$ Commented Nov 30, 2017 at 15:32