Relationship between the Picard-Fuchs differential equation and the hypergeometric differential equation Consider the Picard-Fuchs differential equation:
$$\frac{d^2 \Omega}{dJ^2} + \frac{1}{J} \frac{d\Omega}{dJ} + \frac{31J - 4}{144J^2 (1 - J)^2} \Omega = 0.$$
The author of this article claims that the solutions to this equation are those to the hypergeometric differential equation
$$J(1 - J) \frac{d^2 \Omega}{dJ^2} + [c - (a + b + 1)J] \frac{d\Omega}{dJ} - ab\Omega = 0$$
with $a = 11/12$, $b = 11/12$, $c = 4/3$, multiplied by $J^{1/6} (1 - J)^{3/4}$.
My question is: how does one know that? How can one look the Picard-Fuchs differential equation above and know that its solutions are related to the solutions of the hypergeometric differential equation? Is it because the Picard-Fuchs differential equation is related to Riemann's differential equation?
 A: Let $\Omega=J^p(1-J)^qX$ ,
Then $\dfrac{d\Omega}{dJ}=J^p(1-J)^q\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)X$
$\dfrac{d^2\Omega}{dJ^2}=J^p(1-J)^q\dfrac{d^2X}{dJ^2}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X=J^p(1-J)^q\dfrac{d^2X}{dJ^2}+2J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X$
$\therefore J^p(1-J)^q\dfrac{d^2X}{dJ^2}+2J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X+\dfrac{1}{J}\left(J^p(1-J)^q\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)X\right)+\dfrac{31J-4}{144J^2(1-J)^2}J^p(1-J)^qX=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X+\dfrac{1}{J}\dfrac{dX}{dJ}+\left(\dfrac{p}{J^2}-\dfrac{q}{J(1-J)}\right)X-\left(\dfrac{1}{36J^2}-\dfrac{23}{144J}-\dfrac{23}{144(1-J)}-\dfrac{3}{16(1-J)^2}\right)X=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p+1}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{p^2}{J^2}-\dfrac{(2p+1)q}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X-\left(\dfrac{1}{36J^2}-\dfrac{23}{144J(1-J)}-\dfrac{3}{16(1-J)^2}\right)X=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p+1}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{36p^2-1}{36J^2}-\dfrac{144(2p+1)q-23}{144J(1-J)}+\dfrac{16q(q-1)+3}{16(1-J)^2}\right)X=0$
Choose $p=\dfrac{1}{6}$ and $q=\dfrac{3}{4}$ , the ODE becomes
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{4}{3J}-\dfrac{3}{2(1-J)}\right)\dfrac{dX}{dJ}-\dfrac{121}{144J(1-J)}X=0$
$J(1-J)\dfrac{d^2X}{dJ^2}+\left(\dfrac{4}{3}-\dfrac{17}{6}J\right)\dfrac{dX}{dJ}-\dfrac{121}{144}X=0$
