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\begin{document}
\MTversion{lmvtt}
Let $(X,Y)$ be two functions of a variable $a$. If they obey  the differential system $(VI_{\nu,n})$:
\begin{align*}  a\frac{d}{da} X &= \nu
  X - (1 - X^2)\frac{2n a}{1 - a^2}\frac{aX+Y}{1+a XY} \\
a\frac{d}{da} Y &= -(\nu+1) Y + (1-Y^2)\frac{2n a}{1 - a^2}\frac{X+aY}{1+a XY}
\end{align*}
then the quantity $q=a\frac{aX+Y}{X+aY}$ satisfies as function of $b=a^2$ the $P_{VI}$
differential equation:
\begin{equation*}
\begin{split}
  \frac{d^2 q}{db^2} = \frac12\left\{\frac1q+\frac1{q-1}
  +\frac1{q-b}\right\}\left(\frac{dq}{db}\right)^2 - \left\{\frac1b+\frac1{b-1}
  +\frac1{q-b}\right\}\frac{dq}{db}\\+\frac{q(q-1)(q-b)}{b^2(b-1)^2}\left\{\alpha+\frac{\beta
    b}{q^2} + \frac{\gamma (b-1)}{(q-1)^2}+\frac{\delta
    b(b-1)}{(q-b)^2}\right\}
\end{split}
\end{equation*}
with parameters $(\alpha,\beta,\gamma,\delta) = (\frac{(\nu+n)^2}2,
\frac{-(\nu+n+1)^2}2, \frac{n^2}2, \frac{1 - n^2}2)$.
\end{document}