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How does one find the solution of

$\dfrac{dy}{dx}\left( 1-\left( 1-t\right) x-x^{2}\right) -\left( 1+h\left( 1+t\right) +x\right) y=0\quad ?$

where $h$ is an integer constant and $t$ is a real constant between $0$ and $1$.

$($ In Roger Apéry, Interpolations de Fractions Continues et Irrationalité de certaines Constantes, Bull. section des sciences du C.T.H.S., n.º3, p.37-53, the solution is

$y=(1-x)^{-1-h}(1+tx)^{h}.)$

Note: The sequence $(v_{h,n})$ in $y=f_{h}(x)=\displaystyle\sum_{n\ge 0}v_{h,n}x^n$ satisfies a recurrence related to $\log (1+t)$.


Added: Copy of the original with the equation and solution

alt text


Addendum 2: I transcribe the comment in the 1st answer: "the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation."

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    @Qiaochu Yuan: There is a typo in the original.2010-10-06

1 Answers 1

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If the given solution is correct then the posted differential equation is wrong. Instead it should be as follows, with the corrected terms underlined:

$y^{\prime }\ \left( 1-\left( 1-t\right) x - \underline{tx^{2}}\right) -\left( 1+h\left( 1+t\right) + \underline{tx}\right)\ y\ =\ 0$

which of course is trivially integrable since

$ \frac{y'}y\ =\ \frac{1+h}{1-x}\ +\: \frac{ht}{1+tx} $

Update: the corrected differential equation above agrees with the recurrence in your excerpt so there is clearly a typo in the printed differential equation.

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    @Americo: Yes, see my update.2010-10-04