TITLE: Solving a differential equation related to $\log (1+t)$
QUESTION [5 upvotes]: 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


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."

REPLY [7 votes]: 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.