# Solving a differential equation related to $\log (1+t)$

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."

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In the standard way? The variables are separated. – Qiaochu Yuan Oct 4 '10 at 12:58
What is $y'$? Is it $dy/dx$ or $dy/dt$, or what? And $h$ is just a constant? – Hans Lundmark Oct 4 '10 at 13:14
In your note, $f$ has $n$ as a subindex, but $n$ is also the index of the following sum... one of the two is wrong. – Mariano Suárez-Alvarez Oct 4 '10 at 13:31
To find a solution, you can propose one of the form $\sum_{n\geq0}a_nx^n$, put it in the equation and find what conditions the coefficients $a_n$ have to satisfy: you'll get a recurrence equation. – Mariano Suárez-Alvarez Oct 4 '10 at 13:32
The equation as written does not have the solution given (assuming h and t are parameters, which is what they look like). For instance, if h=0 the equation still depends essentially on t, whereas the putative solution does not. – whuber Oct 4 '10 at 14:10

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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Is there a typo in the original? I added a copy of the original. – Américo Tavares Oct 4 '10 at 15:38
@Américo : The typo is likely in the differential equation itself, because you can easily check that the coefficients of the MacLaurin series for the given solution y satisfy the recurrence. In other words, of the three items in the text--the recurrence, the differential equation, and its solution--the first and last are consistent with each other but inconsistent with the middle one. – whuber Oct 4 '10 at 17:05
Thanks for your clarification. +1 (one cannot repeat the vote) for that! – Américo Tavares Oct 4 '10 at 17:21
@Americo: Yes, see my update. – Bill Dubuque Oct 4 '10 at 17:25