\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011(2011), No. 02, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/02\hfil Differential recurrence equations]
{Initial-value problems for first-order differential recurrence 
equations with auto-convolution}

\author[M. C\^irnu\hfil EJDE-2011/02\hfilneg]
{Mircea C\^irnu}

\address{Mircea C\^irnu \newline
Department of Mathematics III, Faculty of Applied Sciences,
University Politehnica of Bucharest, Romania}
\email{cirnumircea@yahoo.com}

\thanks{Submitted September 10, 2010. Published January 4, 2011.}
\subjclass[2000]{11B37, 34A12}
\keywords{Differential recurrence equations; discrete auto-convolution;
\hfill\break\indent algebraic recurrence equations}

\begin{abstract}
 A differential recurrence equation consists of a sequence  of
 differential equations, from which must be determined by
 recurrence a sequence of unknown functions. In this article, we
 solve two initial-value problems for some new types of nonlinear
 (quadratic) first order homogeneous differential recurrence
 equations, namely with discrete auto-convolution and  with
 combinatorial auto-convolution of the unknown functions. In
 both problems, all initial values  form a geometric progression,
 but in the second problem the first initial value is exempted 
 and has a prescribed form.
 Some preliminary results showing the importance of the
 initial conditions are obtained by reducing the differential
 recurrence equations to algebraic type.
 Final results about solving the considered initial
 value problems, are shown by mathematical induction. However,
 they can also be shown by changing the unknown
 functions, or by the generating function method.
 So in a remark, we give a proof of the first theorem by the
 generating  function method.
 Different cases of first order differential
 recurrence  equations and their solutions are presented,
 including those from a previous work. Applications of the
 equations considered here will be given in subsequent articles.
\end{abstract}


\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
%\allowdisplaybreaks


\section{Introduction}

We consider \emph{first-order differential recurrence equations} of the
form
$$
 G_n\big(x_n'(t) \big), x_n(t), x_{n-1}(t), \dots, x_0(t)\big) = 0,
  \quad n=0,1,2, \dots,
$$
with unknowns  $x_0(t),  x_1(t),  \dots,  x_n(t), \dots $,
complex-valued differentiable functions defined on an open
interval $I$ of  real numbers, the functions $G_n$  being given.
For $t_0\in I $, is called \emph{Cauchy initial-value  problem}
for such an equation, the determination of  its solutions
$x_n(t)$, with given initial values $x_n(t_0)$,  $n =
0,1,2,\dots$.

In this article we solve some initial-value problems for
\emph{first order homogeneous differential recurrence equations
with (discrete) auto-convolution}
\begin{equation} \label{e1}
x_n'(t) = a(t) \sum_{k=0}^n x_k(t) x_{n-k}(t),\quad
 \forall t\in I,\; n = 0,1,2,\dots,
\end{equation}
and \emph{with combinatorial auto-convolution}
\begin{equation} \label{e2}
y_n'(t) = a(t) \sum_{k=0}^n \binom{n}{k}
 y_k(t) y_{n-k}(t), \quad \forall t\in I,\; n=0,1,2,\dots,
\end{equation}
where $a(t)\not\equiv 0$ is a given integrable function.

The second type of equation reduces to the first by change
of unknown functions $ y_n(t) = n! x_n(t)$, $n = 0,1,2,\dots$.

Equations \eqref{e1} and \eqref{e2} are considered for the first time, 
they being different from those verified by special functions, 
from de Branges equation (see \cite{k1}) or other types of differential 
recurrence equations, studied so far (see, for example, \cite{d1,f1}).

The discret convolution or Cauchy product of numerical sequences 
and its inverse-discrete deconvolution (see \cite{c5} for these definitions) 
were formely used by the author in a series of papers
\cite{c1,c2,c3} for solving numerical difference, 
differential and polynomial equations. 

In the following we denote
\begin{equation} \label{e3}
 A(t) = \int a(t) {\rm d} t,\quad
 B(t) = 1 + x_0(t) A(t_0) - x_0(t_0) A(t)\neq  0, \quad
 \forall t\in I  \,.
\end{equation}
Obviously,
\begin{equation}
\label{e4}
B(t_0) = 1, \quad B'(t) = - x_0(t_0) a(t), \quad
 \forall t\in I .
\end{equation}

\section{Algebraic recurrence equation  with auto-convolution}

\begin{lemma} \label{lem1}
Let $b_n\neq  0$, $n=0,1,2,\dots $, be a sequence of real
or complex numbers. Following statements are equivalent:
\begin{itemize}
\item[(i)]  %\label{e5}
$(n+1)b_0b_n = \sum_{k=0}^n b_k b_{n-k}$, $n=0,1,2,\dots$;

\item[(ii)]  
$(n-1)b_0b_n = \sum_{k=1}^{n-1} b_k b_{n-k}$, $n=2,3,\dots$;

\item[(iii)] $b_n =\frac{b_1}{b_0} b_{n-1}$, $n=2,3,\dots$;

\item[(iv)] The numbers $b_n$, $n=0,1,2,\dots $, are in geometric
progression, namely
\begin{equation} \label{e6}
b_n =b_0 \Big[\frac{b_1}{b_0}\Big]^n
= \frac{b_1^n}{b_0^{n-1}}, \quad n=0,1,2,\dots \,.
\end{equation}
\end{itemize}
\end{lemma}


\begin{proof}
 $(i) \Leftrightarrow (ii)$ and  $(iii) \Leftrightarrow (iv)$ are
obvious.

$(iv) \Rightarrow (i)$ If the sequence $b_n$  is given by formula \eqref{e6}, we have
$$
\sum_{k=0}^n b_k b_{n-k} = \sum_{k=0}^n \frac{b_1^k}{b_0^{k-1}}
\frac{b_1^{n-k}}{b_0^{n-k-1}} = \sum_{k=0}^n \frac{b_1^n}{b_0^{n-2}}
= (n+1) \frac{b_1^n}{b_0^{n-2}} = (n+1) b_0 b_n,
$$
$n=0,1,2,\dots$,
hence the sequence $b_n$ satisfies (i) in Lemma \ref{lem1}.

$(ii)\Rightarrow (iv)$ (\emph{First proof by induction})
For $n=2$, from (ii) we obtain $b_2=\frac{b_1^2}{b_0}$.
For $n\geq 2$, we suppose that $b_k=\frac{b_1^k}{b_0^{k-1}}$,
for $k=0,1,\dots ,n-1$. Then from (ii) results
\begin{align*}
b_n&=\frac{1}{(n-1)b_0}\sum_{k=1}^{n-1}b_kb_{n-k}
=\frac{1}{(n-1)b_0}\sum_{k=1}^{n-1}\frac{b_1^k}{b_0^{k-1}}
 \frac{b_1^{n-k}}{b_0^{n-k-1}}\\
&=\frac{1}{(n-1)b_0}\sum_{k=1}^{n-1}\frac{b_1^n}{b_0^{n-2}}
=\frac{1}{(n-1)b_0}(n-1)\frac{b_1^n}{b_0^{n-2}}
=\frac{b_1^n}{b_0^{n-1}}.
\end{align*}
In conformity with induction axiom, formula \eqref{e6} is true
 for every $n=0,1,2,\dots$.

$(i)\Rightarrow (iv)$ (\emph{Second proof by the generating function
method, \cite{w1}}) 
Denoting $G(z) = \sum_{n=0}^\infty b_n z^n $
generating function of the numerical sequence $b_n$, given by a
formal series, item (i) in Lemma \ref{lem1} gives a differential equation
with the successive forms
\begin{gather*}
 b_{_0}\big[z G(z) \big]' = G^2(z),\quad
 b_0 z G'(z) + b_0 G(z) = G^2(z), \\
\frac{b_0G'(z)}{G(z)\big[G(z) - b_0 \big]} = \frac{1}{z},\quad
\frac{G'(z)}{G(z) - b_0} - \frac{G'(z)}{G(z)} = \frac{1}{z} .
\end{gather*}
 Integrating results  in $\ln\big| \frac{G(z) - b_0}{G(z)} \big|
= \ln \big|\widetilde k z \big|$,
hence $\frac{G(z)-b_0}{G(z)} = k z $, so
$G(z) =\frac{b_0}{1-kz} = b_0 \sum_{k=0}^\infty k^n z^n $, where
$\widetilde k$ and  $k = \pm  \widetilde k $ are arbitrary constants.
 It results $b_n = b_0 k^n$, $n=0,1,2,\dots$. For $n=1 $,
we have $b_1 = b_0 k $, hence $k = \frac{b_1}{b_0}$ and
$b_n = \frac{b_1^n}{b_0^{n-1}}$, $n=0,1,2,\dots$.
\end{proof}


\begin{corollary} \label{coro1}
 For $a \neq  0$  a given number, the sequence $b_n \neq  0$,
$n=2,3,\dots  $, is solution of the equation
\begin{equation} \label{e7}
 (n-1)b_0 b_n = a \sum_{k=1}^{n-1} b_k b_{n-k}, \; n=2,3,\dots,
\end{equation}
if and only if
\begin{equation} \label{e8}
b_n = \frac{a^{n-1}b_1^n}{b_0^{n-1}}, \; n = 1,2,\dots \,.
\end{equation}
\end{corollary}

\begin{proof}
Making the change of variables $b_0 = a \widetilde b_0$,
$b_n = \widetilde b_n$, $n =1,2,\dots $, the equation \eqref{e7}
 reduces to
$(n-1)\widetilde b_0 \widetilde b_n = \sum_{k=1}^{n-1}
\widetilde b_k \widetilde b_{n-k}$, $n=2,3,\dots$.
In conformity with Lemma \ref{lem1}, we have
$b_n = \widetilde b_n = \frac{\widetilde b_1^n}{\widetilde b_0^{n-1}}
= \frac{a^{n-1} b_1^n}{b_0^{n-1}}$, $n=1,2,\dots$.
\end{proof}


\begin{corollary}  \label{coro2}
 The sequence $b_n \neq  0$, $n=0,1,2, \dots  $, is solution
of the equation
$$
 (n+1) b_0 b_n = \sum_{k=0}^n \binom{n}{k}
 b_k b_{n-k}, \quad n=0,1,2,\dots,
$$
 if and only if
$$
 b_n = \frac{n! b_1^n}{b_0^{n-1}},\quad n=0,1,2,\dots  \,.
$$
\end{corollary}

\begin{proof}   Making the change of variables
  $b_n = n! \widetilde b_n$, $n=0,1,2,\dots  $, the considered
equation reduces to $ (n+1) \widetilde b_0 \widetilde b_n
 = \sum_{k=0}^n \widetilde b_k \widetilde b_{n-k}$, $n=0,1,2,\dots $,
and Corollary \ref{coro2} follows from Lemma \ref{lem1}.
\end{proof}

\noindent\textbf{Remark.}
 Other results, related to those of Lemma \ref{lem1} and its 
 Corollary \ref{coro2}, were given in 
 \cite[Theorem 1.1 and Corollary 1.2]{c4} . 

\section{First initial-value problem}

\begin{lemma} \label{lem2}
The functions
\begin{equation} \label{e9}
 x_n(t) = \frac{x_n(t_0)}{B^{n+1}(t)},\quad
 \forall t\in I,\; n=0,1,2,\dots  \,,
\end{equation}
are solutions of  \eqref{e1} if and only if their initial
values $x_n(t_0)\neq  0$ $n=0,1,2,\dots $, are in geometric
progression.
\end{lemma}

\begin{proof}
If the functions $x_n(t)$ are given by formula \eqref{e9},
the equation \eqref{e1} takes successively the form
\begin{gather*}
 -\frac{(n+1)x_n(t_0)B'(t)}{B^{n+2}(t)}
=a(t)\sum_{k=0}^{n}\frac{x_k(t_0)}{B^{k+1}(t)}
\frac{x_{n-k}(t_0)}{B^{n-k+1}(t)},\\
\frac{(n+1)x_n(t_0)x_0(t_0)a(t)}{B^{n+2}(t)}
=\frac{a(t)}{B^{n+2}(t)}\sum_{k=0}^{n}x_k(t_0)x_{n-k}(t_0),\\
(n+1)x_0(t_0)x_n(t_0)=\sum_{k=0}^{n}x_k(t_0)x_{n-k}(t_0)\,.
\end{gather*}
In conformity with Lemma \ref{lem1}, the last equality is true if and only
if the values $x_n(t_0)$, $n=0,1,2,\dots $, are in geometric
progression.
\end{proof}

\begin{theorem} \label{thm1}
The differentiable functions $ x_n(t)$, $n=0,1,2,\dots$,
with initial values $ x_n(t_0) \neq  0$, $n=0,1,2, \dots$,
in  geometric progression, are solutions of  \eqref{e1} if and only
if they are given by \eqref{e9}.
\end{theorem}

\begin{proof}
We suppose that the functions $x_n(t)$  are solutions of  \eqref{e1}
and have their initial values $x_n(t_0)$, $n=0,1,2,\dots $,
are in geometric progression, hence
\begin{equation} \label{e10}
x_n(t_0)=x_0(t_0)\big[\frac{x_1(t_0)}{x_0(t_0)}\big]^n
=\frac{x_1^{n}(t_0)}{x_0^{n-1}(t_0)},\;n=0,1,2,\dots \,.
\end{equation}
For $n=0$, the equation \eqref{e1} has the form
$ x_0'(t) =a(t) x_0^2(t)$, hence $\frac{x_0'(t)}{x_0^2(t)} = a(t)$.
By integration, we obtain  $-\frac{1}{x_0(t)} = A(t) + C_0 $,
hence $ x_0(t) =- \frac{1}{A(t) + C_0} $, where $C_0$  is an arbitrary
constant.  For $t=t_0$, it results in $C_0 = -\frac{1+x_0(t_0)
A(t_0)}{x_0(t_0)}  $; therefore,
$$
x_0(t) = - \frac{1}{A(t)-\frac{1+x_0(t_0)A(t_0)}{x_0(t_0)}}
= \frac{x_0(t_0)}{1+x_0(t_0)A(t_0)-x_0(t_0)A(t)}\,.
$$
In conformity with \eqref{e3},
\begin{equation} \label{e11}
x_0(t)=\frac{x_0(t_0)}{B(t)} \,.
\end{equation}
For $n=1$, the equation \eqref{e1} has the form
$ x_1'(t) = 2a(t) x_0(t) x_1(t)$, hence
$$
\frac{x_1'(t)}{x_1(t)} =-2\frac{B'(t)}{B(t)}.
$$
By integration, we obtain $x_1(t)=\frac{C_1}{B^2(t)}$,
with $C_1$ an arbitrary constant. For $t=t_0$, it results
$C_1=x_1(t_0)$, hence
\begin{equation} \label{e12}
 x_1(t) = \frac{x_1(t_0)}{B^2(t)} \,.
\end{equation}
 For $n \geq 2$, equation \eqref{e1} has the form
$$
 x_n'(t) = 2a(t)  x_0(t) x_n(t) + a(t)
\sum_{k=1}^{n-1} x_k(t) x_{n-k}(t),
$$
hence, using the relation $a(t)=- \frac{B'(t)}{x_0(t_0)}$,
obtained from \eqref{e4},
$$
 x_n'(t) + 2\frac{B'(t)}{B(t)} x_n(t)
= a(t) \sum_{k=1}^{n-1} x_k(t) x_{n-k}(t),
$$
with general solution
\begin{equation} \label{e13}
\begin{aligned}
x_n(t) 
&= \exp\Big(-2\int\frac{B'(t)}{B(t)}{\rm d}t \Big)
\Big[\int\exp\Big(2\int\frac{B'(t)}{B(t)}{\rm d}t \Big)a(t)
\sum_{k=1}^{n-1} x_n(t) x_{n-k}(t){\rm d}t+C_n \Big] \\
&= \frac{1}{B^2(t)}\Big[C_n - \frac{1}{x_0(t_0)}\int B^2(t) B'(t)
\sum_{k=1}^{n-1} x_k(t) x_{n-k}(t){\rm d}t \Big],
\end{aligned}
\end{equation}
where $C_n $ is an arbitrary constant.

  For $n=2$, from \eqref{e12}  and \eqref{e13}, it results
\begin{equation} \label{e14}
\begin{aligned}
x_2(t)&=\frac{1}{B^2(t)}\Big[ C_2 - \frac{1}{x_0(t_0)}
 \int B^2(t) B'(t) x_1^2(t){\rm d}t \Big] \\
&= \frac{1}{B^2(t)}\Big[C_2-\frac{x_1^2(t_0)}{x_0(t_0)}
  \int \frac{B'(t)}{B^2(t)}{\rm d}t   \Big]
 = \frac{1}{B^2(t)}\Big[C_2 + \frac{x_1^2(t_0)}{x_0(t_0) B(t)} \Big].
\end{aligned}
\end{equation}
  From \eqref{e14} for $t=t_0$ and \eqref{e10} for $n=2$, it
results
$$
x_2(t_0)=C_2+\frac{x_1^2(t_0)}{x_0(t_0)}=\frac{x_1^2(t_0)}{x_0(t_0)},
$$
 hence $C_2 = 0$, and \eqref{e14} becomes
\begin{equation} \label{e15}
 x_2(t) =  \frac{x_1^2(t_0)}{x_0(t_0) B^3(t)}.
\end{equation}

     For $n \geq  2 $ fixed and $k=0,1,2, \dots, n-1$, we suppose that
$$
x_k(t) = \frac{x_1^k(t_0)}{x_0^{k-1} (t_0) B^{k+1}(t)}\,.
$$
Then
\begin{align*}
 \sum_{k=1}^{n-1} x_k(t) x_{n-k}(t)
&= \sum_{k=1}^{n-1} \frac{x_1^k(t_0)}{x_0^{k-1}(t_0) B^{k+1}(t)}
 \frac{x_1^{n-k}(t_0)}{x_0^{n-k-1}(t_0) B^{n-k+1}(t)} \\
& =  \sum_{k=1}^{n-1}  \frac{x_1^n(t_0)}{x_0^{n-2}(t_0) B^{n+2}(t)}
 = \frac{(n-1)x_1^n(t_0)}{x_0^{n-2} (t_0) B^{n+2}(t)}
\end{align*}
and \eqref{e13} becomes
\begin{equation} \label{e16}
\begin{aligned}
  x_n(t)
&= \frac{1}{B^2(t)}\Big[C_n -\frac{(n-1) x_1^n(t_0)}{x_0^{n-1}(t_0)}
  \int\frac{B'(t)}{B^n(t)}{\rm d}t \Big]\\
& =\frac{1}{B^2(t)} \Big[C_n+\frac{x_1^{n}(t_0)}{x_0^{n-1}
 (t_0)B^{n-1}(t)} \Big] \,.
\end{aligned}
\end{equation}
From which, for $t=t_0$ and \eqref{e10}, results
$x_n(t_0)=C_n+\frac{x_1^{n}(t_0)}{x_0^{n-1}(t_0)}
=\frac{x_1^{n}(t_0)}{x_0^{n-1}(t_0)}$, hence $C_n=0$, and \eqref{e16}
 becomes
\begin{equation} \label{e17}
x_n(t)= \frac{x_1^n(t_0) }{x_0^{n-1}(t_0) B^{n+1}(t)}, \quad
n=0,1,2,\dots \,.
\end{equation}
According to induction axiom,  \eqref{e17}
is satisfied for any natural number $n$.
 From \eqref{e10} and \eqref{e17}, it results \eqref{e9}.

 Reciprocally, if the functions $x_n(t)$ are given by \eqref{e9},
with the initial values in geometric progression, then we
have \eqref{e17}, hence
\begin{align*}
x_n'(t)&= - \frac{(n+1)x_1^n(t_0)B'(t)}{x_0^{n-1}(t_0)B^{n+2}(t)}
= \frac{(n+1)x_1^n(t_0)a(t)}{x_0^{n-2}(t_0)B^{n+2}(t)}
=a(t)\sum_{k=0}^{n}\frac{x_1^n(t_0)}{x_0^{n-2}(t_0)B^{n+2}(t)}\\
&=a(t)\sum_{k=0}^{n}\frac{x_1^k(t_0)}{x_0^{k-1}(t_0)B^{k+1}(t)}
 \frac{x_1^{n-k}(t_0)}{x_0^{n-k-1}(t_0)B^{n-k+1}(t)}
=a(t)\sum_{k=0}^{n}x_k(t)x_{n-k}(t);
\end{align*}
therefore, the functions $x_n(t)$ satisfy  \eqref{e1}.
This also results by Lemma \ref{lem2}.
\end{proof}


\noindent\textbf{Remark.}
 Theorem \ref{thm1} can also be demonstrated using the
generating function $ G(t,z) = \sum_{n=0}^\infty x_n(t) z^n$,
of the sequence of functions $x_n(t)$, $n=0,1,2,\dots $, given by
a formal series. Then \eqref{e1} is equivalent to equation
$\frac{\partial}{\partial t} G(t,z) = a(t) G^2(t,z) $, with solution
$G(t,z) = \frac{1}{u(z) - A(t)}$, where $u(z)$ is an arbitrary
function. For $z=0$, we obtain $x_0(t) =G(t,0) = \frac{1}{u(0)
- A(t)}$. Let  $v(z) = u(0) - u(z)$, with $v(0)=0$. Using
geometric series, we have
\begin{align*}
G(t,z) &= \frac{1}{u(0) - A(t) - v(z)} \\
&=\frac{1}{u(0)-A(t)}\frac{1}{1-\frac{v(z)}{u(0)-A(t)}}
 = \sum_{n=0}^\infty \frac{v^n(z)}{\big(u(0) - A(t) \big)^{n+1}}\,.
\end{align*}
The choice $v(z)=Cz$, where $C$ is an arbitrary constant, gives
$$
G(t,z)=\sum_{n=0}^{\infty}\frac{C^n}{\big(u(0)-A(t)\big)^{n+1}}z^n,
$$
hence
\begin{equation} \label{e18}
x_n(t)=\frac{C^n}{\big(u(0)-A(t)\big)^{n+1}},\quad
n=0,1,2,\dots \,.
\end{equation}
For $t=t_0$ and $n=0$, from \eqref{e18} it results
$x_0(t_0)=\frac{1}{u(0)-A(t_0)}$, hence
$u(0)=\frac{1+x_0(t_0)A(t_0)}{x_0(t_0)}$, and
\begin{equation} \label{e19}
u(0)-A(t)=\frac{1+x_0(t_0)A(t_0)}{x_0(t_0)}-A(t)
=\frac{B(t)}{x_0(t_0)}.
\end{equation}
For $t=t_0$ and $n=1$, from \eqref{e18} results
$x_1(t_0)=\frac{C}{\big(u(0)-A(t_0)\big)^2}=Cx_0^2(t_0)$, hence
\begin{equation} \label{e20}
C=\frac{x_1(t_0)}{x_0^2(t_0)}.
\end{equation}
 From \eqref{e18}, \eqref{e19} and \eqref{e20} it results
\begin{equation} \label{e21}
 x_n(t)=\frac{x_1^n(t_0)}{x_0^{2n}(t_0)}
\frac{x_0^{n+1}(t_0)}{B^{n+1}(t)}
=\frac{x_1^n(t_0)}{x_0^{n-1}(t_0)B^{n+1}(t)}, \quad
n=0,1,2,\dots \,.
\end{equation}
For $t=t_0$, from \eqref{e21} we obtain \eqref{e10},
hence  \eqref{e21} takes
the form \eqref{e9} and the initial values $x_n(t_0)$, $n=0,1,2,\dots $,
are in geometric progression. This last statement also follows
by Lemma \ref{lem2}. The reciprocal affirmation results both by direct
calculation as above or by Lemma \ref{lem2}.

\begin{corollary} \label{coro3}
The differentiable functions $y_n(t)$, $n=0,1,2,\dots $, whose
initial values are
$$
y_n(t_0)=\frac{n!y_1^n(t_0)}{y_0^{n-1}(t_0)}, \quad
n=0,1,2,\dots,
$$
are solutions of the differential recurrence equation with
combinatorial auto-\\ convolution \eqref{e2} if and only if they are
given by
$$
y_n(t)=\frac{y_n(t_0)}{B^{n+1}(t)},\quad \forall  t \in I, \;
n=0,1,2,\dots \,.
$$
\end{corollary}

\section{Second initial-value problem}

\begin{lemma} \label{lem3}
If $A(t_0)\neq  0$ and
\begin{equation} \label{e22}
 x_0(t_0) = \frac{x_1^2(t_0)}{x_2(t_0)} - \frac{1}{A(t_0)} \neq  0,
\end{equation}
then the functions
\begin{equation}  \label{e23}
  x_0(t) = \frac{x_0(t_0)}{B(t)}, \quad
  x_n(t) = \frac{x_n(t_0) A^{n-1}(t)}{ A^{n-1}(t_0) B^{n+1}(t)},\quad
 \forall t\in I,\; n=1,2,\dots \,.
\end{equation}
are solutions of  \eqref{e1} if and only if the initial
values $x_n(t_0) \neq  0$, $n=1,2,\dots $, are in geometric
progression.
\end{lemma}

\begin{proof}
If the functions $x_n(t)$ are given by formula \eqref{e23},
then  \eqref{e1} is obviously satisfied for $n=0$ and $n=1$.
For $n=2,3,\dots $, it takes successively the form
\begin{align*}
&\frac{x_n(t_0)}{A^{n-1}(t_0)}\frac{(n-1)A^{n-2}(t)
 A'(t)B^{n+1}(t)-(n+1)B^n(t)B'(t)A^{n-1}(t)}{B^{2n+2}(t)}\\
&=\frac{2a(t)x_0(t_0)x_n(t_0)A^{n-1}(t)}{A^{n-1}(t_0)B^{n+2}(t)}\\
&\quad +a(t)\sum_{k=1}^{n-1}\frac{x_k(t_0)A^{k-1}(t)}{A^{k-1}
 (t_0)B^{k+1}(t)}\frac{x_{n-k}(t_0)A^{n-k-1}(t)}{A^{n-k-1}
 (t_0)B^{n-k+1}(t)},
\end{align*}
\begin{align*}
&\frac{x_n(t_0)}{A^{n-1}(t_0)}\frac{(n-1)A^{n-2}(t)
 a(t)B^{n+1}(t)+(n+1)x_0(t_0)B^n(t)a(t)A^{n-1}(t)}{B^{2n+2}(t)}\\
&=\frac{2a(t)x_0(t_0)x_n(t_0)A^{n-1}(t)}{A^{n-1}(t_0)B^{n+2}(t)}
+\frac{a(t)A^{n-2}(t)}{A^{n-2}(t_0)B^{n+2}(t)}
\sum_{k=1}^{n-1}x_k(t_0)x_{n-k}(t_0),
\end{align*}
\begin{align*}
&\frac{x_n(t_0)}{A(t_0)}\big[(n-1)B(t)+(n+1)x_0(t_0)A(t)\big]\\
&=\frac{2x_0(t_0)x_n(t_0)A(t)}{A(t_0)}
 +\sum_{k=1}^{n-1}x_k(t_0)x_{n-k}(t_0),
\end{align*}
\[
(n-1)\frac{x_n(t_0)}{A(t_0)}\big[B(t)+x_0(t_0)A(t)\big]
=\sum_{k=1}^{n-1}x_k(t_0)x_{n-k}(t_0).
\]
 From \eqref{e4} and \eqref{e22} it results
\begin{equation} \label{e24}
 B(t) + x_0(t_0) A(t) =  1 + x_0(t_0) A(t_0)
= \frac{x_1^2(t_0) A(t_0)}{x_2(t_0)},\quad \forall t\in I;
\end{equation}
 therefore,  \eqref{e1} is still equivalent to the relations
\begin{gather*}
 (n-1)\frac{x_n(t_0)}{A(t_0)}\frac{x_1^2(t_0)A(t_0)}{x_2(t_0)}
=\sum_{k=1}^{n-1}x_k(t_0)x_{n-k}(t_0),\\
(n-1)x_0(t_0)x_n(t_0)=\frac{x_0(t_0)x_2(t_0)}{x_1^2(t_0)}
 \sum_{k=1}^{n-1}x_k(t_0)x_{n-k}(t_0).
\end{gather*}
By Corollary \ref{coro1}, for $b_n=x_n(t_0)$,
$n=0,1,2,\dots $ and $a=\frac{x_0(t_0)x_2(t_0)}{x_1^2(t_0)}$,
the last equality is equivalent to the relation
$$
x_n(t_0)=\frac{x_0^{n-1}(t_0)x_2^{n-1}(t_0)}{x_1^{2n-2}(t_0)}
\frac{x_1^n(t_0)}{x_0^{n-1}(t_0)}
=\frac{x_2^{n-1}(t_0)}{x_1^{n-2}(t_0)}=x_1(t_0)
\Big[\frac{x_2(t_0)}{x_1(t_0)}\Big]^{n-1},
$$
where $n=1,2,\dots$;
hence with the fact that the initial values $x_n(t_0)$, $n=1,2,\dots $,
are in geometric progression.
\end{proof}

\begin{theorem} \label{thm2}
If $A(t_0) \neq  0 $,  while $x_0(t_0)$ is given by  \eqref{e22},
then the differentiable functions $ x_n(t)$, $n=0,1,2,\dots $, with
initial values $x_n(t_0)$, $n=1,2,\dots $, in geometric progression,
are solutions of  \eqref{e1} if and only if they are given
 by  \eqref{e23}.
\end{theorem}

\begin{proof}
We suppose that the functions $x_n(t)$, $n=0,1,2,\dots $,  are
solutions of  \eqref{e1} and the initial values
$x_n(t_0)$, $n=1,2,\dots $, are in geometric progression, hence
\begin{equation} \label{e25}
 x_n(t_0)=x_1(t_0)\Big[\frac{x_2(t_0)}{x_1(t_0)}\Big]^{n-1}
=\frac{x_2^{n-1}(t_0)}{x_1^{n-2}(t_0)},\quad
n=1,2,\dots \,.
\end{equation}
 As was shown in the proof of Theorem \ref{thm1}, the functions
$x_0(t), x_1(t), x_n(t) $  for $n\geq 2$, and $x_2(t)$  are
given by \eqref{e11}, \eqref{e12}, \eqref{e13} and \eqref{e14}.
 From \eqref{e14} and \eqref{e3} we have
\begin{equation} \label{e26}
\begin{aligned}
  x_2(t) &= \frac{C_2 x_0(t_0) B(t) + x_1^2(t_0)}{x_0(t_0) B^3(t)}\\
& =  \frac{C_2 x_0(t_0)\big[1 + x_0(t_0) A(t_0) \big]
 - C_2 x_0^2 (t_0) A(t) + x_1^2(t_0)  }{x_0(t_0) B^3(t)} \,.
\end{aligned}
\end{equation}
    We take
\begin{equation} \label{e27}
 C_2 = - \frac{ x_1^2(t_0)}{x_0(t_0) \big[1 + x_0(t_0) A(t_0) \big]}
=  -  \frac{x_2(t_0)}{x_0(t_0) A(t_0)}\, .
\end{equation}
the above equality resulting from  \eqref{e24},
which in turn resulted from \eqref{e22}.
 From \eqref{e26} and \eqref{e27} it results
\begin{equation}  \label{e28}
 x_2(t) = - \frac{C_2 x_0(t_0) A(t)}{ B^3(t)}
= \frac{x_2(t_0) A(t)}{A(t_0) B^3(t)}\, .
\end{equation}
 From \eqref{e13}, for $n=3$, \eqref{e12} and \eqref{e28}, it results
\begin{equation} \label{e29}
\begin{aligned}
x_3(t)
&= \frac{1}{B^2(t)}\Big[C_3-\frac{2}{x_0(t_0)}
  \int B^2(t) B'(t)x_1(t)x_2(t)dt\Big]\\
& = \frac{1}{B^2(t)}\Big[C_3- \frac{2x_1(t_0)x_2(t_0)}{x_0(t_0)A(t_0)}
\int \frac{A(t)B'(t)} {B^3(t)}dt \Big] \,.
\end{aligned}
\end{equation}
 Using \eqref{e3} and \eqref{e4}, we obtain
$A'(t)=a(t)= - \frac{B'(t)}{x_0(t_0)}$; hence
\begin{equation} \label{e30}
\Big[ \frac{A(t)}{B(t)}\Big]'= \frac{A'(t)B(t)-A(t)B'(t)}{B^2(t)}
= - \frac{[B(t)+x_0(t_0)A(t)]B'(t)}{x_0(t_0)B^2(t)}\,.
\end{equation}
Using again  \eqref{e24}, formula \eqref{e30} takes the form
$$
\Big[ \frac{A(t)}{B(t)} \Big]'
= - \frac{x_1^2(t_0)A(t_0)B'(t)}{x_0(t_0)x_2(t_0)B^2(t)},
$$
from which we obtain
\begin{equation} \label{e31}
\frac{B'(t)}{B^2(t)}= - \frac{x_0(t_0)x_2(t_0)}{x_1^2(t_0)A(t_0)}
\Big[ \frac{A(t)}{B(t)} \Big]'\,.
\end{equation}
 From \eqref{e29} and \eqref{e31}, it results
\begin{equation}  \label{e32}
\begin{aligned}
x_3(t)
&=\frac{1}{B^2(t)}\Big[C_3+ \frac {2x_2^2(t_0)}{x_1(t_0)A^2(t_0)}
 \int \frac{A(t)}{B(t)} \Big[ \frac{A(t)}{B(t)} \Big]'dt\Big]\\
&=\frac{1}{B^2(t)}\Big[C_3+\frac{x_2^2(t_0)A^2(t)}{x_1(t_0)A^2(t_0)
 B^2(t)}\Big] \,.
\end{aligned}
\end{equation}
 From which for $t=t_0$ and \eqref{e25} for $n=3$, it
results
$$
x_3(t_0)=C_3+\frac{x_2^2(t_0)}{x_1(t_0)}
=\frac{x_2^2(t_0)}{x_1(t_0)},
$$
 hence $C_3=0$ and formula \eqref{e32} becomes
\begin{equation} \label{e33}
x_3(t)=\frac{x_2^2(t_0)A^2(t)}{x_1(t_0)A^2(t_0)B^4(t)}\,.
\end{equation}
  For $n\geq 3$ fixed and $k=1,2,\dots,n-1$, we suppose that
\begin{equation} \label{e34}
 x_k(t) =  \frac{x_2^{k-1}(t_0) A^{k-1}(t)}{x_1^{k-2}(t_0) A^{k-1}(t_0)
B^{k+1}(t)}.
\end{equation}
    Then
\begin{equation}
\begin{aligned}
& \sum_{k=1}^{n-1} x_k(t) x_{n-k}(t) \\
&= \sum_{k=1}^{n-1} \frac{ x_2^{k-1}(t_0) A^{k-1}(t)}{x_1^{k-2}(t_0)
 A^{k-1}(t_0) B^{k+1}(t)}
 \frac{x_2^{n-k-1}(t_0) A^{n-k-1}(t)}{ x_1^{n-k-2}(t_{_0})
 A^{n-k-1}(t_0) B^{n-k+1}(t)}\\
&= \frac{(n-1) x_2^{n-2}(t_0) A^{n-2}(t)}{ x_1^{n-4}(t_0)
A^{n-2}(t_0)B^{n+2}(t)}\, .
\end{aligned} \label{e35}
\end{equation}
 From \eqref{e13} and \eqref{e35} it results
\begin{equation} \label{e36}
x_n(t) =  \frac{1}{B^2(t)}\Big[C_n-\frac{(n-1) x_2^{n-2}
(t_{_0})}{x_0(t_0)  x_1^{n-4}(t_0) A^{n-2}(t_0)}
\int \frac{A^{n-2}(t)B'(t)}{B^n(t)} {\rm d}t \Big]\,.
\end{equation}
 From \eqref{e31} and \eqref{e36}, it results
\begin{equation} \label{e37}
\begin{aligned}
x_n(t) &= \frac{1}{B^2(t)} \Big[C_n+\frac{(n-1) x_2^{n-1}
 (t_0)}{x_1^{n-2}(t_0) A^{n-1}(t_0)}
 \int \Big[\frac{A(t)}{B(t)}\Big]^{n-2} \Big[\frac{A(t)}{B(t)}\Big]'
{\rm d}t\Big] \\
&=\frac{1}{B^2(t)}\Big[C_n+\frac{x_2^{n-1}
(t_0)A^{n-1}(t)}{x_1^{n-2}(t_0)A^{n-1}(t_0) B^{n-1}(t)}\Big] \, .
\end{aligned}
\end{equation}
 From this euqality for $t=t_0$ and \eqref{e25}, it
results $x_n(t_0)=C_n+\frac{x_2^{n-1}(t_0)}{x_1^{n-2}(t_0)}
=\frac{x_2^{n-1}(t_0)}{x_1^{n-2}(t_0)}$, hence $C_n=0$ and
formula \eqref{e37} becomes
\begin{equation} \label{e38}
x_n(t)=\frac{x_2^{n-1}(t_0)A^{n-1}(t)}{x_1^{n-2}(t_0)A^{n-1}
(t_0)B^{n+1}(t)}.
\end{equation}
In conformity with induction axiom,  formula \eqref{e38} is
satisfied for every natural number $n \geq 1$.
 For $t=t_0$, from \eqref{e38} it results
$x_n(t_0)=\frac{x_2^{n-1}(t_0)}{x_1^{n-2}(t_0)}$,
 hence  \eqref{e38} reduces to second formula \eqref{e23}.

   Reciprocally, if the functions $x_n(t)$, $n=0,1,2,\dots $,
are given by \eqref{e23}, and the initial values $x_n(t_0)$,
$n=1,2,\dots $, are in geometric progression, then we have
$x_0(t)=\frac{x_0(t_0)}{B(t)}$ and
$$
x_n(t)=\frac{x_n(t_0)A^{n-1}(t)}{A^{n-1}(t_0)B^{n+1}(t)}
=\frac{x_2^{n-1}(t_0)A^{n-1}(t)}{x_1^{n-2}(t_0)A^{n-1}(t_0)B^{n+1}(t)},
$$
 hence for $n=1,2,\dots $, using \eqref{e24}, we have
\begin{align*}
& x_n'(t)-2a(t)x_0(t)x_n(t)\\
&=\frac{x_n(t_0)\big[(n-1)A^{n-2}(t)A'(t)B^{n+1}(t)
 -(n+1)A^{n-1}(t)B^n(t)B'(t)\big]}{A^{n-1}(t_0)B^{2n+2}(t)}\\
&\quad -\frac{2x_0(t_0)x_n(t_0)a(t)A^{n-1}(t)}{A^{n-1}(t_0)B^{n+2}(t)}\\
&=\frac{x_n(t_0)a(t)A^{n-2}(t)}{A^{n-1}(t_0)B^{n+2}(t)}
\big[(n-1)B(t)+(n+1)x_0(t_0)A(t)
 -2x_0(t_0)A(t)\big]\\
&=\frac{(n-1)x_2^{n-1}(t_0)a(t)A^{n-2}(t)}{x_1^{n-2}(t_0)A^{n-1}
(t_0)B^{n+2}(t)}\big[B(t)+x_0(t_0)A(t)\big] \\
&=\frac{(n-1)x_2^{n-2}(t_0)a(t)A^{n-2}(t)}{x_1^{n-4}(t_0)A^{n-2}
(t_0)B^{n+2}(t)}\\
&=a(t)\sum_{k=1}^{n-1}\frac{x_2^{k-1}(t_0)A^{k-1}(t)}{x_1^{k-2}
(t_0)A^{k-1}(t_0)B^{k+1}(t)} \frac{x_2^{n-k-1}
(t_0)A^{n-k-1}(t)}{x_1^{n-k-2}(t_0)A^{n-k-1}(t_0)B^{n-k+1}(t)}\\
&=a(t)\sum_{k=1}^{n-1}x_k(t)x_{n-k}(t)\,;
\end{align*}
therefore, the functions $x_n(t)$ given by \eqref{e23} satisfy
 \eqref{e1}. This also results by Lemma \ref{lem3}.
\end{proof}


\noindent\textbf{Remark.}
   From \eqref{e22} it results that in hypotheses of the
Theorem \ref{thm2}, those of Theorem \ref{thm1} are not satisfied. The solutions
of the second initial values problem are different from those of
the first problem, because in the proof of Theorem \ref{thm1} all arbitrary
constants $C_n$, $n\geq 2$, that arise in solving the differential
equations are zero, while in the proof of Theorem \ref{thm2}, the constant
$C_2$ is non-zero, given by formula \eqref{e27}.


\begin{corollary}  \label{coro4}
 If $A(t_0)\neq  0$, and
$y_0(t_0)=\frac{y_1^2(t_0)}{y_2(t_0)}-\frac{1}{A(t_0)}\neq  0$,
then the differentiable functions $y_n(t)$, $n=0,1,2,\dots $, with
$y_n(t_0)=\frac{n!y_2^{n-1}(t_0)}{y_1^{n-2}(t_0)}$, $n=1,2,\dots $,
are solutions of the recurrence equation with combinatorial
auto-convolution \eqref{e2} if and only if they are
 $$
y_0(t)=\frac{y_0(t_0)}{B(t)}, \quad
y_n(t)=\frac{y_n(t_0)A^{n-1}(t)}{A^{n-1}(t_0)B^{n+1}(t)}, \quad
t\in I, \; n=1,2,\dots \,.
$$
\end{corollary}

\section{Examples}

We give some examples that illustrate the above results,
including those from \cite{c4}.

(1)  $x_n'(t) = \sum_{k=0}^n x_k(t) x_{n-k}(t)$,
$n=1,2,\dots $. Here $a(t) = 1 $, so $A(t) = t$.

(a)   For the initial values $ x_n(0) = 1$, $n=0,1,2,\dots$,
we have $B(t) = 1 + x_0 (0) A(0) -x_0(0)A(t) = 1 -t$, hence
we obtain the solutions $ x_n(t) = \frac{x_n(0)}{B^{n+1}(t)} =
\frac{1}{(1-t)^{n+1}}$, $n=0,1,2,\dots$. Because,
$A(0) = 0$, second initial values problem can not be considered.

(b)  If $x_n(1) = \frac{1}{2^n} $, we obtain $x_n(t) =
\frac{1}{2^n(2-t)^{n+1}}$, $n=0,1,2,\dots $. The second
initial values problem can not be considered when
$x_n(1) = \frac{1}{2^n}$, $n=1,2,\dots  $, because
$x_0(1) = \frac{x_1^2(1)}{x_2(1)} - \frac{1}{A(t)} = 0$.

(c)    If $x_n(1) = \frac{1}{2^{n-1}}$,  $n=0,1,2,\dots  $,
then $x_0(1) = 2$, $B(t) =1+x_0(t) A(t) - x_0(t) A(t) = 3-2t $,
hence $x_n(t) = \frac{1}{2^{n-1} (3-2t)^{n+1}}$, $n=0,1,2,\dots $.
The second initial values problem when $x_n(t) =
\frac{1}{2^{n-1}}$,  $n=1,2,\dots  $, can also be considered and
will be given in the next example.

(d)\   Let $x_n(1) = \frac{1}{2^{n-1}}$,  $n=1,2,\dots  $,
and $x_0(1) = \frac{x_1^2(1)}{x_2(1)} - \frac{1}{A(1)} = 1 $.
Then $B(t)=2-t $, hence
$ x_0(t) =\frac{x_0(1)}{B(t)} =\frac{1}{2-t}$ and
$ x_n(t) =\frac{x_n(1) A^{n-1}(t)}{A^{n-1}(1) B^{n+1}(t)} =
\frac{t^{n-1}}{2^{n-1}(2-t)^{n+1}}$, $ n=1,2,\dots$.

(2) $y_n'(t) = \sum_{k=0}^n \binom{n}{k} y_k(t) y_{n-k}(t)$,
$n=0,1,2,\dots$.


(a) If $y_n(0) = n! $, then $y_n(t) = \frac{n!}{(1-t)^{n+1}}$,
$n=0,1,2,\dots$.

(b) If $y_n(1) = \frac{n!}{2^n}$, then $ y_n(t) =
\frac{n!}{2^n(2-t)^{n+1}}$,  $n=0,1,2,\dots$.

(c) If $y_n(1) = \frac{n!}{2^{n-1}} $, then $y_n(t) =
\frac{n!}{2^{n-1}(3-2t)^{n+1}}$, $n=0,1,2,\dots$.

(d) If  $y_0(1) =1$ and $ y_n(1) =
\frac{n!}{2^{n-1}}$, $n=1,2,\dots $, then
$y_0(t) = \frac{1}{2-t}$ and
$y_n(t) = \frac{n!t^{n-1}}{2^{n-1}(2-t)^{n+1}}$, $n=1,2,\dots$.

(3)  $x_n'(t) = e^t \sum_{k=o}^n x_k(t) x_{n-k}(t)$,
$n=0,1,2,\dots$. Here $a(t) = A(t) = e^t$.

If $x_n(0) =1$, $n=0,1,2,\dots $, we obtain the solutions
$x_n(t) = \frac{1}{(2-e^t)^{n+1}}$, $n=0,1,2,\dots$. The second
problem can not be considered when $x_n(0) = 1$, $n=1,2,\dots  $,
because $x_0(0) = \frac{x_1^2(0)}{x_2(0)} - \frac{1}{A(0)} =0$.

(4)   $x_n'(t)= \sin t\sum_{k=0}^n x_k(t) x_{n-k}(t)$,
$n=0,1,2,\dots $. Here $ A(t) = -\cos t $.

(a) Let $x_n(0) = 1$, $n=0,1,2,\dots $. Then
$B(t) = 1 +x_0(0) A(0) - x_0(0) A(t) = \cos t $,
hence $x_n(t) = \frac{1}{\cos^{n+1}t},\; n=0,1,2,\dots$.

(b) If $x_n(0)=1$, $n=1,2,\dots $, and $x_0(0) =
\frac{x_1^2(0)}{x_2(0)} - \frac{1}{A(0)}=2$, then
$B(t) = 2\cos t -1$, hence
$x_0(t) = \frac{x_0(0)}{B(t)} = \frac{2}{2\cos t-1} $ and
$$
x_n(t) = \frac{x_n(0) A^{n-1}(t)}{ A^{n-1}(0)
B^{n+1}(t)} = \frac{\cos^{n-1}t}{(2\cos t - 1)^{n+1}}, \quad
n=1,2,\dots.
$$

(c) For $x_n\big(\frac{\pi}{2} \big) = 1$, $n=0,1,2,\dots $, we
obtain $x_n(t) = \frac{1}{(1+\cos t)^{n+1}}$, $n=0,1,2,\dots$.
The second problem can not be considered, because
$A\big(\frac{\pi}{2} \big) = 0$.

(5)  $x_n'(t) = \frac{1}{t}\sum_{k=0}^n x_k(t) x_{n-k}(t)$,
$x_n(1) =1$, $n=0,1,2,\dots $. Then
$x_n(t) = \frac{1}{(1-\ln t)^{n+1}}$, $ n=0,1,2,\dots$. The
second problem can not be considered, because $A(1) = 0 $.

(6)   $x_n'(t) = \frac{1}{t^2}\sum_{k=0}^n x_k(t)
x_{n-k}(t)$, $n=0,1,2,\dots$. Here $a(t) = \frac{1}{t^2}$,
so $A(t) = -\frac{1}{t}$.

(a) If $x_n(1) = 1 $, then $B(t) = \frac{1}{t} $, hence
$x_n(t)= t^{n+1}$, $n=0,1,2,\dots$.

(b) Let $x_n(1) = 1$, $n=1,2,\dots$, and
$x_0(1) = \frac{x_1^2(1)}{x_2(1)} - \frac{1}{A(1)} =2$.
Then $B(t) =\frac{2-t}{t}$, hence  the solutions are
$x_0(t) = \frac{x_0(1)}{B(t)} = \frac{2t}{2-t}$,
$x_n(t) = \frac{x_n(1) A^{n-1}(t)}{A^{n-1}(1)B^{n+1}(t)} =
\frac{t^2}{(2-t)^{n+1}}$, $n=1,2,\dots$.

(c) If $x_n(2) = (-1)^{n+1} 4 $, then $B(t) = \frac{3t-4}{t}$,
hence $x_n(t) = \frac{(-1)^{n+1}4t^{n+1}}{(3t-4)^{n+1}}$,
$n=0,1,2,\dots $.

(d) Let  $x_n(2)=(-1)^{n+1} 4$, $n=1,2,\dots$,
$x_0(2) =\frac{x_1^2(2)}{x_2(2)} - \frac{1}{A(2)} = -2$.
 Then $B(t)=\frac{2(t-1)}{t}$, hence the solutions are
$x_0(t) =\frac{t}{1-t}$, $x_n(t) = \frac{t^2}{(1-t)^{n+1}}$,
$n=1,2,\dots $.

(7)  $y_n'(t) = \frac{1}{t^2}\sum_{k=0}^n
\binom{n}{k}  y_k(t) y_{n-k}(t)$, $n=0,1,2,\dots $.

(a) If $y_n(1) = n!$, then $y_n(t) = n! t^{n+1}$, $n=0,1,2,\dots $.

(b) If $ y_0(1) = 2$, and $y_n(1) = n!$, $n=1,2,\dots  $, then
$y_0(t) = \frac{2t}{2-t}$, and $y_n(t) = \frac{n!t^2}{(2-t)^{n+1}}$,
$n=1,2,\dots $.

(c) If $y_n(2) = (-1)^{n+1} n! 4$, then
$y_n(t) = \frac{(-1)^{n+1} n! 4 t^{n+1}}{(3t-4)^{n+1}}$,
$n=0,1,2,\dots$.

(d) If  $y_0(2) = -2 $, and $y_n(2) = (-1)^{n+1} n! 4$,
$n=1,2,\dots  $, then $ y_0(t) = \frac{t}{1-t}$, and
$ y_n(t) = \frac{n! t^2}{(1-t)^{n+1}}$, $n=1,2,\dots $.

(8)  $tz_n'(t) + z_n(t) = \sum_{k=0}^n z_k(t) z_{n-k}(t)$,
$n=0,1,2,\dots $. We make the change of unknown functions
$x_n(t) = t z_n(t)$, $n=0,1,2,\dots $.

(a) If $z_n(1) = 1$, then  $z_n(t) = t^n$, $n=0,1,\dots$. This
example was given in  \cite[Theorem 4.1 (a)]{c4}.

(b) Let $z_0(1) = 2$, $z_n(1) = 1$, $n=1,2,\dots$. Then
$z_0(t) =\frac{2}{2-t}$, $z_n(t) =\frac{t}{(2-t)^{n+1}}$,
$ n=1,2,\dots $.

(c) If $z_n(2) = (-1)^{n+1} 2$, then
 $z_n(t) = \frac{(-1)^{n+1} 4 t^n}{(3t-4)^{n+1}}$,
$n=0,1,2,\dots$.

(d) Let $z_0(2) = -1$, and $z_n(2) = (-1)^{n+1} 2$, $n=1,2,\dots $.
Then $z_0(t) = \frac{1}{1-t}$ and
$z_n(t) =\frac{t}{(1-t)^{n+1}}$, $n=1,2,\dots $. This example was
given in  \cite[theorem 4.1 (b)]{c4}.

(9)   $t u_n'(t) + u_n(t) = \sum_{k=0}^n \binom{n}{k}
 u_k(t) u_{n-k}(t)$, $n=0,1,2,\dots $.

(a) If $u_n(1) = n!$, $n=0,1,2,\dots $. Then
$u_n(t) = n! t^n$, $n=0,1,2,\dots$. This example was given
 in  \cite[corollary of theorem 4.1 (a)]{c4}.

(b) If $u_0(1) = 2$, $u_n(1) = n!$, $n=1,2,\dots  $, then
$u_0(t) =\frac{2}{2-t}$, $u_n(t) = \frac{n!t}{(2-t)^{n+1}}$,
  $n=1,2,\dots $.

(c) If $u_n(2) = (-1)^{n+1} n! 2$, $n=0,1,2,\dots  $, then
$u_n(t) = \frac{(-1)^{n+1} n! 4 t^n}{(3t-4)^{n+1}}$,
$n=0,1,2,\dots$.

(d) If  $u_0(2) = -1$ and $u_n(2) = (-1)^{n+1} n! 2$,
$n=1,2,\dots$, then $u_0(t) = \frac{1}{1-t}$, and
$u_n(t) = \frac{n!t}{(1-t)^{n+1}}$, $n=1,2,\dots$. This example was
given in  \cite[corollary of theorem 4.1 (b)]{c4}, with some mistakes
corrected here.


\subsection*{Acknowledgments}
 The proof of the Theorem \ref{thm1} by the generating function method,
included in the remark in section 3, was suggested by the anonymous
referee, to whom I want to give many thanks.


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\end{document}