How to solve this particular difference equations? Considere a real polynomial function $P_n\left(x\right)$ of two variable, when $n$ is a discrete variable and $x$ a continuous variable
where $P_0\left(x\right)=1$
and satisfies the following recurrence relations: 
$$P_{n+1}\left(x\right)=\left(x+1\right)\cdot P_n\left(x\right)+x\cdot\frac{d}{dx}\left(P_n\left(x\right)\right).$$
How could I find the explicit solution $P_n\left(x\right)$ for this expression?
And what name do these types of equations receive?
 A: Write the polynomial as: $$P_n(x)=c_{n,0}x^0+c_{n,1}x^1+\cdots+c_{n,n}x^n,$$
apply the recurrence, and you will get a relation among the coefficients.
That is
$$
\eqalign{
  & \sum\limits_{0\, \le \;k\,\left( { \le \,n + 1} \right)} {c_{\,n + 1,\,k} x^{\,k} }  = \left( {x + 1} \right)\sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k} x^{\,k} }  + x{d \over {dx}}\sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k} x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k} x^{\,k} }  + \sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k} x^{\,k + 1} }  + \sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {k\,c_{\,n,\,k} x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k} x^{\,k} }  + \sum\limits_{1\, \le \;k\,\left( { \le \,n} \right)} {c_{\,n,\,k - 1} x^{\,k} }  + \sum\limits_{0\, \le \;k\,\left( { \le \,n} \right)} {k\,c_{\,n,\,k} x^{\,k} }  \cr} 
$$
or
$$
c_{\,n + 1,\,k}  = \left( {1 + k} \right)c_{\,n,\,k}  + c_{\,n,\,k - 1} 
$$
with the understanding that when the index is negative
the coefficient is null.
Since the recursion for Stirling N. of 2nd kind is
$$
\left\{ \matrix{
  n \cr 
  m \cr}  \right\} = m\left\{ \matrix{
  n - 1 \cr 
  m \cr}  \right\} + \left\{ \matrix{
  n - 1 \cr 
  m - 1 \cr}  \right\}
$$
then you get
$$
c_{\,n,\,k}  = \left\{ \matrix{
  n + 1 \cr 
  k + 1 \cr}  \right\}
$$
and your polynomials are related to the Touchard Polynomials
$$
P_{\,n} (x) = {1 \over x}T_{\,n + 1} (x)
$$
A: The coefficients of the polynomials are the Stirling numbers of the second kind. The OEIS triangular sequence A008277 has a lot of information about the sequence and the polynomials are almost Touchard polynomials, except $\;x\;p_n(x) = T_{n+1}(x).\;$  The equation you have where each $\;p_{n+1}(x)\;$ is determined by the previous $\;p_n(x)\;$ is usually called a difference-differential equation.
Note that the recurrence equation can be directly translated into a recurrence relation for the polynomial coefficients which is the usual one for the Stirling numbers of the second kind. This is due to $\;(x+1)\;s(n,k)\;x^k = s(n,k)\;x^k + s(n,k)\;x^{k+1},\;\;$ and $\;x \frac{d}{dx} s(n,k)\;x^k = k\; s(n,k)\;x^k.$
A: This is really a comment where we show what the first few polynomials look like.
Let's start with $P_0(x)=1$ and see what we can see: 
$$P_{1}\left(x\right)=\left(x+1\right)\cdot P_0\left(x\right)+x\cdot\frac{d}{dx}\left(P_0\left(x\right)\right)=(x+1)\cdot 1+x\cdot0=x+1$$
Coefficients being: $[1,1]$
$$P_{2}\left(x\right)=\left(x+1\right)\cdot P_1\left(x\right)+x\cdot\frac{d}{dx}\left(P_1\left(x\right)\right)=(x+1)^2+x=x^2+3x+1$$
Coefficients being: $[1,3,1]$
$P_{3}\left(x\right)$
$=\left(x+1\right) P_2\left(x\right)+x\frac{d}{dx}\left(P_2\left(x\right)\right)=(x+1)^3+x(x+1)+2x^2+3x=x^3+6x^2+7x+1 $
Coefficients being $[1,6,7,1]$
Compare with this table: Stirling Numbers of the Second Kind. 
