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?
On
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 $\;x \frac{d}{dx} s(n,k)\;x^k = k\; s(n,k)\;x^k.$
On
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.
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 $$ P_{\,n} (x) = {1 \over x}T_{\,n + 1} (x) $$