If $P(x)-P'(x) = x^n\;,$ Where $n$ is a positive integer. Then $P(0)$ 
$P(x)$ be a polynomial such that $P(x)-P'(x) = x^n\;,$ Where $n$ is a positive integer. Then $P(0)$

$\bf{My\; Try::}$ Let $P(x)=y\;,$ Then equation convert into $\displaystyle \frac{dy}{dx}-y=-x^n.$
Which is First order linear differential equation of order and degree $1$
Whose solution is  $\displaystyle y\cdot e^{-x} = -\int x^n\cdot e^{-x}dx$
Now How can I solve after that, Help me
Thanks
 A: First, we can rewrite $P(x)$ as
$$\begin{align}
P(x) &= \sum_{k=1}^\infty (P^{(k-1)}(x) - P^{(k)}(x))\\
&= (P(x) - P'(x)) + (P'(x) - P''(x)) + (P''(x) - P^{(3)}(x)) + \cdots
\end{align}
$$
Since $P(x)$ is a polynomial, this series terminate after finitely many terms.
Using $P(x) - P'(x) = x^n$, we get
$$\begin{align}P(x) 
&= \sum_{k=1}^\infty \frac{d^{k-1}}{dx^{k-1}}(x^n) \\
&= x^n + (x^n)' + (x^n)'' + \cdots\\
&= x^n + n x^{n-1} + n(n-1) x^{n-2} + \cdots +n!\\
\end{align}$$
This means $P(0)$ is $n!$.
A: Let $P(x)=\displaystyle\sum_{k=0}^ma_kx^k$, then $\forall x,P(x)-P'(x)=x^m+\displaystyle\sum_{k=0}^{m-1}{\left(a_k-(k+1)a_{k+1}\right)x^k}=x^n$.
Hence $a_n=1,m=n$ and $\forall k\in\{0,\dots,m-1\},a_k=(k+1)a_{k+1}$. Thus :
$$a_n=1$$
$$a_{n-1}=na_n=n$$
$$a_{n-2}=(n-1)a_{n-1}=n(n-1)$$
$$\vdots$$
$$a_{1}=n!$$
$$\boxed{P(0)=a_0=a_1=n!}$$
A: The solution for a given $n$ can be built recursively.  Indeed $$P_n(x)=x^n+nP_{n-1}(x)$$
(Pf:  it is easily checked, by differentiation, that this is a solution.  To see that the solution is unique it suffices to note that we could also have read off the coefficients of $P_n(x)$ recursively, starting from $a_n=1$)
In particular:  $P_n(0)=nP_{n-1}(0)=n!P_1(0)$
As it is readily verified that $P_1(x)=x+1$ we have $P_1(0)=1$  whence $P_n(0)=n!$
A: I follow your idea.
We have $$(\exp(-x)P(x))^{\prime}=\exp(-x)(P^{\prime}(x)-P(x))=-x^{n}\exp(-x)$$
Hence
$$\exp(-x)P(x)=P(0)-\int_0^x t^n\exp(-t)dt$$
Now let $x\to +\infty$. We have $P(x)\exp(-x)\to 0$, and $\int_0^{+\infty}t^n\exp(-t)dt=n!$, and we are done. 
