Discrete math Exponential Generating Series Solve the recurrence $y_{n+1} = 2y_{n} + n$ using exponential generating series. The given condition is $y_{0} = 1$. It is also noted that the equation is equivalent to $y_{n} = 2y_{n-1} +n -1$. I know that $Y(x) = \sum_{n = 0}^{\infty}[\frac{y_nx^n}{n!}]$, but I'm not sure what to do with it from there. The professor only went over a very simple example in class, and we don't have a text book.
 A: A related problem. Here is how you advance
$$ \sum_{n=0}^{\infty}y_{n+1}\frac{x^n}{n!} = 2\sum_{n=0}^{\infty}y_{n}\frac{x^n}{n!}+\sum_{n=0}^{\infty}\frac{n}{n!}x^n $$
$$\implies \sum_{n=1}^{\infty}y_{n}\frac{x^{n-1}}{(n-1)!} = 2 F(x) + x e^{x} $$
$$\implies \sum_{n=1}^{\infty}n y_{n}\frac{x^{n-1}}{n(n-1)!} = 2F(x)+xe^{x}. $$

$$\implies F'(x) = 2F(x)+xe^{x}. $$

Now, to obtain $F(x)$ you need to solve the above first order differential equation. Note that you will have a constant and you can find it by exploiting the initial condition $y_0$. Here is the final answer

$$ F(x) =\left( 1- \left( 1+x \right) {{\rm e}^{-x}}+y_{{0}
} \right) {{\rm e}^{2\,x}}.$$ 

A: Define $\displaystyle{{\cal Y}\left(z\right) \equiv \sum_{n = 0}^{\infty}y_{n}z^{n}\,,\quad z \in {\mathbb C}\,,\quad
\left\vert z\right\vert < {1 \over 2}}$
Notice that
$$
{1 \over 1 - z}
=
\sum_{n = 0}^{\infty}z^{n}\,,
\quad
{1 \over \left(1 - z\right)^{2}}
=
\sum_{n = 1}^{\infty}nz^{n - 1}\,,
\quad
{1 \over \left(1 - z\right)^{3}}
=
{1 \over 2}\sum_{n = 2}^{\infty}n\left(n - 1\right)z^{n - 2}\,,
$$
$$------------------------------$$
\begin{align}
\sum_{n = 0}^{\infty}y_{n + 1}z^{n}
&= 
\sum_{n = 0}^{\infty}\left(2y_{n} + n\right)z^{n}
\\[3mm]
\overbrace{\quad\sum_{n = 1}^{\infty}y_{n}z^{n - 1}\quad}
^{z^{-1}\left[\vphantom{\Large A}\,{\cal Y}\left(z\right)\ -\ y_{0}\,\right]}
&= 
2\quad\overbrace{\sum_{n = 0}^{\infty}y_{n}z^{n}}^{{\cal Y}\left(z\right)}\
+\
z\sum_{n = 1}^{\infty}nz^{n - 1}
\\[3mm]
\left(z^{-1} - 2\right){\cal Y}\left(z\right)
&=
z^{-1}y_{0}
+
z\,{1 \over \left(1 - z\right)^{2}}
\end{align}
\begin{align}
{\cal Y}\left(z\right)
&=
{1 \over 1 - 2z}
+
{z^{2} \over \left(1 - 2z\right)\left(1 - z\right)^{2}}
\end{align}
Now, expand ${\cal Y}\left(z\right)$ in powers of $z$.
