Using generating function to solve non-homogenous recurrence relation The given recurrence relation is:
$$
a_{n}  + 2a_{n-2} = 2n + 3
$$
with initial  conditions: $$ a_{0}=3$$ $$a_{1}=5$$
I know
$$G(x) = a_0x^0 + a_1x^1 + \sum_{n=2}^{\infty} a_nx^n $$
and
$$ a_{n}  = -2a_{n-2} + 2n + 3$$
Therefore:
$$G(x) = 3 + 5x + \sum_{n=2}^{\infty} (-2a_{n-2} + 2n + 3).x^n $$
$$G(x) = 3 + 5x - 2.G(x).x^2 + \sum_{n=2}^{\infty}  (2n + 3).x^n $$
I don't know how to go about solving $$\sum_{n=2}^{\infty}  (2n + 3).x^n $$
And thus I can't calculate G(x) and the sequence a(n). Can someone please guide me how can I solve this?
EDIT 1:
From what I have got to know:
$$G(x) (1 + 2.x^2) = 3 + 5x + 2x/(1-x)^2 + 3/(1-x) $$
$$G(x)  = \frac{3 + 5x + 2x/(1-x)^2 + 3/(1-x)}{1 + 2.x^2} $$
How can I convert it into partial fractions?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[10px,#ffd]{a_{n}  + 2a_{n - 2} = 2n + 3\,\qquad a_{0} = 3,\quad a_{1} = 5}:\ {\Large ?}}$

With $\ds{\verts{z} < 1}$:
\begin{align}
\sum_{n = 2}^{\infty}\pars{a_{n}  + 2a_{n - 2}}z^{n} & =
2\sum_{n = 2}^{\infty}nz^{n} + 3\sum_{n = 2}^{\infty}z^{n}
\\[2mm]
\pars{-a_{0} - a_{1}z + \sum_{n = 0}^{\infty}a_{n}z^{n}}  +
2z^{2}\sum_{n = 0}^{\infty}a_{n}z^{n} & =
2z\,\partiald{}{z}\sum_{n = 2}^{\infty}z^{n} + 3\sum_{n = 2}^{\infty}z^{n}
\\[2mm]
-3 - 5z + \pars{2z^{2} + 1}\sum_{n = 0}^{\infty}a_{n}z^{n} & =
2z\,\partiald{}{z}\pars{z^{2} \over 1 - z} +
3\,{z^{2} \over 1 - z}
\end{align}
$$
\sum_{n = 0}^{\infty}a_{n}z^{n} =
{3 - z \over \pars{1 - z}^{2}\pars{1 + 2z^{2}}}
\implies
a_{n} =
\bracks{z^{n}}{3 - z \over \pars{1 - z}^{2}\pars{1 + 2z^{2}}}
$$

\begin{align}
a_{n} & =
\braces{3\bracks{z^{n}} - \bracks{z^{n - 1}}}
{1 \over \pars{1 - z}^{2}\pars{1 + 2z^{2}}}
\\[5mm] & =
\braces{3\bracks{z^{n}} - \bracks{z^{n - 1}}}
\sum_{i = 0}^{\infty}\pars{i + 1}z^{i}
\sum_{j = 0}^{\infty}\pars{-2z^{2}}^{j}
\\[5mm] & =
3\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}
\pars{i + 1}\pars{-2}^{j}\bracks{i + 2j = n}
\\[1mm] & -
\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\pars{i + 1}\pars{-2}^{j}
\bracks{i + 2j = n - 1}
\\[5mm] & =
3\sum_{j = 0}^{\infty}\pars{n - 2j + 1}\pars{-2}^{j}
\bracks{n - 2j \geq 0}
\\[1mm] & -
\sum_{j = 0}^{\infty}\pars{n - 2j}\pars{-2}^{j}
\bracks{n - 1 - 2j \geq 0}
\\[5mm] & =
3\sum_{j = 0}^{\left\lfloor n/2\right\rfloor}
\pars{n + 1 - 2j}\pars{-2}^{j} -
\sum_{j = 0}^{\left\lfloor\pars{n - 1}/2\right\rfloor}
\pars{n - 2j}\pars{-2}^{j}
\end{align}
You just need to perform the sums.
