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?

Decompose it:
$$\sum_{n\ge 2}(2n+3)x^n=2\sum_{n\ge 2}nx^n+3\sum_{n\ge 2}x^n=2x\sum_{n\ge 2}nx^{n-1}+3\sum_{n\ge 2}x^n\;.$$
That last summation is just a geometric series, so you can easily get its closed form, and $\sum_{n\ge 2}nx^{n-1}$ is the derivative of a geometric series, so it also should not be a major problem.
