How to find the closed form formula for this recurrence relation $ x_{0} = 5 $
$ x_{n} = 2x_{n-1} + 9(5^{n-1})$
I have computed: $x_{0} = 5, x_{1} = 19, x_{2} = 83, x_{3} = 391, x_{4} = 1907$, but cannot see any pattern for the general $n^{th}$ term.

If you're not familiar with the method that Phira explained, divide both sides by $2^n$:
$$
\dfrac{x_{n}}{2^n} = \dfrac{x_{n-1}}{2^{n-1}} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
Call $\dfrac{x_{n}}{2^n} = s_n$:
$$
s_n = s_{n-1} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
If we keep expanding $s_{n-1}$ in the RHS recursively, we get:
$$
s_n = s_0 + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{0} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{1} \cdots + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-2} + \dfrac{9}{2}\left(\dfrac{5}{2}\right)^{n-1}
$$
Where $s_0 = \dfrac{x_0}{2^0} = x_0$.
This means that:
$$
s_n = x_0 + \sum_{k=1}^{n}\dfrac{9}{2}\left(\dfrac{5}{2}\right)^{k-1}
$$
This is a geometric series:
$$
s_n = 3 \left(\dfrac{5}{2}\right)^n + 2
$$
And therefore:
$$
x_n = 3 \cdot 5^n + 2 \cdot 2^n
$$
