Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3 
Thanks in advance!
 A: Let $2x_{n+2} = 3 \, x_{n+1} + 8 \, x_{n} + 3 \, x_{n-1}$, with $x_{0} = -1,
x_{1} = 3$ and $x_{2} = 3$ for which letting $x_{n} = p^{n}$ the equation 
$$2 \, p^{3} - 3 \, p^{2} - 8 \, p - 3 = 0$$ 
is obtained. This equation can be factored to $(p-3)(p+1)(2p+1) = 0$ and leads to the roots $p \in \{ 3, -1, -1/2 \}$. From this the general form of $x_{n}$ is
$$x_{n} = a_{0} \, 3^{n} + a_{1} \, (-1)^{n} + a_{2} \, \left( - \frac{1}{2} \right)^{n}.$$ 
Applying the initial conditions yields:
\begin{align}
-1 &= a_{0} + a_{1} + a_{2} \\
3 &= 3 \, a_{0} - a_{1} - \frac{a_{2}}{2} \\
3 &= 9 \, a_{0} + a_{1} + \frac{a_{2}}{4}
\end{align}
which leads to $a_{0} = \frac{1}{2}$, $a_{1} = - \frac{3}{2}$, $a_{2} = 0$.
The resulting sequence is generated by
\begin{align}
x_{n} = \frac{3}{2} \, \left( 3^{n-1} + (-1)^{n-1} \right)  
\end{align} 
A: Hint: Look for solutions of the form $\lambda^n$. Note also that linear combinations of solutions are also solutions.
A: A general way to solve such is using generating functions. Define:
$\begin{align*}
   g(z)
     &= \sum_{n \ge 0} x_n z^n
\end{align*}$
Write your recurrence shifted (subtraction in indices gets messy), multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
   2 x_{n + 3}
     &= 3 x_{n + 2} + 8 x_{n + 1} + 3 x_n \\
   2 \sum_{n \ge 0} x_{n + 3} z^n
     &= 3 \sum_{n \ge 0} x_{n + 2} z^n 
          + 8 \sum_{n \ge 0} x_{n + 1} z^n
          + 3 \sum_{n \ge 0} x_n z^n \\
   2 \frac{g(z) - x_0 - x_1 z - x_2 z^2}{z^3}
     &= 3 \frac{g(z) - x_0 - x_1 z}{z^2}
          + 8 \frac{g(z) - x_0}{z}
          + 3 g(z)
\end{align*}$
Solve this for $g(z)$ (substitute the initial values), express as partial fractions:
$\begin{align*}
   g(z)
     &= \frac{1 - 5 z}{1 - 2 z - 3 z^2} \\
     &= \frac{1 - 5 z}{(1 + z) (1 - 3 z)} \\
     &= - \frac{3}{2 (1 + z)}
          + \frac{1}{2 (1 - 3 z)}
\end{align*}$
We want the coefficient of $z^n$ of this. But this is just two geometric series:
$\begin{align*}
   [z^n] g(z)
     &= - \frac{3}{2} (-1)^n + \frac{1}{2} \cdot 3^n \\
     &= \frac{3^n - (-1)^n}{2}
\end{align*}$
