Finding a solution to the recurrence relation $a_n = 5a_{n−2} − 4a_{n−4}$ Find the solution to $$a_n = 5a_{n−2} − 4a_{n−4}$$ with $$a_0 = 3$$
$$a_1 = 2$$ $$a_2 = 6$$ $$a_3 = 8$$
My answer: Observe that the degree of recurrence is 4.  Hence, the characteristic equation is: $x^4 - 5x^2 +4 = 0$.  Solving for $y=x^2$ we get $y_1=1$, $y_2 = 4$.  Hence, the solutions are $x_1 = -1$, $x_2 = 1$, $x_3 = -2$, $x_4 = 2$.  This implies that $a_n = \alpha_1(-1)^n + \alpha_2 + \alpha_3 (-2)^n + \alpha_4(2^n)$.  It remains to solve the initial conditions for $\alpha_i-s : \alpha_1 = 1, \alpha_2 = 1, \alpha_3 = 0, \alpha_4 = 1$.
Where should I go from here to ensure the answer is fully answered and complete?  Thank you!

Just for kicks, use generating functions on this one. Define $A(z) = \sum_{n \ge 0} a_n z^n$. Your recurrence is:
$$
a_{n + 4} = 5 a_{n + 2} - 4 a_n \quad a_0 = 3, a_1 = 2, a_2 = 6, a_3 = 8
$$
Using properties of ordinary generating afunctions:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2 - a_3 z^3}{z^4}
  = 5 \frac{A(z) - a_0 - a_1 z}{z^2} - 4 A(z)
$$
Solving for $A(z)$, written as partial fractions:
$$
A(z) = \frac{1}{1 - z} + \frac{1}{1 + z} + \frac{1}{1 - 2 z}
$$
Expanding the geometric series:
$$
a_n = 1 + (-1)^n + 2^n
$$
