How to find the values of arbitrary constants from the initial conditions of the recurrence relation? I have recurrence relation  $x_{n+4}-3x_{n+3}+4x_{n+2}-8x_{n}=0,x_0=0$.
Here I tried to solve it using the characteristic equation method. I got roots $t=-1,t=2$.
General solution looks like this $a_n=(-1)^nC_1+2^nC_2$ but how can I find $C_1$ and $C_2$?
I tried to write general solution using the initial condition $a_0=C_1+C_2=0$ but it didn't work, since I can't compose a system for further solution with only one initial condition.
Can I find $a_1$ somehow to build system to find $C_1$ and $C_2$?
 A: You need more initial conditions than what is given; any choice of $x_1$, $x_2$, and $x_3$ (together with your known condition of $x_0 = 0$) would give rise to a (generally) different sequence, and so knowledge of at least the first four terms is necessary to uniquely determine the sequence.
Also: you might want to check your characteristic polynomial again, as I suspect there should be more than two roots in this case. Indeed, I would guess there are four roots (some roots may be repeated) since the recurrence implies that the sequence is unique after the first four terms are chosen (four degrees of freedom, four roots, four unknown coefficients).
A: So actually, it is more complicated than that. If you put in $x_n = a^n$ and then divide through by $a^n$, you will get
$$
a^4-3a^3+4a^2-8 = 0,
$$
which indeed is solved by $-1,2$ as you claim, but also has 2 complex solutions $1 \pm i\sqrt{3}$, so the general form of the solution would look something like
$$
x_n = A(-1)^n + B 2^n + C 2^n e^{-i\pi n/3} + D2^n e^{i\pi n/3}
$$
so you would need 4 initial conditions to determine the recurrence completely.
This is in general the rule: $n$-th degree recurrences require $n$ initial conditions to be fully determined.
