i have problem with Recursion equation. It look like this: $x(k+2) + 2x(k+1) + x(k) = 0 $ , $x(0) = 3$, $x(1) = 6$
then i' m gettin: $x^2 + 2x + 1 = 0$
zero value are: $\Delta = 0$ , $x_{0} = -1$
$x(k) = C_{1}x_{1}^{k} + C_{2}x_{2}^{k}$
for $x(0) = 3$ i get
$C_{1} + C_{2} = 3$
The question is: Am i doing it right? In formula above $x_{1} , x_{2}$ are not the same. The question is how should i do it for $x(1)$ because it by putt for both $x_{1} , x_{2}$ $-1$ i will got $0=0$. What am i doing wrong?
//------------SECOND CASE-------------------------
$x(k+2) + 3x(k+1) + x(k) = 0, x(0) = 1, x(1) = 3$
$x^{2} + 3x + 1 = 0$
$\Delta = 1$
$r_{1} = -2$ $r_{2} = -1$
In this case i'm using this:
$x(k) = C_{1}r_{1}^{k} + C_{2}r_{2}^{k}$
right??
When you have a double root $r$, the general form is
$$ x_k = C_1 r^k + C_2 kr^k $$
In this case, we have $r = -1$, so
$$ 3 = x_0 = C_1 $$ $$ 6 = x_1 = -C_1 - C_2 $$
which yields $C_1 = 3, C_2 = -9$, and
$$ x_k = 3(-1)^k-9k(-1)^k $$
Substitution yields $\{x_k\}_{k=0}^\infty = 3, 6, -15, 24, -33, \ldots$ which satisfies the recurrence.
Further multiplicities yield terms of the form $k^2 r^k, k^3 r^k, \ldots$ You may see a resemblance to differential equations (if you're familiar with those), in which a double root in the characteristic equation yields a solution of the form $f(x) = C_1 e^{rx} + C_2 xe^{rx}$ (and further multiplicities yield similar terms).