Two sequences $y_t$ and $z_t $ satisfy
$$y_t = ay_{t-1} + bz_{t-1}$$ $$z_t = cy_{t-1} + dz_{t-1}$$
Where $a = 6$, $b = -20$, $c = -17$ and $d = -12$. From the two given equations above, a second-order difference equation of the following form can be derived: $$y_{t+1} + ny_t + ry_{t-1} = 0$$ Where $n$ and $r$ are constants. The auxiliary equation $$m^2+nm+r=0$$ Has two solutions $m=h$ and $m=k$ where $h>k$. To 2 decimal places, what is the value of h?
From the statement, I tried to find the value of H , however about the results. can someone help me?
Two sequences are given:
First from the first equation we express zt-1
Then we replace t-1 with t:
And finally we put the last two equations into the second equation:
After rearranging we get:
We multiply everything by -20 and rearrange:
So n = 6 and r = -412. Hence the auxiliary equation is
Solving this quadratic equation we get two answers for m: m =17.52 and m = -23.52 Since h>k the value of h is 17.52. Am I right?
you can write this as matrix equation $$x_{n+1} = \pmatrix{6&-20\\-17&-12}x_n = Ax_n. $$ the eigenvalues of $A$ are $-3 \pm \sqrt{421}=17.518, -23.518$
so the value of $h$ is $17.518$