Question

(t) (6) Solve the difference equation c(k) + 3c(k-1) + 2c(k-2) = r(k) where r(k)=1 for k>0: (Assume all initial conditions are zero.) (2) Find c(2) and c(4)

Name: t 6 solve the difference equation ck 3ck 1 2ck 2 rk where rk1 for k0 assume all initial conditions are zero 2 find c2 and c4 46338
Uploaded: 2022-06-13T10:26:14-08:00
Duration: 3 min 44 s
Channel: Shaiju T
Description: t 6 solve the difference equation ck 3ck 1 2ck 2 rk where rk1 for k0 assume all initial conditions are zero 2 find c2 and c4 46338

          (t) (6) Solve the difference equation
c(k) + 3c(k-1) + 2c(k-2) = r(k)
where r(k)=1 for k>0:
(Assume all initial conditions are zero.)
(2) Find c(2) and c(4)

Added by Malik A.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Shaiju T

06/13/2022

Step 1

We are also given that all initial conditions are zero. This means $c(k) = 0$ for $k \le 0$. So, $c(0) = 0$ and $c(-1) = 0$. Step 2: Find c(1) For $k=1$: $c(1) + 3c(1-1) + 2c(1-2) = r(1)$ $c(1) + 3c(0) + 2c(-1) = 1$ (since $r(1)=1$ for $k>0$) Since $c(0)=0$ and Show more…

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(t) (6) Solve the difference equation c(k) + 3c(k-1) + 2c(k-2) = r(k) where r(k)=1 for k>0: (Assume all initial conditions are zero.) (2) Find c(2) and c(4)