Question

form the difference equation from yn=a+b3^n

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Answer 1

Step-by-step explanation:

called a recurrence relation (rr) or a difference equation (ÎE). ... The unknown (to be solved for ) is yn, the n—th term of the ... Don't confuse this with the fact that the polar form of (2i)n is.

Answer 2

Answer:

$y_{n+2}-4y_{n+1}+3y_{n}=0$

Step-by-step explanation:

Given equation,

$y_n = a+b(3^n)$

since the equation contains two arbitarily constants so we need first and second differance both.

$\Delta y_n = \Delta(a)+b\Delta(3)^n\\y_{n+1}-y_{n} =b(3^{n+1}-3^n)=2b(3^n)......(1)$

taking second difference as

$\Delta^2 y_n = \Delta^2(a)+b\Delta^2(3)^n\\\\y_{n+2}-2y_{1+n}+y_n =2b(3^{n+1}-3^n)=4b(3^n)......(2)$

From (2) and (1),

$\\\frac{y_{n+2}-2y_{1+n}+y_n }{y_{n+1}-y_n } =2,$

So, difference equation after rearranging above eq will be

$y_{n+2}-4y_{n+1}+3y_{n}=0$.