Math, asked by madhuni2806, 5 months ago

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

Answers

Answered by Anonymous
6

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.

Answered by MasterKaatyaayana2
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.

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