Solve the recurrence relation an = 5an-1 – 6an-2 with initial conditions a0 = 1, a1 = 4.

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Solution:

Given recurrance relation is

an = 5an-1 – 6an-2

Here, degree = 2

The characteristics equation is

r2 = 5r – 6

or, r2 – 5r + 6 = 0

or, r2 – 2r – 3r + 6 = 0

or, r(r-2) – 3(r-2) = 0

or, (r-3)(r-2) = 0

or, r = 3, 2

or, r1 = 3 and r2 = 2

Now, The general solution of homogenuous equation is:

an = A1r1n + A2r2n

or, an = A13n + A2 2n  ———–(i)

Given condition are

a0 = 1, a1 = 4

From equation (i), when condition a0 = 1:

1 = A1 + A2 . 30

or, A1 = 1 – A2    —————(ii)

From equation (i), when condition a1 = 4:

4 = A1 . 21 + A2 . 31

or, 4 = 3 . A1 + 2 . A2 ————-(iii)

Fromequation (ii) and (iii), We get

4 = 3 (1 – A2) + 2 A2

or, 4 = 3 – 3 A2 + 2 A2

or, 4-3 = – A2

∴ A2 = -1

Also,

A1 = 1 – (-1) = 2

From equation (i), the general solution is

\(a_n = 2 . 3^n – 1 . 2^n \)

This is the required solution.

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