We begin by looking at the general solution of the (homogeneous linear) differen

Question

We begin by looking at the general solution of the (homogeneous linear) difference equation and then finding the one that satisfies the starting conditions The creative step is to make an inspired guess of the form of some solutions of the difference equation: for some constant r. Substituting in the difference equation and reorganizing gives 0 =rn+2 - rn+1 -rn = rn (r2 -r -1). Therefore = rn is a solution if r is a root of the quadratic equation The smaller of those roots is r1 = , and the larger is r2 = . (you want to use exact expressions for those roots). It's clear that any sequence of the form Where A and B are arbitrary numbers, also satisfy the difference equation. Find the particular values of A and B which make satisfy the starting conditions. thus A = ,and B = .

Explanation / Answer

A s you got r1 and r2 right

you got A and B values wrong

we know f0 = f1 = 1

for f0 = A (r1)^0 + B(r2)^0

==> 1 = A(1) + B(1)

==> A + B = 1

f1 = 1 = A (r1) + B(r2)

==> 1 = A (1/2)( 1- sqrt 5) + B(1/2) (1+ sqrt 5)

==> since A+B = 1 ==> b = 1-A

1 = (1/2) (A+B) + sqrt 5 (B-A)/2

1 = (1/2)(1) + sqrt5 (B-A)'/2

==> B - A = 1/sqrt 5

==> B = (1/2) ( 1 + 1/sqrt 5)

A = (1/2) (1- 1/sqrt 5)

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