Asked by Francesca
Solve the recurrence relation a_n = -2a_n-1 + 15a_n-2, n ≥ 2, given a₀ = 1, a₁ = -1.
x² + 2x - 15, the distinct roots 3 and -5, so a_n = C₁(3^n) + C₂(-5)^n. The initial condition gives a₀ = 1 = C₁ - C₂, a₁ = -1 = 3C₁ - 5C₂. We obtain C₁ = C₂ = 1/2 and so a_n = 1/2(3^n) + 1/2(-5)^n.
My question is how does C₁ = C₂ = 1/2 can some please how do you derive to this answer because I'm confused.Thank you for any help.
x² + 2x - 15, the distinct roots 3 and -5, so a_n = C₁(3^n) + C₂(-5)^n. The initial condition gives a₀ = 1 = C₁ - C₂, a₁ = -1 = 3C₁ - 5C₂. We obtain C₁ = C₂ = 1/2 and so a_n = 1/2(3^n) + 1/2(-5)^n.
My question is how does C₁ = C₂ = 1/2 can some please how do you derive to this answer because I'm confused.Thank you for any help.
Answers
Answered by
Count Iblis
There wasa typo in the equations derived from the initial conditions. You should have:
The initial condition gives
a₀ = 1 = C₁ + C₂,
a₁ = -1 = 3C₁ - 5C₂
It then easily follows that
C₁ = C₂ = 1/2
The initial condition gives
a₀ = 1 = C₁ + C₂,
a₁ = -1 = 3C₁ - 5C₂
It then easily follows that
C₁ = C₂ = 1/2
Answered by
Francesca
Sorry I still don't get it. Can someone please explain?