Long Term of a Difference Equation Given Initial Conditions

Question

I am trying to figure out how to answer the following question

Solve the difference equation $F(N) = F(N - 1) + 6F(N - 2)$

if the initial conditions are $F(0) = 0$ and $F(1) = 2$

The long term behavior of the sequence $F(N)$ is that it is roughly exponential: $F(N) = c * W ^ N $ for some constants c and W. What are they?

Any suggestions would be great. Thank you.

Answer 1

Assume that the difference equation has solution

$$F(N) = c \lambda ^N$$ for $c,\lambda \neq 0$. Then Substitute this into your difference equation and yield a quadratic function

$$\lambda^2 = \lambda +6.$$

This has solution

$$\lambda = \dfrac{1\pm 5}{2} = 3,-2$$

Thus, the solution is a linear combination of both

$$F(N) = c_1(-2)^N + c_2 3^N$$

Use the initial conditions to solve for the $c_i$.

Answer 2

The standard technique is to consider solutions of the form $F(N)=\lambda^{N}$, and plugging that in, you get the equation $\lambda^{2}-\lambda-6=0$ and so the solutions are $\lambda=-2$ or $\lambda=3$, and thus $F(N)=A.(-2)^{N}+B.3^{N}$, and then you could find the values of $A,B$ from the given values of $F(0),F(1)$.