Question

Find the solution to the following lhcc recurrence:

an=−1an−1+6an−2 for n≥2 with initial conditions a0=3,a1=4 . The
solution is of the form: an=α1(r1)n+α2(r2)n for suitable constants
α1,α2

Answer 1

To solve the recurrence relation, we use the characteristic equation derived from it, find its roots, and use them to express the general solution. The initial conditions are then used to find the specific constants for the solution.

Step-by-step explanation:

The student is asking to solve a linear homogeneous recurrence relation with constant coefficients, given the recurrence relation an = -1 * an-1 + 6 * an-2 for n ≥ 2 and initial conditions a0 = 3, a1 = 4. The solution is of the form an = α1(r1)n + α2(r2)n where α1 and α2 are constants to be determined, and r1 and r2 are the roots of the characteristic equation derived from the recurrence relation.

First, we set up the characteristic equation by replacing an with r^n and solving for r: r^n = -r^(n-1) + 6r^(n-2). This simplifies to the quadratic equation r^2 + r - 6 = 0, which factors to (r + 3)(r - 2) = 0. Thus, the roots are r1 = -3 and r2 = 2.

With the roots identified, we can now express the general solution as an = α1(-3)^n + α2(2)^n. We use the initial conditions to solve for α1 and α2. Substituting n = 0 gives 3 = α1 + α2, and substituting n = 1 gives 4 = -3α1 + 2α2. Solving these equations simultaneously gives us the values for α1 and α2.