$r^{ 3 }− 7r − 6 = 0$
$(r + 1)(r 2 − r − 6) = 0$.
Continue factoring and we get $(r + 1)(r − 3)(r + 2) = 0$.
We know our characteristic roots: r1 = −1, r2 = 3, r3 = −2.
Our general solution using Theorem 3 is: an = $α1(−1)^{n} + α2(3)^{n } + α3(−2)^{n}$ .
Find α1, α2, α3 by using the initial conditions.
For a0 = 9
$9 = α1(−1)^0 + α2(3)^ 0 + α3(−2)^0$
9 = α1 + α2 + α3
For a1 = 10
$10 = α1(−1)^{1} + α2(3 )^{1} + α3(−2)^{1}$
10 = −α1 + 3α2 − 2α3
For a2 = 32
$32 = α1(−1)^{2} + α2(3)^{2} + α3(−2)^{2}$
32 = α1 + 9α2 + 4α3
Solve the system of equations:
9 = α1 + α2 + α3
10 = −α1 + 3α2 − 2α3
32 = α1 + 9α2 + 4α3
Our solution is $an = 8(−1)^{n} + 4(3)^{n }+ (−3)(−2)^{n}$
