The recurrence an=6an−1−9an−2a_n = 6a_{n-1} - 9a_{n-2}an=6an−1−9an−2 has a repeated root r=3r = 3r=3. What is the general solution?
an=c1⋅3na_n = c_1 \cdot 3^nan=c1⋅3n
an=(c1+c2n)⋅3na_n = (c_1 + c_2 n) \cdot 3^nan=(c1+c2n)⋅3n
an=c1⋅9na_n = c_1 \cdot 9^nan=c1⋅9n
an=(c1+c2n2)⋅3na_n = (c_1 + c_2 n^2) \cdot 3^nan=(c1+c2n2)⋅3n