Given the recurrence an=3an−1−2an−2a_n = 3a_{n-1} - 2a_{n-2}an=3an−1−2an−2 with a0=2,a1=3a_0 = 2, a_1 = 3a0=2,a1=3, find the general solution form.
an=c1(1)n+c2(2)na_n = c_1(1)^n + c_2(2)^nan=c1(1)n+c2(2)n
an=c1(2)n+c2(3)na_n = c_1(2)^n + c_2(3)^nan=c1(2)n+c2(3)n
an=c1(1)n+c2(n)(2)na_n = c_1(1)^n + c_2(n)(2)^nan=c1(1)n+c2(n)(2)n
an=c1(−1)n+c2(2)na_n = c_1(-1)^n + c_2(2)^nan=c1(−1)n+c2(2)n