Determine the general solution of an=4an−1−3an−2a_n = 4a_{n-1} - 3a_{n-2}an=4an−1−3an−2.
an=A(1n)+B(3n)a_n = A(1^n) + B(3^n)an=A(1n)+B(3n)
an=A(3n)+B(4n)a_n = A(3^n) + B(4^n)an=A(3n)+B(4n)
an=A(2n)+B(2n)a_n = A(2^n) + B(2^n)an=A(2n)+B(2n)
an=A(4n)+B(1n)a_n = A(4^n) + B(1^n)an=A(4n)+B(1n)