I studying Discrete maths during this semester and I need your help. I have been trying to solve one non-homogeneous recurrence relation and read many-many guides how to do this, but I haven't found the way.
However, it's here: $A_n = -3A_{n-1} +4A_{n-2}+ 3.2^n$
Anyways, the characteristic equation here gives $x_1=1$ and $x_2=-4$ as results, but I'm in need to find the special thingy somehow with this one $3.2^n$.
I hope you understand, what I exactly need.
Let $B_n = A_n - A_{n-1}$. Then $$ B_n = 3.2^n - 4B_{n-1}. $$ Let $C_n = B_n/(-4)^n$. Then $$ C_n = (-0.8)^n + C_{n-1}. $$ Therefore $$ C_n = \sum_{k=1}^n (-0.8)^k + C_0 = -\tfrac{4}{9} (1-(-0.8)^n) + C_0 = C_0-\tfrac{4}{9} + \tfrac{4}{9} (-0.8)^n. $$ This implies that $$ B_n = (B_0-\tfrac{4}{9})(-4)^n + \tfrac{4}{9} \cdot 3.2^n. $$ Therefore $$ \begin{align*} A_n &= \sum_{k=1}^n B_k + A_0 \\ &= \sum_{k=1}^n \left[(B_0-\tfrac{4}{9})(-4)^k + \tfrac{4}{9} \cdot 3.2^k\right] + A_0 \\ &= -0.8(B_0-\tfrac{4}{9}) [1-(-4)^n] + \tfrac{64}{99} (3.2^n-1) + A_0 \\ &= 0.8(B_0-\tfrac{4}{9})(-4)^n + \tfrac{64}{99} 3.2^n -0.8(B_0-\tfrac{4}{9}) - \tfrac{64}{99} + A_0. \end{align*} $$ Here $B_0 = A_0 - A_{-1}$.
Morale: There is an inhomogeneous solution of the form $C 3.2^n$. Using this Ansatz, we get $$ C 3.2^n = -3 C 3.2^{n-1} + 4 C 3.2^{n-2} + 3.2^n. $$ Dividing by $3.2^{n-2}$, we get $$ C 3.2^2 = -3C 3.2 + 4C + 3.2^2. $$ This simplifies to $15.84 C = 10.24$ and so $C = 10.24/15.84 = 64/99$.
At this point we can apply the usual methods to obtain the solution calculated explicitly above.