I am having troubles in understanding the answer $L=(\rho-1)^2(\rho -3)$ of one my exercises.
Consider the difference equation: $$ y(n+1) - 2 y(n) = 3n + 3^n\tag{1} $$ Determine the general solution of (1).
Answer We use the annihilator method to find the general solution. The lefthand side of (1) can be written as $L(y)(n)$, where $L = \rho - 2$. The right-hand side of (1) is a solution of the difference equation $L_b(y) = 0$, where $L_b=(\rho - 1)^2(\rho -3)$. The general solution of (1) is thus of the form $$ y(n) = c_1 + c_2n + c_3 2^n + c_4 3^n,\quad n \geq 0 \tag{2} $$ where $c_3 \in \mathbb{C}$ and $c_1, c_2$, and $c_4$ have yet to be determined.
How do you come up with $(\rho - 1)^2(\rho -3)$? I have been dazing at this result for hours and I have given up.
Since the answer has found the annihilator of the LHS as $L=\rho -2$, I suppose that they define $(\rho) u(n)=u(n+1)$ rather than $(\rho) u(n)=u(n+1)-u(n)$ or $(\rho) u(n)=u(n)-u(n-1)$ for any discrete function $u(n)$. For example, $$ {(\rho )n=n+1 \\(\rho) a^n=a^{n+1}. } $$ With this definition, let us find the annihilator of the RHS. Let us first assume the annihilator to be of the form $\rho+k$ for some $k\in \Bbb R$. Then, $$ (\rho+k)(3n+3^n)=3n+3+3\cdot 3^n+3nk+k\cdot 3^n $$ which cannot be zero for any values of $k$. Similarly, one can show that no degree-2 annihilator such as $\rho^2+k_1\rho+k_2$ exists for any values of $k_1$ and $k_2$. We therefore investigate a degree-3 annihilator $L(\rho)=\rho^3+k_1\rho^2+k_2\rho+k_3$. $$ { (\rho^3+k_1\rho^2+k_2\rho+k_3)(3n+3^n)= 3n+9+27\cdot 3^n+k_1(3n+6+9\cdot 3^n) \\+ k_2(3n+3+3\cdot 3^n)+k_3(3n+3^n) \\=3n(1+k_1+k_2+k_3)+3^n(27+9k_1+3k_2+k_3)+9+6k_1+3k_2 } $$For $L(\rho)(3n+3^n)=0$ we should have $$ { 1+k_1+k_2+k_3=0 \\ 27+9k_1+3k_2+k_3=0 \\ 9+6k_1+3k_2=0, } $$which has the following solution $$ k_1=-5\quad,\quad k_2=7\quad,\quad k_3, $$ with the following annihilator $$ L(\rho)=\rho^3-5\rho^2+7\rho-3=(\rho-1)^2(\rho-3) \blacksquare $$