Difference equation with the annihilator method $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$ I want to find the general solution of the difference equation $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$ with the annihilator method. I can rewrite the left-hand side as $L(y)(n)$ where $L=\tau^2-4$ and $\tau$ is the backshift operator. But I do not know how to rewrite the right-hand side as a solution of some difference equation. Could someone help me out? Thanks in advance!
The right-hand side (RHS) is the linear superposition of $c_1 2^n$ and $c_2$ for constants $c_1$ and $c_2$. The annihilator for the RHS is $(\tau -2)\tau$, where $\tau$ is the timeshift operator. Like you said, the annihilator for the LHS (if the difference equation was homogenous) is $(\tau^2 - 4)$. Knowing that, we can make the original equation homogenous by multiplying the difference equation by the annihilator of the RHS: $(\tau-2)^2(\tau+2)\tau = 0$. The complete solution is $y(n) = a2^n + bn2^n + c(-2)^n + d$. The annihilator for the LHS corresponds to the homogenous solution $y_h(n) = a2^n + c(-2)^n$, leaving the particular solution $y_p(n) = bn2^n + d$. The coefficients $b$ and $d$ can be solved for, while $a$ and $c$ depend on initial conditions.
The right-hand side (RHS) is the linear superposition of $c_1 2^n$ and $c_2$ for constants $c_1$ and $c_2$. The annihilator for the RHS is $(\tau -2)\tau$, where $\tau$ is the timeshift operator. Like you said, the annihilator for the LHS (if the difference equation was homogenous) is $(\tau^2 - 4)$. Knowing that, we can make the original equation homogenous by multiplying the difference equation by the annihilator of the RHS: $(\tau-2)^2(\tau+2)\tau = 0$. The complete solution is $y(n) = a2^n + bn2^n + c(-2)^n + d$. The annihilator for the LHS corresponds to the homogenous solution $y_h(n) = a2^n + c(-2)^n$, leaving the particular solution $y_p(n) = bn2^n + d$. The coefficients $b$ and $d$ can be solved for, while $a$ and $c$ depend on initial conditions.