In Exercises 25-28, show that the given signal is a solution of the difference equation. Then find the general solution of that difference\({y_k} = {\bf{1}} + k\) equation.

; \({y_{k + {\bf{2}}}} - {\bf{8}}{y_{k + {\bf{1}}}} + {\bf{15}}{y_k} = {\bf{2}} + {\bf{8}}k\)

Use \({y_k} = {k^2}\) in the difference equation.

\(\begin{aligned} {y_{k + 2}} - 8{y_{k + 1}} + 15{y_k} &= \left( {k + 3} \right) - 8\left( {k + 2} \right) + 15\left( {k + 1} \right)\\ &= k + 3 - 8k - 16 + 15k + 15\\ &= 8k + 2\end{aligned}\)

So, the signal \({y_k} = 1 + k\) is the solution of the given difference equation.

The auxiliary equation for the difference equation \({y_{k + 2}} - 8{y_{k + 1}} + 15{y_k} = 8k + 2\) is obtained below:

\(\begin{aligned} {r^{k + 2}} - 8{r^{k + 1}} + 15{r^k} &= 0\\{r^k}\left( {{r^2} - 8r + 15} \right) &= 0\\{r^2} - 5r - 3r + 15 &= 0\\\left( {r - 3} \right)\left( {r - 5} \right) &= 0\\r &= 3,5\end{aligned}\)

So, the general solutions of the auxiliary set are \({3^k}\) and \({5^k}\).

Thegeneral solution of the difference equation is

\(y = {c_1}{\left( 3 \right)^k} + {c_2}{\left( 5 \right)^k} + \left( {1 + k} \right)\).

So, the given signal is the solution of the difference equation, and the general solution of the difference equation is \(y = {c_1}{\left( 3 \right)^k} + {c_2}{\left( 5 \right)^k} + \left( {1 + k} \right)\).