Question

In Exercises 25-28, show that the given signal is a solution of the difference equation. Then find the general solution of that difference equation.

\({y_k} = {\bf{2}} - {\bf{2}}k\); \({y_{k + {\bf{2}}}} - \frac{{\bf{9}}}{{\bf{2}}}{y_{k + {\bf{1}}}} + {\bf{2}}{y_k} = {\bf{2}} + {\bf{3}}k\)

Short answer

The given signal is the solution of the difference equation, and the general solution of the difference equation is \(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( 4 \right)^k} + \left( {2 - 2k} \right)\).

Step-by-step solution

Substitute \({\bf{1}} + k\) for \({y_k}\) in the difference equation

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

\(\begin{aligned} {y_{k + 2}} - \frac{9}{2}{y_{k + 1}} + 2{y_k} &= \left( {2 - 2\left( {k + 2} \right)} \right) - \frac{9}{2}\left( {2 - 2\left( {k + 1} \right)} \right) + 2\left( {2 - 2k} \right)\\ &= - 2k - 2 - 9 + 9k + 9 + 4 - 4k\\ &= 3k + 2\end{aligned}\)

So, the signal \({y_k} = 2 - 2k\) is the solution of the given difference equation.

Solve the auxiliary equation for the difference equation

The auxiliary equation for the difference equation \({y_{k + 2}} - \frac{9}{2}{y_{k + 1}} + 2{y_k} = 2 + 3k\) is obtained below:

\(\begin{aligned} {r^{k + 2}} - \frac{9}{2}{r^{k + 1}} + 2{r^k} = 0\\{r^k}\left( {{r^2} - \frac{9}{2}r + 2} \right) &= 0\\2{r^2} - 9r + 4 &= 0\\2{r^2} - 8r - r + 4 &= 0\\\left( {2r - 1} \right)\left( {r - 4} \right) &= 0\\r &= \frac{1}{2},\,4\end{aligned}\)

So, the general solutions of the auxiliary set are \({\frac{1}{2}^k}\) and \({4^k}\).

Find the general solution

The general solution of the difference equation is

\(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( 4 \right)^k} + \left( {2 - 2k} \right)\).

So, the given signal is the solution of the difference equation, and the general solution of the difference equation is \(y = {c_1}{\left( {\frac{1}{2}} \right)^k} + {c_2}{\left( 4 \right)^k} + \left( {2 - 2k} \right)\).