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.

\({y_k} = {\bf{2}}k - {\bf{4}}\); \({y_{k + {\bf{2}}}} + \frac{{\bf{3}}}{{\bf{2}}}{y_{k + {\bf{1}}}} - {y_k} = {\bf{1}} + {\bf{3}}k\)

Short answer

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

Step-by-step solution

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

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

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

So, the signal \({y_k} = 2k - 4\) is the solution of the given the difference equation.

Solve the auxiliary equation for the difference equation

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

\(\begin{aligned} {r^{k + 2}} + \frac{3}{2}{r^{k + 1}} - {r^k} &= 0\\{r^k}\left( {{r^2} + \frac{3}{2}r - 1} \right) &= 0\\2{r^2} + 3r - 2 &= 0\\2{r^2} + 4r - r - 2 &= 0\\\left( {r + 2} \right)\left( {2r - 1} \right) &= 0\\r &= - 2,\,\frac{1}{2}\end{aligned}\)

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

Find the general solution

The general solution of the difference equation is

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

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