Question

Verify that the signals in Exercises 1 and 2 are solutions of the accompanying difference equation.

\({{\bf{3}}^k}\), \({\left( { - {\bf{3}}} \right)^k}\); \({y_{k + {\bf{2}}}} - {\bf{9}}{y_k} = {\bf{0}}\)

Short answer

\({3^k}\), \({\left( { - 3} \right)^k}\) are the solution of the difference equation \({y_{k + 2}} + 2{y_{k + 1}} - 8{y_k} = 0\).

Step-by-step solution

Check the given difference equation for \({{\bf{2}}^k}\)

If \({3^k}\) is the solution,

\({y_{k + 2}} = {3^{k + 2}}\), \({y_k} = {3^k}\).

By the difference equation, you get

\(\begin{aligned} {3^{k + 2}} - 9\left( {{3^k}} \right) &= 0\\{3^k}\left( {{3^2} - 9} \right) &= 0\\{3^k}\left( {9 - 9} \right) &= 0.\end{aligned}\)

So, \({3^k}\) is the solution of the given difference equation.

Check the given difference equation for \({\left( { - {\bf{4}}} \right)^k}\)

If \({\left( { - 3} \right)^k}\) is the solution,

\({y_{k + 2}} = {\left( { - 3} \right)^{k + 2}}\), \({y_k} = {\left( { - 3} \right)^k}\).

By the difference equation, you get

\(\begin{aligned} {\left( { - 3} \right)^{k + 2}} - 9\left[ {{{\left( { - 3} \right)}^k}} \right] &= 0\\{\left( { - 3} \right)^k}\left( {{{\left( { - 3} \right)}^2} - 9} \right) &= 0\\{\left( { - 3} \right)^k}\left( {9 - 9} \right) &= 0.\end{aligned}\)

So, \({\left( { - 3} \right)^k}\) is the solution of the difference equation.