Question

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

\({{\bf{2}}^k}\),\({\left( { - {\bf{4}}} \right)^k}\);\({y_{k + {\bf{2}}}} + {\bf{2}}{y_{k + {\bf{1}}}} - {\bf{8}}{y_k} = {\bf{0}}\)

Short answer

\({2^k}\), \({\left( { - 4} \right)^k}\) are the solutions 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\({2^k}\)is the solution,

\({y_{k + 2}} = {2^{k + 2}}\),\({y_{k + 1}} = {2^{k + 1}}\)and\({y_k} = {2^k}\).

By the difference equation, you get

\(\begin{aligned}{c}{2^{k + 2}} + 2\left( {{2^{k + 1}}} \right) - 8\left( {{2^k}} \right) &= 0\\{2^k}\left( {{2^2} + {2^2} - 8} \right) &= 0\\{2^k}\left( {8 - 8} \right) &= 0.\end{aligned}\)

So,\({2^k}\)is the solution of the given difference equation.

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

If\({\left( { - 4} \right)^k}\)is the solution,

\({y_{k + 2}} = {\left( { - 4} \right)^{k + 2}}\), \({y_{k + 1}} = {\left( { - 4} \right)^{k + 1}}\) and \({y_k} = {\left( { - 4} \right)^k}\).

By the difference equation, you get

\(\begin{aligned} {\left( { - 4} \right)^{k + 2}} + 2\left[ {{{\left( { - 4} \right)}^{k + 1}}} \right] - 8\left[ {{{\left( { - 4} \right)}^k}} \right] &= 0\\{\left( { - 4} \right)^k}\left( {{{\left( { - 4} \right)}^2} + 2\left( { - 4} \right) - 8} \right) &= 0\\{\left( { - 4} \right)^k}\left( {16 + 8 - 8} \right) &= 0.\end{aligned}\)

So, \({\left( { - 4} \right)^k}\) is the solution of the difference equation.