Question

Show that the signals in Exercises 3-6 form a basis for the solution set of the accompanying difference equation.

\({{\bf{3}}^k}\), \(k{\left( { - {\bf{3}}} \right)^k}\) \({y_{k + {\bf{2}}}} + {\bf{6}}{y_{k + {\bf{1}}}} + {\bf{9}}{y_k} = {\bf{0}}\)

Short answer

\({3^k}\) and \(k{\left( { - 3} \right)^k}\) form a basis of the solution set for the difference equation.

Step-by-step solution

Check for \({\left( { - {\bf{3}}} \right)^k}\)

Substitute \({y_k} = {\left( { - 3} \right)^k}\) in thedifference equation \({y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} &= {\left( { - 3} \right)^{k + 2}} + 6{\left( { - 3} \right)^{k + 1}} + 9{\left( { - 3} \right)^k}\\ &= {\left( { - 3} \right)^k}\left[ {{{\left( { - 3} \right)}^2} + 6\left( { - 3} \right) + 9} \right]\\ &= {\left( 3 \right)^k}\left[ {9 - 18 + 9} \right]\\ &= 0\end{aligned}\)

Check for \(k{\left( { - {\bf{3}}} \right)^k}\)

Substitute\({y_k} = k{\left( { - 3} \right)^k}\)in the difference equation\({y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 6{y_{k + 1}} + 9{y_k} &= \left( {k + 2} \right){\left( { - 3} \right)^{k + 2}} + 6\left( {k + 1} \right){\left( { - 3} \right)^{k + 1}} + 9k{\left( { - 3} \right)^k}\\ &= {\left( { - 3} \right)^k}\left[ {\left( {k + 2} \right){{\left( { - 3} \right)}^2} + 6\left( {k + 1} \right)\left( { - 3} \right) + 9k} \right]\\ &= {\left( 3 \right)^k}\left[ {9k + 18 - 18k - 18 + 9k} \right]\\ &= 0\end{aligned}\)

Thus, for all\(k\),\({\left( { - 3} \right)^k}\)is in the solution set H.

Check whether the solution is the basis of the difference equation

For all k, in n-dimensional vector space, the dimension of H is 2. So, \({3^k}\) and \(k{\left( { - 3} \right)^k}\) forms a basis of the solution set for the difference equation.