Question

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

${\mathbf{3}}^k$, ${\left(-\mathbf{3}\right)}^k$; $y_{k+\mathbf{2}}-\mathbf{9}{y}_k=\mathbf{0}$

Short answer

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

Step-by-step solution

Check the given difference equation for ${\mathbf{2}}^k$

If $3^k$ is the solution,

$y_{k+2}=3^{k+2}$, $y_k=3^k$.

By the difference equation, you get

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

So, $3^k$ is the solution of the given difference equation.

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

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

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

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