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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.

\({y_k} = {\bf{2}} - {\bf{2}}k\); \({y_{k + {\bf{2}}}} - \frac{{\bf{9}}}{{\bf{2}}}{y_{k + {\bf{1}}}} + {\bf{2}}{y_k} = {\bf{2}} + {\bf{3}}k\)

Short Answer

Expert verified

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

Step by step solution

01

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

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

\(\begin{aligned} {y_{k + 2}} - \frac{9}{2}{y_{k + 1}} + 2{y_k} &= \left( {2 - 2\left( {k + 2} \right)} \right) - \frac{9}{2}\left( {2 - 2\left( {k + 1} \right)} \right) + 2\left( {2 - 2k} \right)\\ &= - 2k - 2 - 9 + 9k + 9 + 4 - 4k\\ &= 3k + 2\end{aligned}\)

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

02

Solve the auxiliary equation for the difference equation

The auxiliary equation for the difference equation \({y_{k + 2}} - \frac{9}{2}{y_{k + 1}} + 2{y_k} = 2 + 3k\) is obtained below:

\(\begin{aligned} {r^{k + 2}} - \frac{9}{2}{r^{k + 1}} + 2{r^k} = 0\\{r^k}\left( {{r^2} - \frac{9}{2}r + 2} \right) &= 0\\2{r^2} - 9r + 4 &= 0\\2{r^2} - 8r - r + 4 &= 0\\\left( {2r - 1} \right)\left( {r - 4} \right) &= 0\\r &= \frac{1}{2},\,4\end{aligned}\)

So, the general solutions of the auxiliary set are \({\frac{1}{2}^k}\) and \({4^k}\).

03

Find the general solution

The general solution of the difference equation is

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

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

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Most popular questions from this chapter

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Consider the polynomials \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}},{p_{\bf{2}}}\left( t \right) = {\bf{1}} - {t^{\bf{2}}}\). Is \(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}}} \right\}\) a linearly independent set in \({{\bf{P}}_{\bf{3}}}\)? Why or why not?

Suppose a nonhomogeneous system of six linear equations in eight unknowns has a solution, with two free variables. Is it possible to change some constants on the equations’ right sides to make the new system inconsistent? Explain.

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.
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