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Chapter 4: Q25E (page 191)

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.

\({y_k} = {k^{\bf{2}}}\); \({y_{k + {\bf{2}}}} + {\bf{3}}{y_{k + {\bf{1}}}} - {\bf{4}}{y_k} = {\bf{7}} + {\bf{10}}k\)

Short Answer

Expert verified

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

Step by step solution

01

Substitute \({k^{\bf{2}}}\) for \({y_k}\) in the difference equation

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

\(\begin{aligned} {y_{k + 2}} + 3{y_{k + 1}} - 4{y_k} &= {\left( {k + 2} \right)^2} + 3{\left( {k + 1} \right)^2} - 4{k^2}\\ &= \left( {{k^2} + 4k + 4} \right) + 3\left( {{k^2} + 2k + 1} \right) - 4{k^2}\\ &= {k^2} + 4k + 4 + 3{k^2} + 6k + 3 - 4{k^2}\\ &= 10k + 7\end{aligned}\)

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

02

Solve the auxiliary equation for the difference equation

The auxiliary equation for the difference equation \({y_{k + 2}} + 3{y_{k + 1}} - 4{y_k} = 10k + 7\) is obtained below:

\(\begin{aligned} {r^{k + 2}} + 3{r^{k + 1}} - 4{r^k} = 0\\{r^k}\left( {{r^2} + 3r - 4} \right) &= 0\\{r^2} + 3r - 4 &= 0\\\left( {r + 4} \right)\left( {r - 1} \right) &= 0\\r &= - 4,1\end{aligned}\)

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

03

Find the general solution

Thegeneral solution of the difference equation is \(y = {c_1}{\left( { - 4} \right)^k} + {c_2} + {k^2}\).

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

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Most popular questions from this chapter

Verify that rank \({{\mathop{\rm uv}\nolimits} ^T} \le 1\) if \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\\5\end{array}} \right]\) and \({\mathop{\rm v}\nolimits} = \left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right]\).

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).
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