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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

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

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鈥� right sides to make the new system inconsistent? Explain.

Let be a linear transformation from a vector space \(V\) \(T:V \to W\)in to vector space \(W\). Prove that the range of T is a subspace of . (Hint: Typical elements of the range have the form \(T\left( {\mathop{\rm x}\nolimits} \right)\) and \(T\left( {\mathop{\rm w}\nolimits} \right)\) for some \({\mathop{\rm x}\nolimits} ,\,{\mathop{\rm w}\nolimits} \)in \(V\).)\(W\)

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)
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