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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.

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

Short Answer

Expert verified

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

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.

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

So, the signal \({y_k} = 2k - 4\) is the solution of the given the difference equation.

02

Solve the auxiliary equation for the difference equation

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

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

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

03

Find the general solution

The general solution of the difference equation is

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

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

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Most popular questions from this chapter

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 \(m \times n\).

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{2}}}\)on the line\(y = {\bf{5}}x\).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

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