/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 In Exercises \(13-16,\) find a b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 13

In Exercises \(13-16,\) find a basis for the solution space of the difference equation. Prove that the solutions you find span the solution set. $$ y_{k+2}-y_{k+1}+\frac{2}{9} y_{k}=0 $$

Short Answer

Expert verified
The basis for the solution space is \( \{ (\frac{2}{3})^k, (\frac{1}{3})^k \} \).

Step by step solution

01

Identify the Characteristic Equation

The given difference equation is \( y_{k+2} - y_{k+1} + \frac{2}{9} y_k = 0 \). To solve this, assume a solution of the form \( y_k = r^k \) where \( r \) is a constant. Substitute \( y_k = r^k \) into the difference equation to get the characteristic equation: \( r^2 - r + \frac{2}{9} = 0 \).
02

Solve the Characteristic Equation

Solve the quadratic characteristic equation \( r^2 - r + \frac{2}{9} = 0 \). Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = \frac{2}{9} \). This gives \( r = \frac{1 \pm \sqrt{1 - \frac{8}{9}}}{2} = \frac{1 \pm \sqrt{\frac{1}{9}}}{2} = \frac{1 \pm \frac{1}{3}}{2} \). Thus, \( r_1 = \frac{2}{3} \) and \( r_2 = \frac{1}{3} \).
03

Form the General Solution

The general solution to the difference equation is a combination of the solutions obtained from the roots of the characteristic equation. Therefore, the general solution is \( y_k = C_1 \left( \frac{2}{3} \right)^k + C_2 \left( \frac{1}{3} \right)^k \), where \( C_1 \) and \( C_2 \) are constants.
04

Define the Basis for the Solution Space

The basis for the solution space of the difference equation is given by the set of independent solutions obtained: \( \{ \left( \frac{2}{3} \right)^k, \left( \frac{1}{3} \right)^k \} \). This set is linearly independent because \( \left( \frac{2}{3} \right)^k \) and \( \left( \frac{1}{3} \right)^k \) are distinct exponential functions of \( k \).
05

Prove that the Solutions Span the Solution Set

Any solution to the difference equation can be expressed as a linear combination of the basis elements. Therefore, any solution \( y_k \) of the form \( y_k = C_1 \left( \frac{2}{3} \right)^k + C_2 \left( \frac{1}{3} \right)^k \) can be represented by the basis \( \{ \left( \frac{2}{3} \right)^k, \left( \frac{1}{3} \right)^k \} \). This shows that the solutions span the solution set.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
The idea of a characteristic equation is central to solving linear difference equations. A difference equation is an equation that defines a sequence recursively: each term of the sequence is defined based on its previous terms. To find a solution to such an equation, we use a characteristic equation, which is a type of algebraic equation derived from the original difference equation.
In the exercise provided, we start by assuming a solution of the form \( y_k = r^k \), where \( r \) is a constant. Substituting \( y_k = r^k \) into the original difference equation, \( y_{k+2} - y_{k+1} + \frac{2}{9} y_k = 0 \), gives us the characteristic equation: \( r^2 - r + \frac{2}{9} = 0 \).

This quadratic equation provides the key to understanding the behavior of solutions, as its roots determine the form of the potential solutions. Solving the characteristic equation using the quadratic formula reveals the values of \( r \) (\( r_1 = \frac{2}{3} \) and \( r_2 = \frac{1}{3} \)), which will be used to construct the general solution.
Solution Space
The solution space of a difference equation is the set of all possible solutions to that equation. In other words, it's all the possible sequences you can get from the equation given different starting conditions. An important notion here is the concept of basis for the solution space. A basis is a set of solutions that spans the entire solution space and are linearly independent from each other.
In our exercise, the roots of the characteristic equation \( r_1 = \frac{2}{3} \) and \( r_2 = \frac{1}{3} \) help us define the basis for the solution space. Specifically, the solutions \( \{ \left( \frac{2}{3} \right)^k, \left( \frac{1}{3} \right)^k \} \) form a basis because:
  • Each member is generated from a distinct root of the characteristic equation, ensuring linearly independence because no combination of one solution can produce the other.
  • Any solution of the original equation can be expressed as a linear combination of these basis solutions.
By using these independent solutions, it is possible to express any other solution in the solution space as a combination of these solutions. This is what it means for a basis to span the solution set.
General Solution
The general solution of a difference equation is a vital concept as it encompasses all possible solutions within a form dependent on constants, which can be adjusted to meet initial conditions or other criteria. This makes the general solution highly adaptable.
For the given difference equation, once we determine the roots \( r_1 = \frac{2}{3} \) and \( r_2 = \frac{1}{3} \) from the characteristic equation, we can construct the general solution by forming a linear combination of terms based on these roots:

\[ y_k = C_1 \left( \frac{2}{3} \right)^k + C_2 \left( \frac{1}{3} \right)^k \]
where \( C_1 \) and \( C_2 \) are arbitrary constants.

This general solution is highly significant because:
  • It provides a formula for constructing any specific solution based on initial conditions or constraints by choosing suitable values for \( C_1 \) and \( C_2 \).
  • It includes the complete set of solutions that the original difference equation can yield.
Therefore, understanding how to derive and manipulate the general solution is crucial in solving any difference equation effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(13-16,\) find a basis for the solution space of the difference equation. Prove that the solutions you find span the solution set. $$ y_{k+2}-7 y_{k+1}+12 y_{k}=0 $$

Let \(M_{2 \times 2}\) be the vector space of all \(2 \times 2\) matrices, and define \(T : M_{2 \times 2} \rightarrow M_{2 \times 2}\) by \(T(A)=A+A^{T},\) where \(A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]\) a. Show that \(T\) is a linear transformation. b. 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 \(T(A)=B\) . c. Show that the range of \(T\) is the set of \(B\) in \(M_{2 \times 2}\) with the property that \(B^{T}=B\) . d. Describe the kernel of \(T\)

Let \(\mathcal{B}=\left\\{1, \cos t, \cos ^{2} t, \ldots, \cos ^{6} t\right\\}\) and \(\mathcal{C}=\\{1, \cos t\) \(\cos 2 t, \ldots, \cos 6 t \\} .\) Assume the following trigonometric identities (see Exercise 37 in Section 4.1 ). \(\cos 2 t=-1+2 \cos ^{2} t\) \(\cos 3 t=-3 \cos t+4 \cos ^{3} t\) \(\cos 4 t=1-8 \cos ^{2} t+8 \cos ^{4} t\) \(\cos 5 t=5 \cos t-20 \cos ^{3} t+16 \cos ^{5} t\) \(\cos 6 t=-1+18 \cos ^{2} t-48 \cos ^{4} t+32 \cos ^{6} t\) Let \(H\) be the subspace of functions spanned by the functions in \(\mathcal{B} .\) Then \(\mathcal{B}\) is a basis for \(H,\) by Exercise 38 in Section \(4.3 .\) a. Write the \(\mathcal{B}\) -coordinate vectors of the vectors in \(\mathcal{C},\) and use them to show that \(\mathcal{C}\) is a linearly independent set in \(H .\) b. Explain why \(\mathcal{C}\) is a basis for \(H\)

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Suppose that \(T\) is a one-to-one transformation, so that an equation \(T(\mathbf{u})=T(\mathbf{v})\) always implies \(\mathbf{u}=\mathbf{v} .\) Show that if the set of images \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\}\) is linearly dependent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Exercises 17 and 18 concern a simple model of the national economy described by the difference equation $$ Y_{k+2}-a(1+b) Y_{k+1}+a b Y_{k}=1 $$ Here \(Y_{k}\) is the total national income during year \(k, a\) is a constant less than \(1,\) called the marginal propensity to consume, and \(b\) is a positive constant of adjustment that describes how changes in consumer spending affect the annual rate of private investment. Find the general solution of equation \((14)\) when \(a=.9\) and \(b=.5 .\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.