Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 3: Problem 4

In each of Problems 1 through 8 determine whether the given pair of functions is linearly independent or linearly dependent. \(f(x)=e^{3 x}, \quad g(x)=e^{3(x-1)}\)

Short Answer

Expert verified
Answer: Linearly dependent.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Find the first derivatives of the given functions

First, we need to find the first derivatives for each function: \(f(x) = e^{3x}\) \(f'(x) = 3e^{3x}\) \(g(x) = e^{3(x-1)}\) \(g'(x) = 3e^{3(x-1)}\)
02

Calculate the Wronskian of the functions

The Wronskian for a pair of functions is defined as: \(W(x) = \det(\begin{bmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{bmatrix})\) We will calculate the Wronskian using the functions and their derivatives: \(W(x) = \det(\begin{bmatrix} e^{3x} & e^{3(x-1)} \\ 3e^{3x} & 3e^{3(x-1)} \end{bmatrix})\) To find the determinant of this matrix, we perform the following operations: \(W(x) = e^{3x} \cdot 3e^{3(x-1)} - e^{3(x-1)} \cdot 3e^{3x}\) Now we can eliminate the repeated term and simplify the equation: \(W(x) = 3e^{3x}e^{3(x-1)} - 3e^{3(x-1)}e^{3x} = 0\)
03

Analyze the result to determine linear independence or dependence

Since the Wronskian is equal to 0, it follows that the given pair of functions \(f(x)=e^{3x}\) and \(g(x)=e^{3(x-1)}\) are linearly dependent.

Key Concepts

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

Differential Equations
Differential equations play a critical role in mathematics, acting as equations that describe the relationship between a function and its derivatives. They are widely used in various fields such as physics, engineering, and economics to model how things change over time. A differential equation can be simple, involving a single derivative, or complex, involving multiple derivatives and variables.
One of the key types of differential equations is the linear differential equation, where the unknown function and its derivatives appear to the first power and do not multiply each other. Understanding whether the functions forming the solutions to these equations are linearly independent or not is vital. This involves concepts such as the Wronskian determinant, which helps in determining solutions' characteristics.
Studying differential equations requires a solid understanding of calculus and how derivatives represent change. It provides a foundational framework for modeling dynamic systems and solving complex real-world problems.
Wronskian Determinant
The Wronskian determinant is a mathematical tool used to determine whether a set of functions are linearly independent. This can be useful, for example, in solving systems of differential equations where knowing the independence of solutions is crucial.
The Wronskian of two functions, say \(f(x)\) and \(g(x)\), is calculated by setting up a determinant of a matrix that includes the functions and their first derivatives:
  • The first row consists of the functions themselves: \(f(x)\) and \(g(x)\).
  • The second row includes their derivatives: \(f'(x)\) and \(g'(x)\).
The Wronskian is given by:\[W(x) = \det\left(\begin{array}{cc} f(x) & g(x) \ f'(x) & g'(x) \end{array}\right)\].
For example, if \(W(x) = 0\) at a particular point, it suggests that the functions may be linearly dependent in that interval. It is an excellent check for independence among solutions to differential equations, helping to ensure we have a complete set of solutions when addressing linear problems.
Linearly Dependent Functions
Linearly dependent functions are those that can be expressed as a linear combination of each other with non-zero coefficients. In simpler terms, if one function can be composed by multiplying another function by a constant, they are linearly dependent. This concept is crucial in understanding the behavior of solutions to differential equations.
For instance, considering the functions \(f(x) = e^{3x}\) and \(g(x) = e^{3(x-1)}\) from the Wronskian calculation, they were found to be linearly dependent because their Wronskian is zero. This means that one is just a shifted version of the other: \(g(x) = e^{-3}f(x)\).
Understanding this dependence helps in solving problems where the independence of solutions is necessary. In the context of a system of equations, if all solutions are linearly dependent, it might indicate redundancies or symmetries in the system. Hence, in mathematics, ensuring functions are independent when needed is fundamental to forming a solid solution basis.

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

Assume that the system described by the equation \(m u^{\prime \prime}+\gamma u^{\prime}+k u=0\) is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Hint: Determine all possible values of \(t\) for which \(u=0\).

In many physical problems the nonhomogencous term may be specified by different formulas in different time periods. As an example, determine the solution \(y=\phi(t)\) of $$ y^{\prime \prime}+y=\left\\{\begin{array}{ll}{t,} & {0 \leq t \leq \pi} \\\ {\pi e^{x-t},} & {t>\pi}\end{array}\right. $$ $$ \begin{array}{l}{\text { satisfying the initial conditions } y(0)=0 \text { and } y^{\prime}(0)=1 . \text { Assume that } y \text { and } y^{\prime} \text { are also }} \\ {\text { continuous at } t=\pi \text { . Plot the nonhomogencous term and the solution as functions of time. }} \\ {\text { Hint: First solve the initial value problem for } t \leq \pi \text { ; then solve for } t>\pi \text { , determining the }} \\ {\text { constants in the latter solution from the continuity conditions at } t=\pi \text { . }}\end{array} $$

A mass of \(5 \mathrm{kg}\) stretches a spring \(10 \mathrm{cm} .\) The mass is acted on by an external force of \(10 \mathrm{sin}(t / 2) \mathrm{N}\) (newtons) and moves in a medium that imparts a viscous force of \(2 \mathrm{N}\) when the speed of the mass is \(4 \mathrm{cm} / \mathrm{sec} .\) If the mass is set in motion from its equilibrium position with an initial velocity of \(3 \mathrm{cm} / \mathrm{sec}\), formulate the initial value problem describing the motion of the mass.

Consider the initial value problem $$ m u^{\prime \prime}+\gamma u^{\prime}+k u=0, \quad u(0)=u_{0}, \quad u^{\prime}(0)=v_{0} $$ Assume that \(\gamma^{2}<4 k m .\) (a) Solve the initial value problem, (b) Write the solution in the form \(u(t)=R \exp (-\gamma t / 2 m) \cos (\mu t-\delta) .\) Determine \(R\) in terms of \(m, \gamma, k, u_{0},\) and \(v_{0}\). (c) Investigate the dependence of \(R\) on the damping coefficient \(\gamma\) for fixed values of the other parameters.

A cubic block of side \(l\) and mass density \(\rho\) per unit volume is floating in a fluid of mass density \(\rho_{0}\) per unit volume, where \(\rho_{0}>\rho .\) If the block is slightly depressed and then released, it oscillates in the vertical direction. Assuming that the viscous damping of the fluid and air can be neglected, derive the differential equation of motion and determine the period of the motion. Hint Use archimedes' principle: An object that is completely or partially submerged in a fluid is acted on by an upward (bouyant) equal to the weight of the displaced fluid.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

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.

Sign-up for free