/*! 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 12 Refer to the equation $$x^{\pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to the equation $$x^{\prime \prime}+b x^{\prime}+c x=0 .$$ What condition(s) must be satisfied to have a periodic solution? If the solution is periodic, what will its period be?

Short Answer

Expert verified
The condition that must be satisfied for the solution to be periodic is \(b^2 - 4c < 0\). If this condition is met, the period of the solution will be \(T = 2\pi / \sqrt{4c - b^2}\).

Step by step solution

01

Formulate and solve the characteristic equation

The characteristic equation of the differential equation, \(x'' + bx' + cx = 0\), is given by \(m^2 + bm + c = 0\). Solve this quadratic equation to get the roots \(m_{1,2} = [-b \pm \sqrt{(b^2 - 4c)}]/2\).
02

Find the condition for complex roots

For a solution to be periodic, the roots must be complex. This happens when the discriminant of the quadratic equation is negative, i.e., \(b^2 - 4c < 0\). Therefore, this is the condition that must be satisfied for the solution to be periodic.
03

Find the real and imaginary parts of the complex roots

The roots of our characteristic equation will then be of the form \(m_{1,2} = -b/2 \pm i \sqrt{4c - b^2}/2\). The real part, \(-b/2\), will account for any damping in the system, while the coefficient of the imaginary part, \(\sqrt{4c - b^2}/2\), is the angular frequency, \(\omega\), of the solution.
04

Determine the period of the solution

The period, \(T\), of the solution is obtained from the angular frequency, \(\omega\), through the relation \(T = 2\pi / \omega\). So, \(T = 2\pi / \sqrt{4c - b^2}.\)

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 91Ó°ÊÓ!

Key Concepts

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

Periodic Solutions
Periodic solutions in differential equations are solutions that repeat after a certain interval. The term "period" refers to the length of this interval. In the context of the given second-order linear homogeneous differential equation, a periodic solution implies that the system's behavior repeats itself over time without any change. For an equation to have a periodic solution, it must exhibit oscillations, typically indicating the presence of imaginary components in the roots of its characteristic equation. This condition is essential because it implies that the solution forms a pair of complex conjugate roots. These roots contribute to generating the sine and cosine components that lead to oscillatory behavior.
  • What does it mean to be periodic? A function is periodic if it repeats at regular intervals over time.
  • This property is pivotal in many physical systems, like springs and pendulums, highlighting significant patterns in natural phenomena.
Recognizing when a solution is periodic involves analyzing the equation's characteristic roots, ensuring they are complex to form a harmonious oscillation, paving the way for predictable and cyclic behavior.
Characteristic Equation
The characteristic equation is a crucial tool for solving linear differential equations. It emerges when converting a differential equation into an algebraic format, allowing the use of algebraic techniques to find solutions. For the differential equation given, the characteristic equation is formed by substituting the derivatives with powers of an unknown variable, typically represented as:\[ m^2 + bm + c = 0 \]Solving this quadratic equation provides insights into the nature of the solutions of the differential equation. The types of solutions largely depend on the nature of the roots (real or complex) of this characteristic equation. By leveraging the characteristic equation, one can determine whether the differential equation will have exponential, oscillatory, or critically damped solutions.
  • The characteristic equation transforms the problem into finding roots, much like finding x-intercepts in algebra.
  • Understanding this concept helps significantly in identifying the type and behavior of solutions, which are pivotal in engineering and physics applications.
Breaking down differential equations into simpler algebraic problems, like the characteristic equation, enhances one's ability to approach complex systems systematically.
Complex Roots
Complex roots arise when the discriminant of a quadratic equation is negative. In engineering and physics, they play a vital role when studying oscillations and waves. For the characteristic equation \[ m^2 + bm + c = 0 \], complex roots occur when \(b^2 - 4c < 0\). This condition ensures that the quadratic formula yields complex conjugate roots rather than real values.
These complex roots are of the form:\[ m_{1,2} = -\frac{b}{2} \pm i\frac{\sqrt{4c - b^2}}{2} \]Understanding complex roots is valuable as they imply an oscillatory solution in the original differential equation, without exponential growth or decay, provided there's no damping resulting from a non-zero real part. The emergence of complex roots signal conditions where the system neither settles to zero nor increases indefinitely but instead maintains a continuous cyclic nature.
  • Complex roots lead to solutions involving sine and cosine functions, which are characteristic of wave and oscillatory patterns.
  • Systems with such roots do not simply stabilize but fluctuate around a central point, exhibiting patterns seen in everyday oscillatory motion, such as pendulums and electrical circuits.
Angular Frequency
Angular frequency, symbolized as \( \omega \), is a measure of how quickly an oscillating system undergoes its cycles. It is a key concept when dealing with periodic solutions derived from complex roots. For the given differential equation, once the complex roots are identified, the imaginary component contributes to determining the angular frequency:\[ \omega = \frac{\sqrt{4c - b^2}}{2} \]Angular frequency is crucial because it directly relates to the period \( T \) of the solution, defined as the time it takes for one complete cycle. The relation \( T = \frac{2\pi}{\omega} \) guides us to calculate the period:- Faster oscillations yield a higher angular frequency and a shorter period.- Conversely, slower oscillations result in a lower angular frequency and a longer period.Angular frequency unifies our understanding of the speed of oscillation across various systems, making it an invaluable concept in engineering, physics, and even music. Recognizing its role in dictating the timing of oscillatory systems allows for better control and prediction in practical applications.

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

Suppose we modify the Volterra predator-prey equations to reflect competition among prey for limited resources and competition among predators for limited resources. The equations would be of the form $$ \left\\{\begin{array}{l} \frac{d x}{d t}=k_{1} x-k_{2} x^{2}-k_{3} x y \\ \frac{d y}{d t}=-k_{4} y-k_{5} y^{2}+k_{6} x y \end{array}\right. $$ where \(k_{1}, k_{2}, \ldots, k_{6}\) are positive constants. Consider the system $$ \left\\{\begin{array}{l} \frac{d x}{d t}=x(1-0.5 x-y) \\ \frac{d y}{d t}=y(-1-0.5 y+x) \end{array}\right. $$ (a) Find the equilibrium points. (b) Do a qualitative phase-plane analysis. (In fact, solution trajectories will spiral in toward the non-trivial equilibrium point.)

A 5-gallon urn is filled with chai, a milky spicy tea. The chai in the urn is \(90 \%\) tea and \(10 \%\) milk. Chai is being consumed at a rate of \(1 / 2\) gallon per hour and the urn is kept full by adding a mixture that is \(80 \%\) tea and \(20 \%\) milk. Assume that the chai is well-mixed. (a) Write a differential equation whose solution is \(M(t)\), the number of gallons of milk in the urn at time \(t\). Specify the initial condition. (b) Use qualitative analysis to sketch the solution to the differential equation in part (a). (c) How much milk is in the urn after 2 hours?

Solve the given differential equation. \(2 y^{\prime}-y=1\)

A population of otters is declining. New otters are born at a rate proportional to the population with constant of proportionality \(0.04\), but otters die at a rate proportional to the population with constant \(0.09\). Today, the population is 1000 . A group of people wants to try to prevent the otter population from dying out, so they plan to bring in otters from elsewhere at a rate of 40 otters per year. We ll model the situation with continuous functions. Let \(P(t)\) be the population of the otters \(t\) years after today. (a) Write a differential equation whose solution is \(P(t)\). (b) Solve this differential equation. Your answer should include no unknown constants. (c) According to this model, will the attempt to save the otter population work? Explain your answer. If it won \(\mathrm{t}\) work, at what rate must otters be brought in to ensure the population s survival? If it will work, for how many years must the importation of otters continue?

Elmer takes out a $$\$ 100,000$$ loan for a house. He pays money back at a rate of $$\$ 12,000$$ per year. The bank charges him interest at a rate of \(8.5 \%\) per year compounded continuously. Make a continuous model of his economic situation. Write a differential equation whose solution is \(B(t)\), the balance he owes the bank at time \(t\).

See all solutions

Recommended explanations on Math Textbooks

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.