/*! 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 15 Graph the two basic solutions of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 15

Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common? \(5 \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}-3 y=0\)

Short Answer

Expert verified
The differential equation's solutions show exponential growth or decay, influenced by initial conditions, with all tending to zero as \( x \to -\infty \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \[5 \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}-3 y=0\]
02

Write the Characteristic Equation

For this type of differential equation, we replace \( \frac{d^{2} y}{d x^{2}} \) with \( r^2 \), \( \frac{d y}{d x} \) with \( r \), and \( y \) with 1, giving the characteristic equation:\[ 5r^2 - 2r - 3 = 0 \]
03

Solve the Characteristic Equation

To find the roots, use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 5 \), \( b = -2 \), \( c = -3 \):\[ r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(-3)}}{2(5)} \]Solve:\[ r = \frac{2 \pm \sqrt{4 + 60}}{10} = \frac{2 \pm 8}{10} \]Thus, the roots are \( r_1 = 1 \) and \( r_2 = -\frac{3}{5} \).
04

Write the General Solution

Given that the roots are real and distinct, the general solution is: \[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]Substitute the roots \( r_1 = 1 \) and \( r_2 = -\frac{3}{5} \):\[ y(x) = C_1 e^{x} + C_2 e^{-\frac{3}{5}x} \]
05

Graph the Solution Components

Graph \( y_1(x) = e^x \) and \( y_2(x) = e^{-\frac{3}{5}x} \) separately. These are the basic solutions corresponding to each of the roots.
06

Graph Several Combined Solutions

Select various values for \( C_1 \) and \( C_2 \) (e.g., \( C_1 = 1, C_2 = 0 \); \( C_1 = 0, C_2 = 1 \); \( C_1 = 1, C_2 = 1 \); etc.) and graph \( y(x) = C_1 e^{x} + C_2 e^{-\frac{3}{5}x} \) to show how combinations of these solutions evolve.
07

Identify Common Features of Solutions

All solutions approach zero as \( x \to -\infty \), and tend toward either exponential growth or decay as \( x \to \infty \) depending on the coefficients \( C_1 \) and \( C_2 \). This illustrates stability at negative infinity and variability due to initial conditions affecting growth/decay at positive infinity.

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
In a second-order linear homogeneous differential equation like \[5 \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}-3 y=0\],understanding the characteristic equation is crucial.This equation determines the behavior of the solutions. It is formed by replacing the derivatives with powers of a new variable, usually denoted as \(r\).Here's the transformation:
  • Replace \( \frac{d^{2} y}{d x^{2}} \) with \( r^2 \)
  • Replace \( \frac{d y}{d x} \) with \( r \)
  • Replace \( y \) with \(1\)
The characteristic equation for our problem is \[5r^2 - 2r - 3 = 0\].Solving this equation gives us the roots, which dictate the form of the general solution of the differential equation.
Quadratic Formula
To solve the characteristic equation \[5r^2 - 2r - 3 = 0\],the quadratic formula is our tool of choice. The quadratic formula is \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\).In our case:
  • \(a = 5\)
  • \(b = -2\)
  • \(c = -3\)
Plugging these values into the formula gives:\[r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(-3)}}{2(5)} = \frac{2 \pm \sqrt{4 + 60}}{10} = \frac{2 \pm 8}{10}\].This results in two roots: \(r_1 = 1\) and \(r_2 = -\frac{3}{5}\).These roots are the key to forming the general solution.
General Solution
The general solution of a second-order differential equation is a linear combination of functions, each corresponding to a root of the characteristic equation.For equations with real and distinct roots like ours, the solution form is:\[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\].Given the roots \(r_1 = 1\) and \(r_2 = -\frac{3}{5}\), the general solution becomes: \[y(x) = C_1 e^{x} + C_2 e^{-\frac{3}{5}x}\].
  • \(C_1\) and \(C_2\) are arbitrary constants, determined by initial conditions.
  • This solution shows how different solutions can be composed by varying \(C_1\) and \(C_2\).
This general form captures all possible solutions to our differential equation, considering all boundary or initial conditions.
Exponential Growth and Decay
In the context of differential equations, exponential functions like \(e^{x}\) and \(e^{-\frac{3}{5}x}\)represent growth and decay behaviors.The function \(e^{x}\) implies exponential growth as \(x\) increases, driven by the positive exponent.On the other hand, \(e^{-\frac{3}{5}x}\) shows exponential decay, due to the negative exponent.
  • As \(C_1 e^{x}\) dictates growth, it dominates when \(C_1 > 0\) and the solution grows without bound as \(x\) increases.
  • Meanwhile, \(C_2 e^{-\frac{3}{5}x}\) trends toward zero, showing decay for large \(x\).
As a result, these equations exhibit different behaviors at \(x \to \infty\),making it important to determine \(C_1\) and \(C_2\)'s impact based on initial or boundary conditions.Similarly, as \(x \to -\infty\), both components approach zero, illustrating stability and decay over time.

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

The figure shows a pendulum with length \(L\) and the angle \(\theta\) from the vertical to the pendulum. It can be shown that \(\theta,\) as a function of time, satisfies the nonlinear differential equation $$\frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \sin \theta=0$$ where \(g\) is the acceleration due to gravity. For small values of \(\theta\) we can use the linear approximation \(\sin \theta \approx \theta\) and then the differential equation becomes linear. $$\begin{array}{l}{\text { (a) Find the equation of motion of a pendulum with length }} \\ {1 \mathrm{m} \text { if } \theta \text { is initially } 0.2 \mathrm{rad} \text { and the initial angular velocity }} \\ {\text { is } d \theta / d t=1 \mathrm{rad} / \mathrm{s}}\end{array}$$ $$\begin{array}{l}{\text { (b) What is the maximum angle from the vertical? }} \\\ {\text { (c) What is the period of the pendulum (that is, the time to }} \\\ {\text { complete one back-and-forth swing)? }} \\ {\text { (d) When will the pendulum first be vertical? }} \\ {\text { (e) What is the angular velocity when the pendulum is }} \\ {\text { vertical? }}\end{array}$$

Solve the differential equation using the method of variation of parameters. \(y^{\prime \prime}+3 y^{\prime}+2 y=\sin \left(e^{x}\right)\)

Let \(L\) be a nonzero real number. (a) Show that the boundary-value problem \(y^{\prime \prime}+\lambda y=0\) , \(y(0)=0, y(L)=0\) has only the trivial solution \(y=0\) for the cases \(\lambda=0\) and \(\lambda<0\) . (b) For the case \(\lambda>0,\) find the values of \(\lambda\) for which this problem has a nontrivial solution and give the corresponding solution.

Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients. \(y^{\prime \prime}+2 y+10 y=x^{2} e^{-x} \cos 3 x\)

A spring has a mass of 1 \(\mathrm{kg}\) and its damping constant is \(c=10 .\) The spring starts from its equilibrium position with a velocity of 1 \(\mathrm{m} / \mathrm{s}\) . Graph the position function for the follow- ing values of the spring constant \(\mathrm{k} : 10,20,25,30,40 .\) What type of damping occurs in each case?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

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.