/*! 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 19 If the roots of the characterist... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 19

If the roots of the characteristic equation are real, show that a solution of \(a y^{\prime \prime}+b y^{\prime}+c y=\) 0 can take on the value zero at most once.

Short Answer

Expert verified
For a second-order linear homogeneous differential equation with real and distinct roots, prove that the solution can take on the value zero at most once. The given differential equation is of the form \(a y'' + b y' + c y = 0\), with a characteristic equation \(ar^2 + br + c = 0\). When the roots of the characteristic equation are real and distinct (\(r_1\) and \(r_2\)), the general solution is \(y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\) where \(C_1\) and \(C_2\) are constants. Suppose the solution takes on the value zero for two distinct points \(x_1\) and \(x_2\). By analyzing the resulting equations and assuming that the solution has two distinct zeros, we derived a contradiction. Therefore, the solution cannot have two distinct zeros, and it can indeed take on the value zero at most once.

Step by step solution

01

Write down the given differential equation

The given second-order linear homogeneous differential equation is: \[a y'' + b y' + c y = 0\]
02

Determine the characteristic equation

The characteristic equation of the given differential equation is: \[ar^2 + br + c = 0\] where r is the characteristic root.
03

Find the solutions for the given roots

The problem states that the roots of the characteristic equation are real. Therefore, we have two real and distinct roots, say \(r_1\) and \(r_2\). The general solution of a second-order linear homogeneous differential equation with real and distinct roots is given by: \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\] where \(C_1\) and \(C_2\) are constants.
04

Prove that the solution can take on the value zero at most once

We need to prove that the solution \(y(x)\) can take on the value zero at most once. Suppose \(y(x) = 0\) for two distinct points \(x_1\) and \(x_2\). Then, we have: \[y(x_1) = C_1 e^{r_1 x_1} + C_2 e^{r_2 x_1} = 0\] \[y(x_2) = C_1 e^{r_1 x_2} + C_2 e^{r_2 x_2} = 0\] From the above equations, we can create another equation: \[C_1 (e^{r_1 x_1} - e^{r_1 x_2}) = C_2 (e^{r_2 x_2} - e^{r_2 x_1})\] Now, let \(k = x_2 - x_1\), which is a nonzero constant since \(x_1\) and \(x_2\) are distinct. Then we can rewrite the above equation as: \[C_1 (e^{r_1 k} - 1) = C_2 (e^{r_2 k} - 1)\] Since the given roots of the characteristic equation are real and distinct, we conclude that \(e^{r_1 k} - 1\) must be nonzero (otherwise, the roots would be equal). Now, there are two possibilities: 1. Case 1: \(C_1 = 0\) \\ In this case, the solution becomes \[y(x) = C_2 e^{r_2 x}\] Since \(r_2\) is real, the exponential function takes the value zero only if \(C_2 = 0\). However, if both \(C_1\) and \(C_2\) are zero, then the solution is trivial, and our assumption that the solution has two distinct zeros is violated. 2. Case 2: \(C_1 \neq 0\) \\ In this case, we can divide both sides of the equation by \(C_1 (e^{r_1 k} - 1)\): \[\frac{C_2}{C_1} = \frac{e^{r_1 k} - 1}{e^{r_2 k} - 1}\] The left-hand side is a constant, while the right-hand side is a function of k. Now if we can show that the right-hand side is not a constant, we will have a contradiction, which implies that a solution cannot have two distinct zeros. The exponentials are continuous functions with respect to k and their derivatives are not equal, as \(r_1 \neq r_2\). Therefore, the right-hand side is not a constant function of k. This contradiction means that a solution cannot have two distinct zeros, which proves that a solution of the given differential equation can take on the value zero at most once.

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.

Homogeneous Differential Equation
In mathematics, a differential equation is termed as 'homogeneous' if every term is a function of the dependent variable and its derivatives. In simpler terms, all the parts of the equation involve the function or its derivatives, without any constant or separate function added. For a second-order linear homogeneous differential equation, we write it in the standard form:
  • \[a y'' + b y' + c y = 0\]
This equation is linear, which means that it can be combined linearly by its solutions. A key feature of homogeneous equations like this one is that if you multiply the function by a constant, the result is still a solution. The goal is usually to find the function \(y(x)\) that satisfies this equation based on the initial conditions provided. Homogeneous differential equations are often solved using characteristic equations, which help determine the nature of their solutions.
Real and Distinct Roots
When solving a homogeneous differential equation, the characteristic equation
  • \[ar^2 + br + c = 0\]
plays a crucial role. This quadratic equation arises from substituting a trial solution of the form \(y = e^{rx}\) into the differential equation. The roots of this characteristic equation, which we denote as \(r_1\) and \(r_2\), determine the form of the solution.If these roots are real and distinct, it means they are two separate, real numbers. This situation leads to a specific type of general solution, characterized by the linear combination of exponential functions:
  • \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
Here, \(C_1\) and \(C_2\) are constants determined by initial conditions. Real and distinct roots indicate different growth or decay rates for the exponential components in the general solution, which helps in determining the solution uniquely based on given conditions.
General Solution
The general solution of a homogeneous differential equation captures all possible solutions and reflects the relationship between the independent variable and the dependent one. Specifically, when the characteristic equation yields real and distinct roots \(r_1\) and \(r_2\), the general solution is expressed as:
  • \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
Each term in this solution represents a particular solution, dependent on coefficients \(C_1\) and \(C_2\), which are constants derived from initial conditions of the problem. The general solution, therefore, is a foundation that encompasses all scenarios possible within the parameters of the differential equation, enabling us to address specific cases by plugging in different values for the constants.
Zero-Crossing
Zero-crossing refers to the points in a solution where the value of the function becomes zero. In the context of the solution for the homogeneous differential equation with real and distinct roots, zero-crossings are insightful for understanding the behavior of the system described by the equation. Considering the general solution:
  • \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
the zero-crossing occurs when \(y(x)\) equals zero. For such an equation with real and distinct roots, the solution can cross zero at most once. This is a crucial limitation. If you suppose that the solution is zero at two distinct points, it results into a contradiction, implying that having more than one zero-crossing is impossible. This behavior stems from the unique characteristics of exponential functions and the distinct nature of the roots, which preclude the possibility of oscillations, thereby limiting the number of zero-crossings.

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

Show that \(A \cos \omega_{0} t+B \sin \omega_{0} t\) can be written in the form \(r \sin \left(\omega_{0} t-\theta\right) .\) Determine \(r\) and \(\theta\) in terms of \(A\) and \(B\). If \(R \cos \left(\omega_{0} t-\delta\right)=r \sin \left(\omega_{0} t-\theta\right),\) determine the relationship among \(R, r, \delta,\) and \(\theta .\)

Consider the initial value problem $$ u^{\prime \prime}+\gamma u^{\prime}+u=0, \quad u(0)=2, \quad u^{\prime}(0)=0 $$ We wish to explore how long a time interval is required for the solution to become "negligible" and how this interval depends on the damping coefficient \(\gamma\). To be more precise, let us seek the time \(\tau\) such that \(|u(t)|<0.01\) for all \(t>\tau .\) Note that critical damping for this problem occurs for \(\gamma=2\) (a) Let \(\gamma=0.25\) and determine \(\tau,\) or at least estimate it fairly accurately from a plot of the solution. (b) Repeat part (a) for several other values of \(\gamma\) in the interval \(0<\gamma<1.5 .\) Note that \(\tau\) steadily decreases as \(\gamma\) increases for \(\gamma\) in this range. (c) Obtain a graph of \(\tau\) versus \(\gamma\) by plotting the pairs of values found in parts (a) and (b). Is the graph a smooth curve? (d) Repeat part (b) for values of \(\gamma\) between 1.5 and \(2 .\) Show that \(\tau\) continues to decrease until \(\gamma\) reaches a certain critical value \(\gamma_{0}\), after which \(\tau\) increases. Find \(\gamma_{0}\) and the corresponding minimum value of \(\tau\) to two decimal places. (e) Another way to proceed is to write the solution of the initial value problem in the form (26). Neglect the cosine factor and consider only the exponential factor and the amplitude \(R\). Then find an expression for \(\tau\) as a function of \(\gamma\). Compare the approximate results obtained in this way with the values determined in parts (a), (b), and (d).

Assume that the system described by the equation \(m u^{\prime \prime}+\gamma u^{\prime}+k u=0\) is critically damped and the initial conditions are \(u(0)=u_{0}, u^{\prime}(0)=v_{0}\), If \(v_{0}=0,\) show that \(u \rightarrow 0\) as \(t \rightarrow \infty\) but that \(u\) is never zero. If \(u_{0}\) is positive, determine a condition on \(v_{0}\) that will assure that the mass passes through its equilibrium position after it is released.

Find the general solution of the given differential equation. In Problems 11 and \(12 \mathrm{g}\) is an arbitrary continuous function. $$ y^{\prime \prime}+4 y^{\prime}+4 y=t^{-2} e^{-2 t}, \quad t>0 $$

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

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.