/*! 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 30 (a) Prove that solutions need no... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 30

(a) Prove that solutions need not be unique for nonlinear initial-value problems by finding two solutions to $$y \frac{d y}{d x}=x, \quad y(0)=0$$ (b) Prove that solutions need not exist for nonlinear initial-value problems by showing that there is no solution for $$y \frac{d y}{d x}=-x, \quad y(0)=0$$

Short Answer

Expert verified
(a) There are two solutions: \(y = x\) and \(y = -x\). (b) No solution exists.

Step by step solution

01

Solve first differential equation using separation of variables

Start with the differential equation \(y \frac{dy}{dx} = x\). This can be solved by separating variables. Rewrite it as \(y \, dy = x \, dx\). Integrate both sides: \(\int y \, dy = \int x \, dx\), which results in \(\frac{1}{2}y^2 = \frac{1}{2}x^2 + C\). Simplifying leads to \(y^2 = x^2 + 2C\).
02

Apply initial condition for first solution

Use the initial condition \(y(0) = 0\). Substitute \(y = 0\) and \(x = 0\) into the equation \(y^2 = x^2 + 2C\) to get \(0 = 0 + 2C\). This implies \(C = 0\). Therefore, one solution is \(y^2 = x^2\), or \(y = x\) and \(y = -x\).
03

Identify multiple solutions for first problem

The previous step revealed that two possible solutions arise from \(y^2 = x^2\): \(y = x\) and \(y = -x\). Both solutions satisfy the differential equation and the initial condition \(y(0) = 0\). Thus, we have non-unique solutions.
04

Analyze second differential equation

Consider \(y \frac{dy}{dx} = -x\). Rewrite it as \(y \, dy = -x \, dx\) and integrate both sides: \(\int y \, dy = \int -x \, dx\), giving \(\frac{1}{2}y^2 = -\frac{1}{2}x^2 + C\), or \(y^2 = -x^2 + 2C\).
05

Apply initial condition for second problem

Using \(y(0) = 0\) leads to \(\frac{1}{2}(0)^2 = -\frac{1}{2}(0)^2 + C\), resulting in \(C = 0\). Therefore, \(y^2 = -x^2\), which requires both sides to be non-negative.
06

Show non-existence of solutions for second problem

The equation \(y^2 = -x^2\) implies \(y^2 = -(x^2)\). Since a square cannot be negative, there is no real \(y\) that can satisfy this equation for any real \(x\). Thus, no solution exists for the initial value problem.

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.

Differential Equations
A differential equation involves relationships between functions and their derivatives. It represents change and is fundamental in modeling real-world phenomena like motion, growth, and decay. Differential equations can be classified as either linear or nonlinear, depending on how the unknown function and its derivatives appear in the equation.
Linear differential equations yield solutions that form straight lines when graphed, while nonlinear differential equations, like the ones discussed in the problem, can produce more complex curves, often leading to intricate and diverse solutions.
Nonlinear initial-value problems are particularly intriguing because they sometimes produce unexpected behaviors such as multiple solutions or even no solutions at all. This complexity arises due to how the dependent variable, such as the function we want to find, is squared or cubed, or involves other nonlinear operations. Understanding nonlinear differential equations opens up insight into complex systems and patterns that occur naturally.
Separation of Variables
The separation of variables is a powerful method used for solving certain types of differential equations. This technique works best for **separable equations**, where variables can be independently represented on either side of the equation.
In our first differential equation, \( y \frac{dy}{dx} = x \), we can separate the variables by rewriting it as \( y \, dy = x \, dx \). By doing so, each side depends on only one variable, allowing us to integrate both sides independently.
The integration results in \( \frac{1}{2}y^2 = \frac{1}{2}x^2 + C \), leading to the expression \( y^2 = x^2 + 2C \). By applying the initial condition, the value(s) of \( C \) is determined. The strength of this method lies in its simplicity, enabling us to find solutions effectively when a differential equation can be made separable.
Existence and Uniqueness of Solutions
The existence and uniqueness of solutions in differential equations are deep concepts that deal with whether a solution can be found, and if so, whether it's unique. These properties are especially crucial for initial-value problems, where a specific starting condition is given.
For the first differential equation, after solving it, we found the solution set \( y = x \) and \( y = -x \). Both equations satisfy the initial condition where \( y(0) = 0 \). Hence, this problem exhibits non-unique solutions. Non-uniqueness means that more than one function fits both the differential equation and the initial condition.
On the other hand, for the second differential equation \( y \frac{dy}{dx} = -x \), we face a non-existence of solutions due to the impossibility of squaring a real number to get a negative number, as posed by the equation \( y^2 = -x^2 \). Therefore, no real functions satisfy both the equation and the initial condition. Thus, exploring existence and uniqueness highlights the diverse nature of solutions and guides us in diagnosing the behavior of differential equations.

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 each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. \(y^{\prime \prime}+4 y=0\) (a) \(\sin 2 x\) and \(\cos 2 x\) (b) \(c_{1} \sin 2 x+c_{2} \cos 2 x\left(c_{1}, c_{2} \text { constants }\right)\)

Fish is a rich source of protein and omega-3 fatty acids. However, some fish, such as shark, swordfish, and tuna, contain high concentrations of methylmercury and should be eaten only in moderation. For example, suppose a man starts a new diet in which he consumes 425 grams of albacore tuna per week (roughly equivalent to three 5 oz cans). His body will absorb about 145 micrograms \((\mu \mathrm{g})\) of methylmercury from the tuna. On the other hand, approximately \(9 \%\) of any given quantity of methylmercury within his system will be eliminated in a week. Assume that there are \(300 \mu \mathrm{g}\) of methylmercury in his body when he begins the diet and that afterwards his tuna consumption is the only source of additional amounts of the compound. Let \(y(t)\) denote the number of micrograms of methylmercury in his body \(t\) weeks after the start of his diet. (a) Find an initial-value problem whose solution is \(y(t)\) [Hint: Model this as a mixing problem.] (b) Find a formula for \(y(t)\). (c) Suppose the man weighs 160 lb. According to U.S. EPA guidelines, the amount of methylmercury within his body should be at or below \(569 \mu \mathrm{g}\) to avoid toxicity. Based upon your solution to part (b), determine whether the man should modify his diet.

Radon-222 is a radioactive gas with a half-life of 3.83 days. This gas is a health hazard because it tends to get trapped in the basements of houses, and many health officials suggest that homeowners seal their basements to prevent entry of the gas. Assume that \(5.0 \times 10^{7}\) radon atoms are trapped in a basement at the time it is sealed and that \(y(t)\) is the number of atoms present \(t\) days later. (a) Find an initial-value problem whose solution is \(y(t).\) (b) Find a formula for \(y(t).\) (c) How many atoms will be present after 30 days? (d) How long will it take for \(90 \%\) of the original quantity of gas to decay?

In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. \(y^{\prime \prime}-y^{\prime}-6 y=0\) (a) \(e^{-2 x}\) and \(e^{3 x}\) (b) \(c_{1} e^{-2 x}+c_{2} e^{3 x}\left(c_{1}, c_{2} \text { constants }\right)\)

State the order of the differential equation, and confirm that the functions in the given family are solutions. (a) \(2 \frac{d y}{d x}+y=x-1 ; y=c e^{-x / 2}+x-3\) (b) \(y^{\prime \prime}-y=0 ; y=c_{1} e^{t}+c_{2} e^{-t}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

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.