/*! 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 20 A more detailed version of Theor... [FREE SOLUTION] | 91Ó°ÊÓ

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A more detailed version of Theorem 1 says that, if the function \(f(x, y)\) is continuous near the point \((a, b)\), then at least one solution of the differential equation \(y^{\prime}=f(x, y)\) exists on some open interval I containing the point \(x=a\) and, moreover, that if in addition the partial derivative \(\partial f / \partial y\) is continuous near \((a, b)\), then this solution is unique on some (perhaps smaller) interval \(J .\) In Problems 11 through 20, determine whether \(e x\) istence of at least one solution of the given initial value problem is thereby guaranteed and, if so, whether uniqueness of that solution is guaranteed. $$ \frac{d y}{d x}=x^{2}-y^{2} ; \quad y(0)=1 $$

Short Answer

Expert verified
Existence and uniqueness of a solution are guaranteed.

Step by step solution

01

Identify the Function and Conditions

The given differential equation is \( \frac{dy}{dx} = x^2 - y^2 \) with the initial condition \( y(0) = 1 \). We need to check the continuity of the function \( f(x, y) = x^2 - y^2 \) and its partial derivative with respect to \( y \).
02

Check Continuity of \( f(x, y) \)

The function \( f(x, y) = x^2 - y^2 \) is a polynomial in both \( x \) and \( y \). Polynomial functions are continuous everywhere, so \( f(x, y) \) is continuous near the point \( (0, 1) \). This satisfies the continuity condition for existence of at least one solution.
03

Differentiate \( f(x, y) \) with Respect to \( y \)

Compute the partial derivative of \( f(x, y) \) with respect to \( y \):\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = -2y. \]
04

Check Continuity of \( \frac{\partial f}{\partial y} \)

The derivative \( \frac{\partial f}{\partial y} = -2y \) is also a polynomial in \( y \) and hence continuous everywhere. Thus, \( \partial f / \partial y \) is continuous near the point \( (0, 1) \).
05

Conclude Existence and Uniqueness

Since \( f(x, y) \) is continuous near \( (0, 1) \), at least one solution to the differential equation exists around the initial condition \( y(0) = 1 \). Additionally, the continuity of the partial derivative \( \frac{\partial f}{\partial y} \) ensures that the solution is unique near \( x = 0 \).

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Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. It shows how a function changes as its variables change, enabling us to describe a wide variety of physical, biological, and chemical processes. In many applications, differential equations are used to model systems over time.

The given equation \( \frac{dy}{dx} = x^2 - y^2 \) is a first-order differential equation since it involves the first derivative of \( y \) with respect to \( x \). This equation tells us how the rate of change of \( y \) depends on both \( x \) and \( y \).

Understanding the nature of these equations is crucial for solving them. Solutions to differential equations are functions themselves, rather than numbers. These solutions provide insights into the behavior of systems described by the equations.
Initial Value Problems
An initial value problem consists of a differential equation along with specified values, often called boundary conditions, that the solution must satisfy at a particular point. For the equation \( \frac{dy}{dx} = x^2 - y^2 \) with the initial condition \( y(0) = 1 \), you are looking for a function \( y(x) \) such that when \( x = 0 \), \( y = 1 \).

Initial value problems are crucial because they allow us to uniquely determine a single solution from the infinite family of solutions to the differential equation. This means that among all functions satisfying the differential equation, only one will pass through the specified initial condition. By ensuring that the function meets this specific point, we can predict how the system behaves beyond just the initial state.
Continuity of Functions
Continuity is an essential concept when it comes to determining the existence of solutions to differential equations. A function is said to be continuous if small changes in the input result in small changes in the output, without any abrupt jumps or gaps.

For instance, the function \( f(x, y) = x^2 - y^2 \) is continuous across all values of \( x \) and \( y \) because it is composed of polynomial terms. Polynomials are inherently continuous functions. Therefore, this continuity implies there will be no sudden changes in the rate of our differential equation's behavior around the given initial point \( (0, 1) \).

This property of continuity is what helps ensure that at least one solution to our differential equation exists in a neighborhood around the initial condition provided.
Partial Derivatives
Partial derivatives are derivatives with respect to one variable while keeping others constant. In the context of differential equations, they help in understanding how a function changes with respect to one variable while observing the impact of others.

Consider the partial derivative \( \frac{\partial f}{\partial y} \) for our function \( f(x, y) = x^2 - y^2 \). When differentiating \( f(x, y) \) with respect to \( y \), we treat \( x \) as a constant, yielding \( \frac{\partial f}{\partial y} = -2y \). This derivative shows the sensitivity of the function to changes in \( y \) specifically.

Since \( \frac{\partial f}{\partial y} \) is continuous (as it is a polynomial), it suggests that any change in \( y \) is gradual, without abrupt alterations. This continuity ensures the uniqueness of the solution to the differential equation around the specified initial condition.

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Most popular questions from this chapter

A tank contains 1000 liters (L) of a solution consisting of \(100 \mathrm{~kg}\) of salt dissolved in water. Pure water is pumped into the tank at the rate of \(5 \mathrm{~L} / \mathrm{s}\), and the mixture - kept uniform by stirring \(-\) is pumped out at the same rate. How long will it be until only \(10 \mathrm{~kg}\) of salt remains in the tank?

The half-life of radioactive cobalt is \(5.27\) years. Suppose that a nuclear accident has left the level of cobalt radiation in a certain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable? (Ignore the probable presence of other radioactive isotopes.)

Find a general solution of each reducible second-order differential equation. Assume \(x, y\) and/or \(y^{\prime}\) positive where helpful (as in Example I1). $$ y^{\prime \prime}+4 y=0 $$

Consider the Clairaut equation $$ y=x y^{\prime}-\frac{1}{4}\left(y^{\prime}\right)^{2} $$ for which \(g\left(y^{\prime}\right)=-\frac{1}{4}\left(y^{\prime}\right)^{2}\) in Eq. (37). Show that the line $$ y=C x-\frac{1}{4} C^{2} $$ is tangent to the parabola \(y=x^{2}\) at the point \(\left(\frac{1}{2} C, \frac{1}{4} C^{2}\right)\). Explain why this implies that \(y=x^{2}\) is a singular solution of the given Clairaut equation. This singular solution and the one-parameter family of straight line solutions are illustrated in Fig. 1.6.10.

Verify that the given differential equation is exact; then solve it. $$ \left(2 x y^{2}+3 x^{2}\right) d x+\left(2 x^{2} y+4 y^{3}\right) d y=0 $$

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