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Chapter 1: Problem 9

Determine a region of the \(x y\)-plane for which the given differential equation would have a unique solution through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ \frac{d y}{d x}=x^{3} \cos y $$

Short Answer

Expert verified
The differential equation has a unique solution for any point in the entire \(xy\)-plane.

Step by step solution

01

Identify the Form of the Differential Equation

The given differential equation is \( \frac{dy}{dx} = x^3 \cos y \). This is a first-order ordinary differential equation.
02

Apply the Existence and Uniqueness Theorem

The Picard-Lindel枚f (or Existence and Uniqueness) Theorem states that if the function \( f(x, y) \) and its partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), are continuous in some region containing the point \((x_0, y_0)\), then there is a unique solution passing through that point.
03

Check Continuity of the Function

The function \( f(x, y) = x^3 \cos y \) is composed of \( x^3 \), which is a polynomial, and \( \cos y \), which is continuous everywhere. Thus, \( f(x, y) \) is continuous for all \( x \) and \( y \).
04

Compute Partial Derivative with Respect to y

Calculate the derivative of the function \( f(x, y) = x^3 \cos y \) with respect to \( y \):\[\frac{\partial f}{\partial y} = x^3 \frac{d}{dy}(\cos y) = -x^3 \sin y\]Since \( \sin y \) is also continuous everywhere, \( \frac{\partial f}{\partial y} \) is continuous for all \( x \) and \( y \).
05

Determine the Region of Uniqueness

Because both \( f(x, y) \) and \( \frac{\partial f}{\partial y} \) are continuous for all real \( x \) and \( y \), there are no restrictions on \( (x_0, y_0) \). Thus, the region where there is a unique solution is the entire \( xy \)-plane.

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Key Concepts

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

Existence and Uniqueness Theorem
The Existence and Uniqueness Theorem is a crucial concept when working with differential equations. This theorem helps us determine when a solution to a differential equation is not only existent but also unique.
  • It ensures that there is a single solution that passes through a given point in a specific region of the plane.
  • The theorem applies if the function and its partial derivatives meet specific continuity criteria.
In the context of the given differential equation, the theorem states that if the right-hand side function, \( f(x, y) = x^3 \cos y \), and its partial derivative with respect to \( y \), \( \partial f/\partial y \), are continuous, then a unique solution exists.
In our exercise, both criteria are met across the entire \( xy \)-plane, thus guaranteeing a unique solution everywhere.
Picard-Lindel枚f Theorem
The Picard-Lindel枚f Theorem is an extension of the Existence and Uniqueness Theorem, providing a method to ensure the conditions are met.
  • It requires the function to be Lipschitz continuous regarding \( y \).
  • This implies that small changes in \( y \) lead to small changes in the function's value.
The theorem guarantees not only existence and uniqueness but also provides a way to construct solutions iteratively. In our differential equation example, it confirms that the function \( x^3 \cos y \) satisfies these properties, helping verify the solution's uniqueness across the plane.
First-order Ordinary Differential Equations
First-order ordinary differential equations (ODEs) represent relationships involving derivatives of a single variable. They capture how quantities change and are foundational in mathematical modeling.
  • These equations involve first derivatives explicitly.
  • They often describe simple dynamic systems or rates of change.
In our exercise, the equation \( \frac{dy}{dx} = x^3 \cos y \) is a first-order ODE. This means it involves the first derivative of \( y \) with respect to \( x \), indicating a rate of change dependent on both \( x \) and \( y \). Understanding how these work allows us to apply existences theorems to analyze and solve them effectively.
Continuity of Functions
Continuity plays a pivotal role in ensuring the solutions to differential equations are consistent and predictable.
  • A function is continuous if small changes in inputs result in small changes in outputs, without sudden jumps.
  • This property allows us to apply mathematical theorems with confidence, including the existence results.
In our context, the function \( f(x, y) = x^3 \cos y \) is composed of \( x^3 \), a polynomial, and \( \cos y \), trigonometrically continuous. This ensures continuity everywhere.
Additionally, its derivative, \( -x^3 \sin y \), maintains this continuity property. These aspects verify the underlying conditions needed for the Existence and Uniqueness Theorem, securing a unique solution throughout the entire plane.

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Most popular questions from this chapter

Determine an interval on which \(y^{2}-2 y=x^{2}-x-1\) defines a solution of \(2(y-1) d y+(1-2 x) d x=0\).

After a mass \(m\) is attached to a spring, it stretches the spring \(s\) units and then hangs at rest in the equilibrium position, as shown in Figure \(1.19(b)\). After the spring/mass system has been set in motion, let \(x(t)\) denote the directed distance of the mass beyond the equilibrium position. Assume that the downward direction is positive and that the motion takes place in a vertical straight line through the center of gravity of the mass. Assume, too, that the only forces acting on the system are the weight \(m g\) of the mass and the restoring force of the stretched spring, which, by Hooke's law, is proportional to its total elongation. Determine a differential equation for the displacement \(x(t)\) for any time \(t\).

Verify that the indicated function is a solution of the given differential equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}+y=0, \quad y=c_{1} \cos (\ln x)+c_{y} \sin (\ln x), x>0 $$

\(x^{2} y^{\prime}-x y^{\prime}+2 y=0 ; \quad y=x \cos (\ln x), x>0\)

Determine by inspection at least one solution for the given differential equation. $$ y^{\prime \prime}=1 $$
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