/*! 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 48 Are the statements true or false... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Are the statements true or false? Give an explanation for your answer. The system of differential equations \(d x / d t=-x+x y^{2}\) and \(d y / d t=y-x^{2} y\) requires initial conditions for both \(x(0)\) and \(y(0)\) to determine a unique solution.

Short Answer

Expert verified
True, initial conditions for both \(x(0)\) and \(y(0)\) are needed for a unique solution.

Step by step solution

01

Analyzing the System

The given system of differential equations is: \[ \frac{d x}{d t} = -x + x y^2 \] \[ \frac{d y}{d t} = y - x^2 y \] This system consists of two coupled first-order differential equations in terms of \(x\) and \(y\). To analyze it, we need to check if it is necessary to have initial conditions for both \(x(0)\) and \(y(0)\).
02

Understanding the Requirement for Unique Solutions

For a system of first-order differential equations, having an initial condition for each equation is generally necessary to determine a unique solution. This is because the initial conditions determine the specific trajectory of the solution paths for the system on the given plane. Without them, multiple solution trajectories could satisfy the differential equations.
03

Applying Existence and Uniqueness Theorem

The existence and uniqueness theorem for ordinary differential equations states that if the functions involved are continuous and satisfy a Lipschitz condition, then a unique solution exists for given initial conditions. Here, both equations have continuous coefficients of \(x\) and \(y\). Therefore, providing initial conditions \(x(0)\) and \(y(0)\) is necessary to determine a unique trajectory of solutions as they influence the system's path uniquely.
04

Conclusion

To find a unique solution to the system \( \frac{d x}{d t} = -x + x y^2 \) and \( \frac{d y}{d t} = y - x^2 y \), it is necessary to specify initial values for both \(x(0)\) and \(y(0)\). Without these initial conditions, the solutions could be non-unique, aligning with the general principles of solving systems of differential equations.

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 91Ó°ÊÓ!

Key Concepts

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

Initial Conditions and Their Importance
Initial conditions play a crucial role in solving differential equations, especially when dealing with first-order systems. For a system of equations like \[ \frac{d x}{d t} = -x + x y^2 \] and \[ \frac{d y}{d t} = y - x^2 y \], initial conditions specify the starting values of the unknown functions at a particular time, typically at time zero such as \(x(0)\) and \(y(0)\).
  • Initial conditions act like a starting point on the graph, guiding the trajectory of the potential solutions.
  • They ensure that the differential equations provide a single, distinct path rather than a multitude of possibilities.

A system without initial conditions could have numerous solutions, each starting from a different point, thus making it impossible to determine one unique path for \(x\) and \(y\). Without these, the problem becomes ill-defined, largely because the system does not know where to begin sua specific solution, leaving it with infinite possibilities. Initial conditions eliminate this ambiguity, providing clarity and precision to the solution.
Existence and Uniqueness Theorem
The Existence and Uniqueness Theorem is a fundamental concept when dealing with differential equations. It assures us whether a solution exists, and if it does, whether it is unique given the initial conditions. Specifically, for a first-order system such as our example equations involving \(x\) and \(y\), the theorem suggests:
  • A solution exists if the functions within the equation are continuous over the region of interest.
  • Uniqueness is guaranteed if these functions satisfy the Lipschitz condition, a specific type of continuity.

For the system \[ \frac{d x}{d t} = -x + x y^2 \] and \[ \frac{d y}{d t} = y - x^2 y \], because the coefficients of \(x\) and \(y\) are continuous and typically satisfy the Lipschitz condition, the theorem ensures a unique solution given the appropriate initial conditions \(x(0)\) and \(y(0)\).
This theorem is powerful because it reassures that our problem is well-posed; not only does a solution exist, but it is also the only one that corresponds with the given initial conditions. This is why specifying initial conditions is compulsory before tackling a system of differential equations, as they lead the way for this unique scenario as assured by the existence and uniqueness theorem.
Understanding First-Order Differential Equations
First-order differential equations, like those in the system \[ \frac{d x}{d t} = -x + x y^2 \] and \[ \frac{d y}{d t} = y - x^2 y \], involve derivatives of the first order, meaning only the first derivative \(\frac{d}{dt}\) appears, not higher derivatives. This specific nature influences the solutions and the manner in which these equations are analyzed.
  • First-order equations typically require initial conditions to solve due to their dependency solely on the first derivative.
  • They are foundational in calculus and serve as a stepping stone to more complex equations.

First-order systems differ from higher-order systems by simplifying the mathematical operations and barriers to solving them — no second derivatives or more complicated transformations are needed. However, they are still indispensable for various practical applications, like modeling dynamic systems in physics or biology.
These simpler equations capture changes in a system with respect to one variable and are thus easier to interpret. Understanding these equations is crucial for grasping more advanced concepts in differential calculus and mathematical modeling.

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

Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. If the tank initially contains 200 liters and 20 liters leak out during the first day, when will the tank be half empty? How much water will there be after 4 days?

Give an example of: A system of differential equations for the profits of two companies if each would thrive on its own but the two companies compete for business. Let \(x\) and \(y\) represent the profits of the two companies.

An aquarium pool has volume \(2 \cdot 10^{6}\) liters. The pool initially contains pure fresh water. At \(t=0\) minutes, water containing 10 grams/liter of salt is poured into the pool at a rate of 60 liters/minute. The salt water instantly mixes with the fresh water, and the excess mixture is drained out of the pool at the same rate (60 liters/minute). (a) Write a differential equation for \(S(t),\) the mass of salt in the pool at time \(t.\) (b) Solve the differential equation to find \(S(t).\) (c) What happens to \(S(t)\) as \(t \rightarrow \infty ?\)

The motion of a mass on the end of a spring satisfies the differential equation $$ \frac{d^{2} s}{d t^{2}}+2 \frac{d s}{d t}+3 s=0 $$ (a) Give the general solution to the differential equation. (b) Solve the differential equation if the initial height is +2 and the initial velocity is +5 (c) How low does the mass at the end of the spring go? How high does it go? (d) How long does it take until the spring stays within 0.1 unit of equilibrium?

An object of mass \(m\) is fired vertically upward from the surface of the earth with initial velocity \(v_{0} .\) We will calculate the value of \(v_{0},\) called the escape velocity, with which the object just escapes the pull of gravity and never returns to earth. since the object is moving far from the surface of the earth, we must take into account the variation of gravity with altitude. If the acceleration due to gravity at sea level is \(g,\) and \(R\) is the radius of the earth, the gravitational force, \(F\), on the object of mass \(m\) at an altitude \(h\) above the surface of the earth is $$F=\frac{m g R^{2}}{(R+h)^{2}}.$$ (a) The velocity of the object (measured upward) is \(v\) at time \(t .\) Use Newton's Second Law of Motion to show that 1 $$\frac{d v}{d t}=-\frac{g R^{2}}{(R+h)^{2}.}$$ (b) Rewrite this equation with \(h\) instead of \(t\) as the independent variable using the chain rule \(\frac{d v}{d t}=\) \(\frac{d v}{d h} \cdot \frac{d h}{d t} .\) Hence, show that $$v \frac{d v}{d h}=-\frac{g R^{2}}{(R+h)^{2}}.$$ (c) Solve the differential equation in part (b). (d) Find the escape velocity, the smallest value of \(v_{0}\) such that \(v\) is never zero.

See all solutions

Recommended explanations on Math Textbooks

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.