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Find all equilibrium points. Give answers as ordered pairs \((x, y).\) $$\begin{aligned} &\frac{d x}{d t}=15 x-5 x y\\\ &\frac{d y}{d t}=10 y+2 x y \end{aligned}$$

Short Answer

Expert verified
The equilibrium points are \((0, 0)\) and \((-5, 3)\).

Step by step solution

01

Understand Equilibrium Points

Equilibrium points occur where the rates of change are zero, meaning \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). We need to set each equation equal to zero and solve for \(x\) and \(y\).
02

Set the First Equation to Zero

Set \(15x - 5xy = 0\). Factor out \(x\), which gives: \[ x(15 - 5y) = 0 \]This equation is satisfied either if \(x = 0\) or \(15 - 5y = 0\).
03

Solve \(x = 0\) Case

If \(x = 0\), substitute this value into the second equation. The second equation simplifies to \(10y = 0\). Therefore, \(y = 0\). The equilibrium point is \((0, 0)\).
04

Solve \(15 - 5y = 0\) Case

Solve for \(y\):\[ 15 - 5y = 0 \]\[ y = 3 \]Substitute \(y = 3\) into the second equilibrium equation.
05

Set the Second Equation to Zero

Set \(10y + 2xy = 0\). Factor out \(y\), which gives:\[ y(10 + 2x) = 0 \]This equation is satisfied either if \(y = 0\) or \(10 + 2x = 0\).
06

Solve for \(10 + 2x = 0\)\, given \(y = 3\)

Since \(y = 3\), substitute this into the second factor and solve for \(x\): \(10 + 2x = 0\)\[ 2x = -10 \]\[ x = -5 \]The equilibrium point is \((-5, 3)\).
07

Identify All Equilibrium Points

Based on steps 3 and 6, the set of equilibrium points are: \((0, 0)\) and \((-5, 3)\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that demonstrate how a particular quantity changes with respect to another quantity, often time. In this context, these equations describe how two variables, typically represented as \(x\) and \(y\), change over time within the system. They are used to model a wide range of real-world phenomena. For example, they can illustrate the dynamics between predators and prey in an ecosystem or the interactions of chemicals in a reaction.

In the exercise, we look at two first-order differential equations:
  • \(\frac{d x}{d t} = 15x - 5xy\)
  • \(\frac{d y}{d t} = 10y + 2xy\)
Each equation depicts a rate of change—how the system's state evolves over time—making differential equations vital for analyzing dynamic systems.
System of Equations
A system of equations involves multiple equations with multiple unknowns that are solved together. Finding solutions that satisfy all the equations together is paramount. In our exercise, the system is made up of:
  • \(\frac{d x}{d t} = 15x - 5xy\)
  • \(\frac{d y}{d t} = 10y + 2xy\)
The solution to this system involves finding the values of \(x\) and \(y\) where both equations equal zero, known as the equilibrium points.

Solving a system of equations like this one can give insight into where a system reaches balance, such that variables no longer change as time progresses. Here, solving the system provided the equilibrium points \((0, 0)\) and \((-5, 3)\). Each point represents a state where the system settles into a steady condition.
Phase Plane Analysis
Phase plane analysis is a graphical method to study the behavior of the system described by differential equations. This method involves plotting the variables \(x\) and \(y\) on a two-dimensional plane to visualize how changes occur over time. By documenting trajectories and equilibrium points on the plane, we can infer the stability and types of behavior other potential solutions may exhibit.

For the given system, we have two equilibrium points: \((0, 0)\) and \((-5, 3)\). Within the phase plane, these points are significant as they help illustrate the potential paths that the system can take over time. The analysis offers information about whether these points are nodes, spirals, saddles, or centers, aiding in understanding the stability and behavior of the system.
Linearization
Linearization is a technique used to approximate nonlinear systems through linear equations, thus simplifying the analysis of differential equations around equilibrium points. This process entails taking the derivative (or the Jacobian matrix for systems) to study the behavior near these points.

In practice, linearization views the system as locally linear at equilibrium. By considering small deviations about these points, we can determine whether they are stable or unstable and predict the system's behavior in proximity to them.
  • Stable Equilibria: The system returns to equilibrium upon disturbance.
  • Unstable Equilibria: Any disturbance leads the system away from equilibrium.
Identifying the nature of each equilibrium point allows us to predict how the actual nonlinear system behaves in its vicinity, providing crucial insights into the system's dynamics.

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Most popular questions from this chapter

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(\lim _{x \rightarrow \infty} g(x)=\infty,\) then \(\lim _{x \rightarrow \infty} f(x)=\infty.\)

Before Galileo discovered that the speed of a falling body with no air resistance is proportional to the time since it was dropped, he mistakenly conjectured that the speed was proportional to the distance it had fallen. (a) Assume the mistaken conjecture to be true and write an equation relating the distance fallen, \(D(t),\) at time \(t,\) and its derivative. (b) Using your answer to part (a) and the correct initial conditions, show that \(D\) would have to be equal to 0 for all \(t,\) and therefore the conjecture must be wrong.

The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of \(Q_{0}\) is injected directly into the blood, \(20 \%\) is left in the blood after 3 hours. (a) Write and solve a differential equation for the quantity, \(Q,\) of the drug in the blood after \(t\) hours. (b) How much of this drug is in a patient's body after 6 hours if the patient is given 100 mg initially?

Explain what is wrong with the statement. The function \(y=x^{2}\) is an equilibrium solution to the differential equation \(d y / d x=y-x^{2}.\)

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