/*! 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 12 The autonomous differential equa... [FREE SOLUTION] | 91Ó°ÊÓ

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The autonomous differential equations represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for \(P(t),\) selecting different starting values \(P(0) .\) Which equilibria are stable, and which are unstable? $$\frac{d P}{d t}=3 P(1-P)\left(P-\frac{1}{2}\right)$$

Short Answer

Expert verified
Equilibria at \(P=0\) and \(P=1\) are unstable; \(P=\frac{1}{2}\) is stable.

Step by step solution

01

Identify Equilibria Points

Equilibria occur where \( \frac{dP}{dt} = 0 \). We set the equation \( 3P(1-P)(P-\frac{1}{2}) = 0 \) and solve for \( P \). This gives the equilibria points \( P = 0, \frac{1}{2}, \text{and } 1 \).
02

Analyze the Sign of \(\frac{dP}{dt}\) Between Equilibria

Examine the sign of \( 3P(1-P)(P-\frac{1}{2}) \) in the intervals defined by the equilibria: \(( -\infty, 0 ), ( 0, \frac{1}{2} ), ( \frac{1}{2}, 1 ), ( 1, \infty )\). Between these points, choose test values to determine the sign:\(P = -0.1\) results in negative, \(P = 0.25\) results in positive, \(P = 0.75\) results in negative, and \(P = 1.1\) results in positive.
03

Create the Phase Line

Using the information from Step 2, sketch a phase line. Mark points \(P = 0\), \(P = \frac{1}{2}\), and \(P = 1\) on the line, with arrows indicating the direction of \(\frac{dP}{dt}\): \(0\rightarrow\frac{1}{2}\) is an increasing region, \(\frac{1}{2}\rightarrow1\) is a decreasing region, and to the right of 1 it increases.
04

Identify Stability of Equilibria

Based on the phase line: \(P = 0\) is unstable (positive derivative to the right), \(P = \frac{1}{2}\) is stable (arrows point toward it), and \(P = 1\) is unstable (arrows point away from it to the left).
05

Sketch Solution Curves for Different Initial Conditions

Draw solution curves \(P(t)\) by considering initial conditions \(P(0) < 0\), \(0 < P(0) < \frac{1}{2}\), \(\frac{1}{2} < P(0) < 1\), and \(P(0) > 1\). Solutions from less than \(0\) approach \(0\), solutions from \([0, \frac{1}{2})\) approach \(\frac{1}{2}\), solutions from \((\frac{1}{2}, 1)\) approach \(\frac{1}{2}\), and solutions greater than \(1\) grow.

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Key Concepts

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

Phase Line Analysis
Phase line analysis is a graphical way to visualize how solutions to differential equations behave. For autonomous differential equations, such as our exercise with \( \frac{dP}{dt} = 3P(1-P)(P-\frac{1}{2}) \), the phase line helps us track growth rates depending on the population density. By representing increases and decreases between equilibrium points, students can sketch the general shape of the solution curves, \( P(t) \), based on different initial populations, \( P(0) \). These lines include arrows showing the direction of population change over time.
  • If the derivative \( \frac{dP}{dt} \) is positive in an interval, \( P(t) \) increases.
  • If the derivative is negative, \( P(t) \) decreases.
  • Phase lines are marked at equilibrium points where the growth rate changes direction.

Using this visual tool, particularly helpful in this exercise, clarifies where the population will stabilize or diverge depending on starting conditions.
Equilibria in Differential Equations
Equilibrium points, in the context of differential equations, are where the rate of change, \( \frac{dP}{dt} \), is zero. These points are significant because they indicate where the population doesn't change; they're the values that solutions naturally move towards or away from.
To find equilibrium points in the given differential equation, we solve \( 3P(1-P)(P-\frac{1}{2}) = 0 \). The equilibria here are \( P = 0 \), \( P = \frac{1}{2} \), and \( P = 1 \). At these points, there's no population change, which suggests the population could stabilize.
These points divide the number line into intervals, helping us understand how the population behaves between and around these equilibria.
Stability Analysis
The stability of each equilibrium point informs us whether small perturbations will return to or deviate from the equilibrium. In stability analysis, the phase line's arrows point towards or away from equilibrium points, showing their stability.
  • An equilibrium is stable if small disturbances cause the system to return to it.
  • It's unstable if disturbances cause the system to move away from it.

In our differential equation, the stability analysis reveals:- \( P = 0 \) is unstable because the direction of the flow is away from it.- \( P = \frac{1}{2} \) is stable since the flow converges toward this point.- \( P = 1 \) is also unstable as the flow diverges away no matter the perturbation direction.
Understanding stability helps predict long-term population behavior without solving the equation explicitly.
Population Growth Models
Population growth models help us understand how population sizes evolve over time based on initial conditions. Autonomous differential equations serve as a tool to model this growth by predicting changes in population density with respect to time. These models can be complex, but with appropriate analyses, they become manageable and insightful.
The given model, \( \frac{dP}{dt} = 3P(1-P)(P-\frac{1}{2}) \), illustrates how growth rates depend nonlinearly on population size. This means that growth doesn’t happen at a constant rate, and several factors influence the population's dynamics simultaneously.
  • If \( P(0) < \frac{1}{2} \), populations grow towards \( \frac{1}{2} \).
  • If \( \frac{1}{2} < P(0) < 1 \), the population decreases towards \( \frac{1}{2} \).
  • Population sizes exceeding \( 1 \) tend to grow indefinitely, indicating no natural checks on growth.

These insights allow students to predict population behavior over time, useful in ecology, sociology, and economics.

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Most popular questions from this chapter

Suppose that a healthy population of some species is growing in a limited environment and that the current population \(P_{0}\) is fairly close to the carrying capacity \(M_{0} .\) You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity \(M_{1}\) considerably less than \(M_{0}\) and, in fact, less than the current population \(P_{0} .\) Starting at some time before the catastrophe, sketch a "before-and-after" curve that shows how the fish population responds to the change in environment.

Use a CAS to explore graphically each of the differential equations. Perform the following steps to help with your explorations. a. Plot a slope field for the differential equation in the given \(x y\) -window. b. Find the general solution of the differential equation using your CAS DE solver. c. Graph the solutions for the values of the arbitrary constant \(C=-2,-1,0,1,2\) superimposed on your slope field plot. d. Find and graph the solution that satisfies the specified initial condition over the interval \([0, b]\) e. Find the Euler numerical approximation to the solution of the initial value problem with 4 subintervals of the \(x\) -interval, and plot the Euler approximation superimposed on the graph produced in part (d). f. Repeat part (e) for \(8,16,\) and 32 subintervals. Plot these three Euler approximations superimposed on the graph from part (e). g. Find the error ( \(y\) (exact) \(-y\) (Euler)) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$\begin{aligned}&y^{\prime}=x+y, \quad y(0)=-7 / 10 ; \quad-4 \leq x \leq 4, \quad-4 \leq y \leq 4\\\&b=1\end{aligned}$$

Show that the curves \(2 x^{2}+3 y^{2}=5\) and \(y^{2}=x^{3}\) are orthogonal.

a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$\frac{d y}{d x}=(y+2)(y-3)$$

Solve the Bernoulli equations $$y^{\prime}-y=-y^{2}$$

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