/*! 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 10 The autonomous differential equa... [FREE SOLUTION] | 91Ó°ÊÓ

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The autonomous differential equations in Exercises \(9-12\) represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for \(P(t),\) selecting different starting values \(P(0)\) (as in Example 5 ). Which equilibria are stable, and which are unstable? $$ \frac{d P}{d t}=P(1-2 P) $$

Short Answer

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

Step by step solution

01

Identify Equilibria

To find the equilibria, we set \( \frac{dP}{dt} = 0 \). This gives us the equation \( P(1 - 2P) = 0 \). Solving this, we get \( P = 0 \) and \( 1 - 2P = 0 \). Solving \( 1 - 2P = 0 \) gives \( P = \frac{1}{2} \). So the equilibria are \( P = 0 \) and \( P = \frac{1}{2} \).
02

Determine Stability

To determine the stability of the equilibria, we analyze the sign of \( \frac{dP}{dt} = P(1-2P) \) around the equilibria \( P = 0 \) and \( P = \frac{1}{2} \). For values of \( P \) slightly greater than 0, \( (1 - 2P) \) is positive, indicating that \( P = 0 \) is unstable. For \( P \) slightly less than \( \frac{1}{2} \), \( (1-2P) \) is positive, while for \( P \) slightly greater than \( \frac{1}{2} \), \( (1-2P) \) is negative, indicating that \( P = \frac{1}{2} \) is stable.
03

Phase Line Analysis

Draw a vertical line representing \( P \) values, marking \( P = 0 \) (unstable) and \( P = \frac{1}{2} \) (stable). For \( P < 0 \), both terms \( P \) and \( 1-2P \) are negative, so \( \frac{dP}{dt} > 0 \), indicating \( P \) increases. For \( 0 0 \), so \( P \) increases towards \( \frac{1}{2} \). For \( P > \frac{1}{2} \), \( \frac{dP}{dt} < 0 \), so \( P \) decreases towards \( \frac{1}{2} \).
04

Sketch Solution Curves

Use the phase line analysis to sketch solution curves. For \( P(0) = 0 \), \( P(t) \) remains 0. For \( 0 < P(0) < \frac{1}{2} \), \( P(t) \) grows asymptotically towards \( \frac{1}{2} \). For \( P(0) > \frac{1}{2} \), \( P(t) \) decays towards \( \frac{1}{2} \). All curves reach \( P = \frac{1}{2} \), regardless of starting position unless \( P(0) = 0 \).

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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 visually intuitive method to understand the behavior of solutions to differential equations, particularly autonomous ones. An autonomous equation has the form \( \frac{dP}{dt} = f(P) \), where the rate of change depends only on \(P\) and not explicitly on time \(t\).

To perform a phase line analysis, begin by identifying equilibrium points, where \( \frac{dP}{dt} = 0 \). Plot these points on a vertical line representing the variable \( P \). Between these points, you mark whether \( \frac{dP}{dt} \) is positive or negative, indicating the direction of \(P\)'s movement.

This simple line can show which equilibria the population will tend towards or away from. It's a handy tool for sketching solution behaviors without solving the differential equation outright, especially when dealing with population growth models.
Population Growth Models
Population growth models are mathematical descriptions of how populations change over time. Although they can vary widely in complexity, simple models are often based on autonomous differential equations. These models can describe how the size of a population, denoted here as \(P(t)\), evolves.

One common form is the logistic model, which can be represented by \( \frac{dP}{dt} = P(1 - 2P) \). This particular equation captures two crucial effects:
  • Population growth at a rate proportional to its size, indicated by the presence of \(P\) in the equation.
  • A limiting factor that restricts growth as \(P\) approaches a certain value, represented by \(1 - 2P\). This is where the concept of carrying capacity comes into play. The population tends towards the equilibrium point \(P = \frac{1}{2}\), showing a natural limit of sustainable growth.
Understanding these models is essential in biology and environmental science, where managing real-life populations effectively is critical.
Equilibrium Stability
Equilibrium stability in differential equations refers to whether solutions of the system tend to return to equilibrium points when slightly disturbed. Equilibrium points are essentially fixed points where the rate of change \( \frac{dP}{dt} = 0 \).

To determine stability, look at the behavior surrounding these points:
  • If solutions are drawn back towards an equilibrium, it is stable.
  • If they move away, the equilibrium is unstable.
In the equation \( \frac{dP}{dt} = P(1 - 2P) \), two equilibrium points are identified: \(P = 0\) and \(P = \frac{1}{2}\).

Upon analyzing \( \frac{dP}{dt} \), we found:
  • \(P = 0\) is unstable, since any small increase in \(P\) results in \(P\) increasing further.
  • \(P = \frac{1}{2}\) is stable, as disturbances lead \(P\) back towards this point for \(P \leq \frac{1}{2}\).
This concept is crucial for understanding how systems react to external and internal changes, predicting long-term population behavior effectively.

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Most popular questions from this chapter

In Exercises \(25-30\) , obtain a slope field and add to it graphs of the solution curves passing through the given points. $$ \begin{array}{l}{y^{\prime}=y \text { with }} \\ {\begin{array}{lll}{\text { a. }(0,1)} & {\text { b. }(0,2)} & {\text { c. }(0,-1)}\end{array}}\end{array} $$

In Exercises 31 and \(32,\) obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {\quad 0 \leq x \leq 4, \quad 0 \leq y \leq 3}\end{array} $$

In Exercises \(1-8\) 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. $$ \frac{d y}{d x}=y^{2}-4 $$

Solve the exponential growth/decay initial value problem for \(y\) as a function of \(t\) thinking of the differential equation as a first-order linear equation with \(P(x)=-k\) and \(Q(x)=0 :\) $$\frac{d y}{d t}=k y \quad(k \text { constant }), \quad y(0)=y_{0}$$

Use a CAS to explore graphically each of the differential equations in Exercises \(21-24 .\) 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 \((y \text { exact })-y(\text { Euler })\) at the specified point \(x=b\) for each of your four Euler approximations. Discuss the improvement in the percentage error. $$ \begin{array}{l}{y^{\prime}=x+y, \quad y(0)=-7 / 10 ; \quad-4 \leq x \leq 4, \quad-4 \leq y \leq 4} \\ {b=1}\end{array} $$

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