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Chapter 8: Problem 65

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d x}=(y-1)(y-2)(y-5)\) (a) \(y(0)=0\), (b) \(y(0)=4\) (c) \(y(0)=3 / 2\), (d) \(y(0)=6\).

Short Answer

Expert verified
The solutions based on initial conditions go as follows: (a) moves down, (b) moves towards 2, (c) moves towards 2 and, (d) moves up indefinitely.

Step by step solution

01

Understanding the Differential Equation

We start by analyzing the given differential equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \). This is an autonomous equation because the right-hand side is a function of \( y \) only. The roots of this polynomial, \( y = 1, 2, \) and \( 5 \), are the equilibrium points, where \( \frac{d y}{d x} = 0 \).
02

Analyzing the Behavior Around Equilibria

For \( y < 1 \), \( \frac{d y}{d x} < 0 \) indicating \( y \) is decreasing. For \( 1 < y < 2 \), \( \frac{d y}{d x} > 0 \), so \( y \) is increasing. Similarly, for \( 2 < y < 5 \), \( \frac{d y}{d x} < 0 \), and for \( y > 5 \), \( \frac{d y}{d x} > 0 \). This shows \( y = 1 \) and \( y = 5 \) are stable equilibria, while \( y = 2 \) is unstable.
03

Drawing the Vector Field

Sketch the vector field based on the sign of \( \frac{d y}{d x} \) identified in Step 2. Draw arrows for increasing functions (positive slopes) and reversed arrows for decreasing functions (negative slopes), showing how \( y \) moves with \( x \).
04

Solving the Differential Equation with Initial Condition (a): \( y(0) = 0 \)

The initial condition \( y(0) = 0 \) shows that \( y < 1 \). From Step 2, we see \( y \) will decrease. Sketch the solution curve initially at \( y = 0 \) going downward over time.
05

Solving the Differential Equation with Initial Condition (b): \( y(0) = 4 \)

The initial condition \( y(0) = 4 \) means \( y \) starts between 2 and 5. According to Step 2, \( y \) will decrease until it approaches the stable equilibrium \( y = 2 \). Sketch the solution curve starting from \( y = 4 \) heading downward toward \( y = 2 \).
06

Solving the Differential Equation with Initial Condition (c): \( y(0) = 3/2 \)

With \( y(0) = 3/2 \), \( y \) starts between 1 and 2. As seen in Step 2, \( y \) will increase towards the equilibrium at \( y = 2 \). Draw the solution starting from \( y = 3/2 \) moving upwards towards \( y = 2 \).
07

Solving the Differential Equation with Initial Condition (d): \( y(0) = 6 \)

Starting with \( y(0) = 6 \), means \( y > 5 \). From Step 2, we know \( y \) increases, confirming it will move further away from 5. Sketch this solution curve by showing \( y \) starting from 6 and moving upwards.
08

Sketching All Solutions on One Graph

Combine the individual solutions onto one graph. Place the \( x \)-axis as the independent variable axis, and \( y \)-axis for the dependent variable. Mark the equilibria points \( y = 1, 2, 5 \) clearly and plot all solutions together, representing each initial condition.

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Key Concepts

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

Vector Field Plot
A vector field plot is a visual representation of the differential equation's behavior. It shows how the dependent variable, \( y \), changes with respect to the independent variable, often denoted as \( x \). To create this plot for the differential equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \), we analyze the equation's structure. Here, we assess the sign of \( \frac{d y}{d x} \) across different ranges of \( y \).

By examining \( y < 1 \), \( 1 < y < 2 \), \( 2 < y < 5 \), and \( y > 5 \), we observe whether \( y \) increases or decreases.
  • For \( y < 1 \), \( \frac{d y}{d x} < 0 \), showing a downward trend for \( y \).
  • At \( 1 < y < 2 \), \( \frac{d y}{d x} > 0 \), illustrating an upward motion.
  • When \( 2 < y < 5 \), \( \frac{d y}{d x} < 0 \), suggesting that \( y \) decreases again.
  • For \( y > 5 \), \( \frac{d y}{d x} > 0 \), meaning \( y \) starts to increase.
In the field plot, arrows display the direction and behavior of \( y \) depending on where it lies relative to these points.
Equilibrium Points
Equilibrium points are values of \( y \) where the rate of change, \( \frac{d y}{d x} \), equals zero. Essentially, these are points at which the system is in balance and there is no change in the value of \( y \). For our differential equation, the equilibrium points occur at the roots of \( (y-1)(y-2)(y-5) = 0 \), which are \( y = 1, 2, \) and \( 5 \).

Knowing whether these points are stable or unstable is important.
  • Stable equilibrium means small perturbations will return to the equilibrium.
  • Unstable equilibrium means that disturbances grow larger as time progresses.
In our case:
  • \( y = 1 \) and \( y = 5 \) are stable equilibria because small changes result in \( y \) returning to these points.
  • \( y = 2 \) is unstable because disturbances result in \( y \) moving away.
These properties help determine the expected behavior of solutions as they approach or diverge from these points.
Autonomous Equation
An autonomous equation is a type of differential equation where the derivative \( \frac{d y}{d x} \) depends solely on the dependent variable \( y \), and not explicitly on the independent variable \( x \). The equation \( \frac{d y}{d x} = (y-1)(y-2)(y-5) \) qualifies as autonomous because \( (y-1)(y-2)(y-5) \) is a function only of \( y \).

The significance of autonomous equations is that their solutions can often be analyzed through qualitative means, focusing on behavior over time rather than exact values.
  • You can determine long-term behavior by looking at equilibrium points and how \( y \) behaves in their vicinity.
  • Solutions are typically visualized through sketches like vector fields or phase lines rather than explicit functions of \( x \).
Overall, autonomous equations simplify the process of understanding a system's dynamics by centering the analysis around \( y \)-values.
Initial Conditions
Initial conditions are specific values set at the start of analyzing a differential equation, typically given in the form of \( y(x_0) = y_0 \). They serve as starting points for finding particular solutions to the differential equation. For our example, we have initial conditions like \( y(0) = 0 \), \( y(0) = 4 \), \( y(0) = \frac{3}{2} \), and \( y(0) = 6 \).

To understand how these initial conditions affect the solution:
  • If \( y(0) = 0 \), since \( 0 < 1 \), \( y \) will move downward due to the negative derivative.
  • For \( y(0) = 4 \), as \( 2 < 4 < 5 \), \( y \) decreases towards \( y = 2 \).
  • With \( y(0) = \frac{3}{2} \), \( y \) starts below 2 and is drawn upward to this equilibrium.
  • When \( y(0) = 6 \), above 5, \( y \) is propelled higher due to the positive derivative.
Graphically, these conditions anchor solution curves to specific trajectories within the vector field, highlighting the dynamic paths \( y \) takes over time.

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Most popular questions from this chapter

In Problems you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{t+1}{t v+t v^{3}} \text { where } y(1)=1 \text { . } $$

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d x}=y^{2}-h $$

In this question we will interpret the recovery rate, \(c\), that appears in the model. Assume that a population of infected individuals is quarantined (that is, they are unable to transmit the disease to others, or to catch it again once they recover). (a) Explain why under these assumptions we expect: $$ \frac{d I}{d t}=-c I $$ (b) Assuming \(I(0)=I_{0}\), find \(I(t)\) by solving \((8.71)\). (c) How long will it take for the number of infected individuals to decrease from \(I_{0}\) to \(I_{0} / 2 ?\) (d) Assume that it takes 7 days for the number of infected individuals to decrease from 50 to \(25 .\) Calculate the recovery rate \(c\) for this disease.

A cell constantly gains or loses small molecules to its environment because the small molecules are able to diffuse through the cell membrane. We will build a model for this process. Suppose a molecule is present in the cell at a concentration \(C(t)\), and present in its environment at a concentration \(C_{\infty}\) (you may assume \(C_{\infty}\) is a constant). One model for the diffusion of molecules across the cell membrane is that the rate at which molecules travel through the membrane is proportional to the difference in concentration between the cell and its surroundings. That is: Rate at which $$ \text { molecules flow out }=k\left(C-C_{\infty}\right) $$ of cell The constant \(k\) is known as the permeability of the membrane: \(k>0\), and \(k\) depends on the surface area of the cell and the chemistry of the membrane, as well as the type of molecule. (a) Starting with a word equation for the amount of small molecules in the cell, show, if the cell volume is \(V\), then: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right) $$ (b) Find the equilibrium of \((8.53)\) and use a graphical analysis to determine whether it is stable or unstable. (c) Suppose that the molecule we are studying is produced within the cell. The cell produces the molecule at a rate \(r\); that is, a quantity \(r\) is produced (added to the cell) in unit time. Explain why the differential equation for the concentration of molecules in the cell should be modified to: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right)+\frac{r}{V} $$ (d) Analyze Equation (8.54) to find the equilibrium value of the cell concentration. Is this equilibrium stable or unstable? You may use a graphical argument or calculate the eigenvalue to determine the equilibrium's stability.

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d N}{d t}=N \ln (2 / N), N>0 $$
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