/*! 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 4 In Problems first solve the equa... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 4

In Problems first solve the equation \(f(x)=0\) to find the critical points of the given autonomous differential equation \(d x / d t=f(x) .\) Then analyze the sign of \(f(x)\) to determine whether each crifical point is stable or unstable, and construct the corresponding phase diagrant for the differential equarion. Next, solve the differential equarion explicitly for \(x(t)\) in terms of t. Finally use either the exact solution or a computer-generated slope field to sketch typical solution curves for the given differential equation, and verify visually the stability of each critical point. $$ \frac{d x}{d t}=3 x-x^{2} $$

Short Answer

Expert verified
Critical points: \( x = 0, 3 \). \( x=0 \) is unstable; \( x=3 \) is stable.

Step by step solution

01

Find Critical Points

To find the critical points, we solve the equation \( f(x) = 0 \) or \( 3x - x^2 = 0 \). Factoring gives us \( x(3-x) = 0 \), so the critical points are \( x = 0 \) and \( x = 3 \).
02

Determine Stability of Critical Points

Analyze the sign of \( f(x) = 3x - x^2 \) in the intervals created by the critical points:- For \( x < 0 \), \( f(x) = 3x - x^2 > 0 \) implies an increase in \( x \).- For \( 0 < x < 3 \), \( f(x) = 3x - x^2 > 0 \) implies an increase in \( x \).- For \( x > 3 \), \( f(x) = 3x - x^2 < 0 \) implies a decrease in \( x \).Thus, \( x = 0 \) is unstable, and \( x = 3 \) is stable. Draw arrows on the phase line indicating the flow towards the stable point \( x = 3 \).
03

Solve the Differential Equation

The differential equation is \( \frac{dx}{dt} = 3x - x^2 \). We rearrange:\[ \frac{dx}{3x - x^2} = dt \]To solve, perform partial fraction decomposition on \( \frac{1}{x(3-x)} \) and integrate:\[ \int \left( \frac{1}{3x} + \frac{1}{3(3-x)} \right) dx = \int dt \]This gives:\[ \frac{1}{3} \ln |x| - \frac{1}{3} \ln |3-x| = t + C \]Solving for \( x \), we have:\[ x = \frac{3}{1 + Ce^{-3t}} \].
04

Sketch Solution Curves

To visually confirm the stability, you can sketch solutions for different initial conditions using either the explicit solution \( x(t) = \frac{3}{1 + Ce^{-3t}} \) or a computer-generated slope field. Solution curves for initial points between 0 and 3 will tend towards \( x = 3 \), demonstrating that \( x = 3 \) is stable, while any trajectory starting close to \( x = 0 \) will move away from it, confirming its instability.

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Key Concepts

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

Critical Points
Critical points in a differential equation are values of the variable, here denoted as \( x \), where the rate of change of \( x \) with respect to time \( t \) is zero. For an equation \( \frac{dx}{dt} = f(x) \), these are the solutions to the equation \( f(x) = 0 \). By solving \( 3x - x^2 = 0 \), we find the critical points by factoring as \( x(3 - x) = 0 \). This reveals that the critical points are \( x = 0 \) and \( x = 3 \).
A critical point can indicate where the dynamic behavior of the system changes, such as from increasing to decreasing or vice versa. These critical points help us understand where a system might be in equilibrium or at a steady state.
Stability Analysis
Stability analysis involves evaluating the behavior of a differential equation around its critical points. We analyze the sign of \( f(x) = 3x - x^2 \) in the intervals created by critical points to determine stability:
  • For \( x < 0 \), \( f(x) > 0 \) indicates \( x \) is increasing.
  • For \( 0 < x < 3 \), \( f(x) > 0 \) which also signifies \( x \) is increasing.
  • For \( x > 3 \), \( f(x) < 0 \) means \( x \) is decreasing.
Hence, \( x = 0 \) is unstable as it does not attract trajectories鈥攊t pushes them away. Conversely, \( x = 3 \) is stable since it attracts nearby trajectories. Understanding the stability of each critical point allows you to predict the long-term behavior of the system at those points.
Phase Diagram
A phase diagram visually represents the stability of the critical points and the behavior of the system between them. In our example, the diagram includes arrows along a line (phase line) indicating the flow or direction of \( x \) over time.
At \( x = 0 \), the phase line shows arrows pointing away from the point, symbolizing its instability.
At \( x = 3 \), arrows point towards it, signifying it is an attractor and thus stable.
These visual aids are invaluable in depicting where solutions of the differential equation are heading without solving the equation explicitly. They provide an overview of the dynamics and can guide further analysis of the equation.
Explicit Solution
The explicit solution of a differential equation explicitly expresses the variable in terms of time or other variables. For the equation \( \frac{dx}{dt} = 3x - x^2 \), this involves separating variables and performing integration.
  • The integrated equation \( \int \left( \frac{1}{3x} + \frac{1}{3(3-x)} \right) dx = \int dt \) yields a complex logarithmic solution:
  • \( \frac{1}{3} \ln |x| - \frac{1}{3} \ln |3-x| = t + C \).
Solving for \(x\) gives us \( x = \frac{3}{1 + Ce^{-3t}} \).
This solution reveals how \( x \) evolves over time, starting from various initial conditions. As \( t \rightarrow \infty \), the solution approaches the stable critical point \( x = 3 \). Explicit solutions allow for precise predictions about the system's behavior, providing a clear mathematical description of how the variable behaves with time.

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Most popular questions from this chapter

A computer with a printer is required for Problems 17 through 24\. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size h. Throughout, primes denote derivatives with respect to \(x\). \(y^{\prime}=x^{2}+y^{2}, y(0)=0 ; 0 \leqq x \leqq 1\)

A population \(P(t)\) of small rodents has birth rate \(\beta=\) (0.001) \(P\) (births per month per rodent) and constant death rate \(\delta\). If \(P(0)=100\) and \(P^{\prime}(0)=8\), how long (in months) will it take this population to double to 200 rodents? (Suggestion: First find the value of \(\delta\).)

As the salt \(\mathrm{KNO}_{3}\) dissolves in methanol, the number \(x(t)\) of grams of the salt in a solution after \(t\) seconds satisfies the differential equation \(d x / d t=0.8 x-0.004 x^{2}\) (a) What is the maximum amount of salt that will ever dissolve in the methanol? (b) If \(x=50\) when \(t=0\), how long will it take for an additional \(50 \mathrm{~g}\) of salt to dissolve?

Consider a population \(P(t)\) satisfying the extinctionexplosion equation \(d P / d t=a P^{2}-b P\), where \(B=a P^{2}\) is the time rate at which births occur and \(D=b P\) is the rate at which deaths occur. If the initial population is \(P(0)=P_{0}\) and \(B_{0}\) births per month and \(D_{0}\) deaths per month are occurring at time \(t=0\), show that the threshold population is \(M=D_{0} P_{0} / B_{0}\).

An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=2 x y^{2}, y(0)=1 ; y(x)=\frac{1}{1-x^{2}} $$
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