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Chapter 2: Problem 1

If the given differential equation is autonomous, identify the equilibrium solution(s). Use a numerical solver to sketch the direction field and superimpose the plot of the equilibrium solution(s) on the direction field. Classify each equilibrium point as either unstable or asymptotically stable. $$ P^{\prime}=0.05 P-1000 $$

Short Answer

Expert verified
The equilibrium solution is \( P = 20000 \), which is asymptotically stable.

Step by step solution

01

Identify the Nature of the Differential Equation

The given differential equation is \( P' = 0.05P - 1000 \). An autonomous differential equation is one where the independent variable (usually \( t \)) does not explicitly appear. In this case, the equation depends only on \( P \), so it is autonomous.
02

Find Equilibrium Solutions

Equilibrium solutions occur when \( P' = 0 \). Set the equation \( 0.05P - 1000 = 0 \). Solving for \( P \), we get \[ P = \frac{1000}{0.05} = 20000 \]. Thus, \( P = 20000 \) is the equilibrium solution.
03

Sketch the Direction Field

Using a numerical solver, such as a computer algebra system or online tool, input the differential equation to generate the direction field. The field should show arrows indicating the slope \( P' = 0.05P - 1000 \) at various points.
04

Plot the Equilibrium Solution

Superimpose the equilibrium solution \( P = 20000 \) onto the direction field. This will appear as a horizontal line, since the derivative \( P' = 0 \) at this value indicates that there is no change in \( P \) when \( P = 20000 \).
05

Classify the Equilibrium Point

To determine stability, consider the derivative with respect to \( P \). We have \( P' = 0.05(P - 20000) \). If \( P < 20000 \), \( P' > 0 \) and \( P \) increases towards 20000. If \( P > 20000 \), \( P' < 0 \) and \( P \) decreases towards 20000. Thus, \( P = 20000 \) is an asymptotically stable equilibrium point.

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Key Concepts

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

Equilibrium Solutions
When dealing with autonomous differential equations like the one in our exercise, finding equilibrium solutions is a key step. Equilibrium solutions are values of the variable where the rate of change, or derivative, equals zero. This means the system is at rest or in a balanced state at these points. In our specific equation, \( P' = 0.05P - 1000 \), setting the derivative to zero helps us find these solutions.
  • Set \( P' = 0 \) to solve for \( P \).
  • Solve the equation \( 0.05P - 1000 = 0 \).
  • This results in the equilibrium solution \( P = \frac{1000}{0.05} = 20000 \).
Thus, \( P = 20000 \) is a steady state where the population does not change over time.
Direction Field
A direction field provides a visual representation of the behavior of solutions to a differential equation. It's a plot that shows small arrows indicating the slope of the solution at various points on a grid. This helps us understand how the solutions to the differential equation change over time.
To sketch the direction field for our differential equation, you can use a numerical solver or graphing tool:
  • Input the equation \( P' = 0.05P - 1000 \).
  • The solver generates a grid with arrows indicating the direction and magnitude of \( P' \) at various points.
  • This visual aid will help you see how solutions behave as \( P \) changes.
The horizon on which you see a constant or altering line helps map anticipated behavior over time. In this case, you'll see that around \( P = 20000 \), the arrows indicate no change, confirming our equilibrium point.
Stability Classification
Stability classification of equilibrium solutions helps us understand how a system behaves if it's disturbed from that point. An equilibrium is considered stable if small perturbations do not lead the system away from the equilibrium. If it returns to equilibrium, it's asymptotically stable; if it moves away, it's unstable.
To classify our equilibrium point at \( P = 20000 \), check adjacent behaviors:
  • For \( P < 20000 \), \( P' = 0.05(P - 20000) > 0 \), indicating \( P \) increases.
  • For \( P > 20000 \), \( P' = 0.05(P - 20000) < 0 \), indicating \( P \) decreases.
Both cases show the system returning towards \( P = 20000 \), declaring \( P = 20000 \) as an asymptotically stable equilibrium point. Such insights are crucial in predicting long-term behavior of the system.
Numerical Solver
A numerical solver is a tool or software that helps in solving differential equations, especially when analytical solutions are tough to obtain. It computes approximate values and visualizations for the equation's behavior.
Here's how you can utilize it to analyze \( P' = 0.05P - 1000 \):
  • Insert the differential equation into a solver like MATLAB, Python's NumPy/SciPy libraries, or an online tool.
  • Let it compute the direction field, showing different trajectories and equilibrium.
  • Superimpose the equilibrium solution on this field to confirm no net change when \( P = 20000 \).
These tools don't just make visualization easier; they also allow for experimenting with different parameters to see varied outcomes, a powerful method for learning and understanding differential equations.

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Most popular questions from this chapter

A 50 -gal tank initially contains 20 gal of pure water. Saltwater solution containing \(0.5 \mathrm{lb}\) of salt for each gallon of water begins entering the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\). Simultaneously, a drain is opened at the bottom of the tank, allowing the salt-water solution to leave the tank at a rate of \(2 \mathrm{gal} / \mathrm{min}\). What is the salt content (lb) in the tank at the precise moment that the tank is full of salt- water solution?

Each have the form \(P(x, y) d x+Q(x, y) d y=0\). In each case, show that \(P\) and \(Q\) are homogeneous of the same degree. State that degree. $$ (\ln x-\ln y) d x+d y=0 $$

Two tanks, Tank I and Tank II, ate filled with \(V\) gal of pure water. A solution containing \(a \mathrm{lb}\) of salt per gallon of water is poured into Tank I at a rate of \(b\) gal per minute. The solution leaves Tank I at a rate of \(b \mathrm{gal} / \mathrm{min}\) and enters Tank II at the same rate (b gal/min). A drain is adjusted on Tank II and solution leaves Tank II at a rate of \(b \mathrm{gal} / \mathrm{min}\). This keeps the volume of solution constant in both tanks ( \(V \mathrm{gal})\). Show that the amount of salt solution in Tank II, as a function of time \(t\), is given by \(a V-a b t e^{-(b / V) t}-a V e^{-(b / V) t}\).

Is it possible to find a function \(f(t, x)\) that is continuous and has continuous partial derivatives such that the functions \(x_{1}(t)=t\) and \(x_{2}(t)=\sin t\) are both solutions to \(x^{\prime}=f(t, x)\) near \(t=0\) ?

Determine which of the equations are exact and solve the ones that are. $$ \frac{x}{\sqrt{x^{2}+y^{2}}} d x+\frac{y}{\sqrt{x^{2}+y^{2}}} d y=0 $$
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