Question

Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume \(t \geq 0\) and \(P \geq 0\). $$P^{\prime}(t)=0.05 P-0.001 P^{2} ; P(0)=10, P(0)=40, P(0)=80$$

Short answer

Question: Sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions for the logistic equation $$P^{\prime}(t)=0.05 P-0.001 P^{2}$$ with initial conditions \(P(0)=10, P(0)=40\), and \(P(0)=80\). Answer: The equilibrium solutions are \(P = 0\) and \(P = 50\). The solutions for each initial condition are as follows: 1. \(P(t) = \frac{50}{4.9\times e^{0.05t} - 0.4}\) 2. \(P(t) = \frac{50}{0.1\times e^{0.05t} + 1}\) 3. \(P(t) = \frac{50}{0.1\times e^{0.05t} - 0.2}\) A simple sketch of the direction field can be created based on these results, showing the equilibrium solutions and the solution curves corresponding to the given initial conditions.

Step-by-step solution

Setting up the equation and initial conditions

The logistic equation is given by: $$P^{\prime}(t)=0.05 P-0.001 P^{2}$$ The initial conditions are: 1. \(P(0) = 10\) 2. \(P(0) = 40\) 3. \(P(0) = 80\) These initial conditions will be used to find the solution curves and a direction field.

Find the equilibrium solutions

To find equilibrium solutions, we need to find the values of P where \(P^{\prime}(t)=0\). Thus, \(0=0.05P - 0.001P^2\) By factoring, we get \(0 = P(0.05-0.001P)\) The equilibrium solutions are \(P=0\) and \(P=50\).

Find an expression for \(P(t)\)

We apply the method of separation of variables to the given logistic equation: $$\frac{dP}{dt} = 0.05P - 0.001P^2$$ Separate these variables: $$\frac{dP}{(0.05P - 0.001P^2)} = dt$$ Integrate both sides: $$\int\frac{1}{0.05P - 0.001P^2}dP=\int dt$$ Let \(u=0.05P-0.001P^2\), then \(-\frac{200}{49}du=PdP.\) We can then rewrite the left integral: $$-\frac{49}{200}\int\frac{1}{u}du=\int dt$$ Integrate: $$-\frac{49}{200}\ln|u|=t+C$$ Where \(C\) is the constant of integration.

Solve for P(t) and substitute the initial conditions

Organize the equation to isolate \(P(t)\), and then substitute the initial conditions one by one. $$\ln|0.001P^2-0.05P|=200t+C$$ Use the initial conditions to find \(C\) for each case. 1. \(P(0) = 10 \implies C = \ln(1)\) 2. \(P(0) = 40 \implies C = \ln(8)\) 3. \(P(0) = 80 \implies C = \ln(9)\)

Final Result

The following solutions result from each initial condition: 1. \(P(t) = \frac{50}{4.9\times e^{0.05t} - 0.4}\) 2. \(P(t) = \frac{50}{0.1\times e^{0.05t} + 1}\) 3. \(P(t) = \frac{50}{0.1\times e^{0.05t} - 0.2}\) The direction field and solution curves can be sketched based on these results and the equilibrium solutions.

Key concepts

Equilibrium Solutions

In the context of logistic equations, equilibrium solutions represent the stable populations where the rate of change is zero. For our given logistic equation, equilibrium solutions are found by setting the derivative of the population, \(P' (t)\), equal to zero. This lets us determine at which population levels the growth ceases to change.

By solving \(0 = 0.05P - 0.001P^2\), we factor and find two solutions: \(P = 0\) and \(P = 50\).

• \(P = 0\) corresponds to the absence of a population—a trivial case.
• \(P = 50\) indicates a stable population where growth is balanced by limiting factors.

Equilibrium solutions are useful because they help determine the behavior of the population over time and are starting points for analyzing the entire system's dynamics.

Direction Field

A direction field provides a visual representation of the behavior of differential equations, helping us understand how solutions evolve over time. For logistic equations, it shows how population changes at different levels of \(P\), allowing us to visualize how the system behaves for varying initial conditions.

In the specific case given, the direction field consists of small directional arrows representing \(P' (t)\). They indicate the slope of the solution curves for small intervals of \(t\).

By sketching this field, we can observe:

• How solutions approach equilibrium points, \(P = 0\) and \(P = 50\).
• Transitions between growth and decay phases of the population.
• Convergence of the solutions toward equilibrium depending on initial conditions.

This graphical representation is extremely useful for getting a quick overview of the system's dynamics without solving the equation analytically.

Separation of Variables

Separation of variables is a technique used to solve differential equations by separating the variables into two sides of the equation. This method is particularly effective for our logistic equation. It allows us to manipulate the equation so that each variable appears on one side, making it integrable.

For the given logistic equation, it is separated as follows:\[\frac{dP}{0.05P - 0.001P^2} = dt\]Integrating both sides leads to logarithmic expressions that can be solved algebraically to find \(P(t)\).

The separation process translates the problem of solving a differential equation into evaluating integrals, which are typically more straightforward to manage. Once completed, we can determine the solution curve of the population over time.

Initial Conditions

Initial conditions are crucial in solving differential equations because they ensure that the solution curve matches the specific scenario described by the problem. For logistic equations, initial conditions specify the starting population levels at time \(t = 0\).

In our case, we have three different scenarios with initial conditions:

• \(P(0) = 10\)
• \(P(0) = 40\)
• \(P(0) = 80\)

Applying each initial condition to the integrated equation allows us to solve for the constant \(C\) in the logarithmic expression, tailoring the solution to each scenario.

This step guarantees that each derived \(P(t)\) solution accurately describes how the population evolves from its initial state, informing us how it aligns with equilibrium and its future trajectory.