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Give the general solution to the logistic differential equation. $$\frac{d P}{d t}=0.05 P\left(1-\frac{P}{2800}\right)$$

Short Answer

Expert verified
The general solution is \( P(t) = \frac{2800Ae^{0.05t}}{1 + Ae^{0.05t}} \) for constant \( A \).

Step by step solution

01

Understand the Logistic Differential Equation

The logistic differential equation is often given in the form \( \frac{dP}{dt} = rP \left(1 - \frac{P}{K}\right) \), where \( r \) is the growth rate and \( K \) is the carrying capacity. In this exercise, \( r = 0.05 \) and \( K = 2800 \).
02

Separate Variables

To solve the equation, we first separate the variables. Rewrite the equation as \( \frac{dP}{P(1-P/2800)} = 0.05 \, dt \). This allows us to integrate both sides separately.
03

Integrate the Variables

Perform partial fraction decomposition on \( \frac{1}{P(1-P/2800)} \) and integrate both sides. We can express \( \frac{1}{P(1-P/2800)} \) as \( \frac{1}{P} + \frac{1/2800}{1-P/2800} \). Thus, integrate to get: \[ \int \left( \frac{1}{P} + \frac{1/2800}{1-P/2800} \right) \, dP = \int 0.05 \, dt \].
04

Solve the Integrals

The integral on the left is \( \ln |P| - \ln |1-P/2800| = 0.05t + C \), where \( C \) is the constant of integration. It can be rewritten as \( \ln \left| \frac{P}{2800 - P} \right| = 0.05t + C \).
05

Exponentiate Both Sides

Exponentiate both sides to solve for \( P \): \( \frac{P}{2800 - P} = e^{0.05t + C} = Ae^{0.05t} \), where \( A \) is a constant.
06

Solve for P

Rearrange to find \( P \) in terms of \( t \): \( P = \frac{2800A e^{0.05t}}{1 + A e^{0.05t}} \). This is the general solution of the logistic differential equation.

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Key Concepts

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

Separation of Variables
Separation of Variables is a method used to solve differential equations by separating variables on different sides of the equation. In the context of the logistic differential equation, our goal is to rearrange the terms so that each differential contains only one variable.

Consider the example equation:
\[ \frac{dP}{dt} = 0.05P \left(1-\frac{P}{2800}\right) \]
We start by rewriting this equation in a separable form:
\[ \frac{dP}{P(1-P/2800)} = 0.05 \, dt \]
This step is crucial as it allows us to integrate both sides independently. The left side of the equation contains all terms involving \( P \), while the right side handles \( t \).

By separating the variables, we clarify the relationship between \( P \) and \( t \) and enable the integration process that follows. This technique is particularly helpful for examining population dynamics using logistic equations.
Partial Fraction Decomposition
Partial Fraction Decomposition breaks down a complicated fraction into simpler, more manageable parts. This decomposition is especially useful when dealing with the logistic differential equation since it allows us to integrate the expression more easily.

In the given logistic differential equation, after separation, we have:
\[ \int \frac{1}{P(1-P/2800)} \, dP \]
By using partial fraction decomposition, we can express this as:
\[ \frac{1}{P} + \frac{1/2800}{1-P/2800} \]
Integrating each part separately becomes straightforward:
  • \( \int \frac{1}{P} \, dP = \ln |P| \)
  • \( \int \frac{1/2800}{1-P/2800} \, dP = -\ln |1-P/2800| \)

The decomposition helps simplify complex expressions, making the integration process intuitive and more approachable.
Carrying Capacity
Carrying Capacity, denoted as \( K \) in the logistic differential equation, represents the maximum population size that an environment can sustainably support. It plays a vital role in modeling real-world phenomena where resources such as food, space, or competition naturally limit population growth.

In our logistic equation:
\[ \frac{dP}{dt} = 0.05P \left(1-\frac{P}{2800}\right) \]
The carrying capacity \( K \) is 2800. This value indicates the threshold population size at which the growth rate slows to zero because the environment can't support more than this population. When \( P \) (the population) approaches \( K \), the term \( 1-\frac{P}{K} \) becomes zero, halting growth.

Understanding carrying capacity is essential when analyzing population models, as it explains why populations don't grow indefinitely but stabilize around a certain level. This concept is fundamental in ecology, conservation, and resource management, where the limitations of the environment must be considered.

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Most popular questions from this chapter

Table 11.7 gives values for a logistic function \(P=f(t)\) (a) Estimate the maximum rate of change of \(P\) and estimate the value of \(t\) when it occurs. (b) If \(P\) represents the growth of a population, estimate the carrying capacity of the population. $$\begin{array}{c|c|c|c|c|c|c|c|c}\hline t & 0 & 10 & 20 & 30 & 40 & 50 & 60 & 70 \\ \hline P & 120 & 125 & 135 & 155 & 195 & 270 & 345 & 385 \\\\\hline\end{array}$$

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Are the statements true or false? Give an explanation for your answer. The solutions of the differential equation \(d y / d x=x^{2}+\) \(y^{2}+1\) are concave up at every point.

(a) A cup of coffee is made with boiling water and stands in a room where the temperature is \(20^{\circ} \mathrm{C}\) If \(H(t)\) is the temperature of the coffee at time \(t,\) in minutes, explain what the differential equation $$\frac{d H}{d t}=-k(H-20)$$ says in everyday terms. What is the sign of \(k ?\) (b) Solve this differential equation. If the coffee cools to \(90^{\circ} \mathrm{C}\) in 2 minutes, how long will it take to cool to \(60^{\circ} \mathrm{C}\) degrees?

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(x)\) is even, then \(f(x)\) is odd.

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