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Give the general solution to the logistic differential equation. $$\frac{d P}{d t}=0.2 P-0.0008 P^{2}$$

Short Answer

Expert verified
The general solution is \(P(t) = \frac{0.2}{0.0008 + Ce^{-0.04t}}\).

Step by step solution

01

Identify the Form of the Logistic Equation

The given differential equation is \(\frac{dP}{dt} = 0.2P - 0.0008P^2\). This is in the form of a logistic growth equation \(\frac{dP}{dt} = rP - kP^2\), where \(r = 0.2\) and \(k = 0.0008\). The logistic equation describes population growth with a carrying capacity.
02

Express in Separable Form

To solve the logistic differential equation, express it in separable form: \(\frac{dP}{P(0.2 - 0.0008P)} = dt\). This separation allows us to integrate both sides.
03

Integrate Both Sides

Integrate the left-hand side with respect to \(P\) and the right-hand side with respect to \(t\):\[\int \frac{dP}{P(0.2 - 0.0008P)} = \int dt.\]We use partial fraction decomposition for the left-hand side integration.
04

Partial Fraction Decomposition

Decompose the fraction for integration: express \(\frac{1}{P(0.2 - 0.0008P)}\) as \(\frac{A}{P} + \frac{B}{0.2 - 0.0008P}\). Solve for \(A\) and \(B\) by setting \[1 = A(0.2 - 0.0008P) + BP,\] which gives \(A = 5\) and \(B = 1250\).
05

Complete the Integration

Using the partial fraction decomposition, integrate both parts:\[\int \frac{5}{P} \, dP + \int \frac{1250}{0.2 - 0.0008P} \, dP = \int dt.\]This results in:\[5 \ln |P| - 1562.5 \ln|0.2 - 0.0008P| = t + C,\]where \(C\) is the constant of integration.
06

Solve for P

Simplify and express \(P\) explicitly:\[\ln \left| \frac{P}{0.2 - 0.0008P} \right| = \frac{t+C}{5}.\]Exponentiating both sides gives:\[\frac{P}{0.2 - 0.0008P} = e^{\frac{t+C}{5}}.\]Rearrange to express \(P\) explicitly.
07

Solve for General Solution

On solving, we get the general solution as:\[P(t) = \frac{0.2}{0.0008 + Ce^{-0.04t}},\]where \(C = e^{\frac{C_1}{5}}\) is a constant determined by initial conditions.

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Key Concepts

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

Population Growth
Population growth can often be modeled using differential equations that describe how the number of individuals in a population changes over time. In most cases, simple exponential growth in an unlimited environment is portrayed inaccurately because environmental limits always exist. Thus, more realistic models like the logistic growth model are used, which consider a decelerating growth rate as the population size reaches environmental limits.

The logistic growth model is particularly useful as it incorporates both the birth and death rates, factoring in competition for limited resources in the environment. It starts with a period of rapid growth, which slows down as the population approaches its carrying capacity. This concept ensures that the population's growth is sustainable, accounting for real-world scenarios where resources are finite.
Carrying Capacity
Carrying capacity is a critical concept when dealing with population growth models, particularly the logistic growth equation. It represents the maximum population size that an environment can sustain indefinitely. In mathematical terms, it's the point at which the growth rate of the population changes from being positive to negative.

In logistic differential equations, the carrying capacity determines the stability of the population. An equilibrium state is reached when the population size is equal to this capacity, causing the growth to halt. In our differential equation model, the carrying capacity is determined by the coefficient values, specifically it's the ratio of the growth rate coefficient to the competition coefficient.

To calculate it from a logistics equation like \(\frac{dP}{dt} = 0.2P - 0.0008P^2\), the carrying capacity can be found by setting the growth term to zero and solving for \(P\), giving \(P = \frac{0.2}{0.0008} = 250\).
Separable Differential Equations
Differential equations are powerful tools for modeling various dynamic systems, and the logistic growth equation is no exception. Separable differential equations are a type that can be rewritten allowing us to separate the variables on each side of the equation, making them easier to integrate and solve. This technique is instrumental in solving logistic differential equations.

In the separable form, the equation is expressed as a product of a function of the dependent variable and a function of the independent variable. This separation is crucial in integration, as it allows us to apply necessary calculus tools effectively. For instance, starting with \(\frac{dP}{dt} = 0.2P - 0.0008P^2\), this can be rewritten as \(\frac{dP}{P(0.2 - 0.0008P)} = dt\), creating a clear path for integration on both sides.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used in calculus to simplify the integration process of rational functions. When you're dealing with separable differential equations, you often encounter expressions that can benefit from this approach, particularly when integrating complex fractions.

In our logistic growth equation, we decompose the term \(\frac{1}{P(0.2 - 0.0008P)}\) into simpler fractions: \(\frac{A}{P} + \frac{B}{0.2 - 0.0008P}\). Finding the constants \(A\) and \(B\) involves algebraic manipulation to match terms, simplifying the integration process considerably.

This decomposition helps to break the complex fraction into parts that are easier to handle, allowing the integrals to be solved using logarithmic functions, ultimately enabling us to solve the logistic differential equation to find the population growth model's general solution. This technique is essential for handling the nonlinear terms present in logistic equations.

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Most popular questions from this chapter

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 \(g(0)=1\) and \(x(x)\) is increasing for \(x \geq 0,\) then \(f(x)\) is also increasing for \(x \geq 0.\)

The differential equation \(d Q / d t=-0.15 Q+25\) represents the quantity of a drug in the body if the drug is metabolized at a continuous rate of \(15 \%\) per day and an IV line is delivering the drug at a constant rate of \(25 \mathrm{mg}\) per hour.

(a) Expand \(A \sin (\omega t+\phi)\) using the trigonometric identity \(\sin (x+y)=\sin x \cos y+\cos x \sin y\) (b) Assume \(A>0 .\) If \(A \sin (\omega t+\phi)=C_{1} \cos \omega t+\) \(C_{2} \sin \omega t,\) show that we must have $$A=\sqrt{C_{1}^{2}+C_{2}^{2}} \text { and } \tan \phi=C_{1} / C_{2}$$

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 $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{35}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?

When a gas expands without gain or loss of heat, the rate of change of pressure with respect to volume is proportional to pressure divided by volume. Find a law connecting pressure and volume in this case.

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