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Chapter 4: Problem 1

Solve the general logistic problem, $$ \frac{d y}{d t}=k y-c y^{2}, \quad y(0)=y_{0} $$ using separation of variables.

Short Answer

Expert verified
The solution of the logistic equation \( \frac{d y}{d t}=k y-c y^{2}, y(0)=y_{0} \) is given by \( y(t) = (y_{0} + \frac{1}{k}) e^{kt} - \frac{1}{k} \).

Step by step solution

01

Rewrite the Differential Equation

First, rewrite the equation in terms of \( y \) and \( t \) before moving towards separating these variables. The given equation can be rearranged as: \[ \frac{1}{k y - c y^{2}} dy = dt \]
02

Integrate Both Sides

The next task is to integrate both sides of the equation. However, the left-hand side can be simplified before proceeding with the integration. A simple tactic here is to use partial fraction decomposition for the left-hand side. Therefore we'll need to write \( \frac{1}{k y - c y^{2}} \) as \( \frac{A}{y} + \frac{B}{y^2} \), where \( A \) and \( B \) are constants that we need to find. To do so, clear the fractions: \( 1 = A y + B k \). Setting \( y = 0 \), we find \( B = - \frac{1}{k} \). So our fraction becomes \( \frac{-1/k}{y} + \frac{A}{y^2} \), and by observing, we find that \( A = -k \). So it simplifies to: \( \frac{k}{y} - \frac{1}{ky} \). Now we can integrate it: \( \int \frac{k}{y} dy - \int \frac{1}{k y^{2}} dy = \int dt \)
03

Resolve the Integrals

Now, perform the integrations: \( k \ln|y| + \frac{1}{ky} = t + C \), where \( C \) is the integration constant.
04

Solve for y

Solving the equation for \( y \) we first get rid of the absolute value sign and put \( y \) in terms of \( e \) using the equation: \( y = Ce^{kt} - \frac{1}{k} \).
05

Apply Initial Condition

After having solved for \( y \), we now substitute \( y(0) = y_{0} \) into the equation to get the value of the constant \( C = y_{0} + \frac{1}{k} \). And substituting back this into the equation we previously derived gives \( y(t) = (y_{0} + \frac{1}{k}) e^{kt} - \frac{1}{k} \).

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Key Concepts

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

Separation of Variables
The method of separation of variables is a technique commonly used to solve differential equations, where the equation is rearranged so that each variable appears on a separate side of the equation. It's particularly useful for ordinary differential equations (ODEs) of the first order.

For the logistic differential equation given, \[\frac{d y}{d t}=k y-c y^{2}\], we aim to separate variables y and t. The separation involves algebraically manipulating the equation so that all terms involving y are on one side and those involving t are on the other. By dividing both sides by \(ky-cy^{2}\), and multiplying by \(dt\), we can rewrite the equation as \(\frac{1}{k y - c y^{2}} dy = dt\).

Once the variables are separated, we integrate both sides of the equation. This essentially means we sum up the infinitesimally small changes in each variable to find a functional relationship between y and t. This step brings us closer to the solution by providing us with a general equation representing the behavior of the system over time.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex fractions into simpler parts that are easier to integrate. In differential equations, this comes in handy when dealing with rational expressions.

In our logistic differential equation, we encounter the term \(\frac{1}{k y - c y^{2}}\), which can be difficult to integrate as is. To simplify this, we express it in terms of its partial fractions: \(\frac{A}{y} + \frac{B}{y^2}\), where A and B are constants.

To determine A and B, we multiply through by the common denominator to clear the fractions, yielding: \(1 = Ay + Bk\). By strategically choosing values for y, such as y = 0, we can solve for B, and by comparing coefficients or substituting other values for y, we can solve for A. The goal is to rewrite the complex fraction in terms of its simpler, component fractions, enabling us to integrate the equation more easily.
Integrating Differential Equations
After using partial fraction decomposition, we can integrate the simplified terms. Integration is the process of finding the antiderivative or the area under the curve represented by the derivative. When we integrate both sides of our separated equation, we find the integral of each term with respect to its own variable.

For our example, we compute \(\int \frac{k}{y} dy - \int \frac{1}{k y^{2}} dy = \int dt\). Solving these integrals, we use the natural logarithm for the \(\int \frac{k}{y} dy\) term and a simple power rule for the \(\int \frac{1}{k y^{2}} dy\) term. This leaves us with \(k \ln|y| + \frac{1}{ky}\) on the left side and \(t + C\), where C is the constant of integration, on the right side.

Eventually, after integrating and rearranging back in terms of y, we obtain the general solution for the differential equation. If an initial condition is given, such as \(y(0) = y_{0}\), we can use it to find the exact solution by determining the integration constant C and hence solving the specific instance of the differential equation.

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Most popular questions from this chapter

Consider a model of a food chain of three species. Assume that each population on its own can be modeled by logistic growth. Let the species be labeled by \(x(t), y(t)\), and \(z(t)\). Assume that population \(x\) is at the bottom of the chain. That population will be depleted by population \(y\). Population \(y\) is sustained by \(x^{\prime}\) s, but eaten by \(z^{\prime}\) s. A simple, but scaled, model for this system can be given by the system $$ \begin{aligned} x^{\prime} &=x(1-x)-x y \\ y^{\prime} &=y(1-y)+x y-y z \\ z^{\prime} &=z(1-z)+y z \end{aligned} $$ a. Find the equilibrium points of the system. b. Find the Jacobian matrix for the system and evaluate it at the equilibrium points. c. Find the eigenvalues and eigenvectors. d. Describe the solution behavior near each equilibrium point. e. Which of these equilibria are important in the study of the population model and describe the interactions of the species in the neighborhood of these point(s).

For \(y^{\prime}=y-y^{2}\), find the general solution corresponding to \(y(0)=y_{0}\). Provide specific solutions for the following initial conditions and sketch them: a. \(y(0)=0.25\), b. \(y(0)=1.5\), and c. \(y(0)=-0.5\).

Consider the family of differential equations \(x^{\prime}=x^{3}+\delta x^{2}-\mu x\) a. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta=0\). b. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta>0\). Hint: Pick a few values of \(\delta\) and \(\mu\) in order to get a feel for how this system behaves.

The Michaelis-Menten kinetics reaction is given by $$ E+S \frac{k_{3}}{k_{1}}+E S \underset{k_{2}}{ } E+P $$ The resulting system of equations for the chemical concentrations is $$ \begin{aligned} \frac{d[S]}{d t} &=-k_{1}[E][S]+k_{3}[E S] \\ \frac{d[E]}{d t} &=-k_{1}[E][S]+\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[E S]}{d t} &=k_{1}[E][S]-\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[P]}{d t} &=k_{3}[E S] \end{aligned} $$ In chemical kinetics, one seeks to determine the rate of product formation \(\left(v=d[P] / d t=k_{3}[E S]\right)\). Assuming that \([E S]\) is a constant, find \(v\) as a function of \([S]\) and the total enzyme concentration \(\left[E_{T}\right]=[E]+[E S] .\) As a nonlinear dynamical system, what are the equilibrium points?

In Equation (3.153), we saw a linear version of an epidemic model. The commonly used nonlinear SIR model is given by $$ \begin{aligned} \frac{d S}{d t} &=-\beta S I \\ \frac{d I}{d t} &=\beta S I-\gamma I \\ \frac{d R}{d t} &=\gamma I \end{aligned} $$ where \(S\) is the number of susceptible individuals, \(I\) is the number of infected individuals, and \(R\) is the number who have been removed from the other groups, either by recovering or dying. a. Let \(N=S+I+R\) be the total population. Prove that \(N=\) constant. Thus, one need only solve the first two equations and find \(R=N-S-I\) afterward. b. Find and classify the equilibria. Describe the equilibria in terms of the population behavior. c. Let \(\beta=0.05\) and \(\gamma=0.2\). Assume that in a population of 100 there is one infected person. Numerically solve the system of equations for \(S(t)\) and \(I(t)\) and describe the solution being careful to determine the units of population and the constants. d. The equations can be modified by adding constant birth and death rates. Assuming these rates are the same, one would have a new system. $$ \begin{aligned} \frac{d S}{d t} &=-\beta S I+\mu(N-S) \\ \frac{d I}{d t} &=\beta S I-\gamma I-\mu I \\ \frac{d R}{d t} &=-\gamma I-\mu R \end{aligned} $$ How does this affect any equilibrium solutions? e. Again, let \(\beta=0.05\) and \(\gamma=0.2\). Let \(\mu=0.1\) For a population of 100 with one infected person, numerically solve the system of equations for \(S(t)\) and \(I(t)\) and describe the solution being careful to determine the units of populations and the constants.
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