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Chapter 2: Problem 16

Consider the initial value problem $$ \frac{d P}{d t}=r(t)\left(1-\frac{P}{P_{e}}\right) P, \quad P(0)=P_{0} . $$ Observe that the differential equation is separable. Let \(R(t)=\int_{0}^{t} r(s) d s\). Solve the initial value problem. Note that your solution will involve the function \(R(t)\).

Short Answer

Expert verified
Question: Given the differential equation \(\frac{dP}{dt} = r(t) \left( 1- \frac{P}{P_e} \right) P\), solve the initial value problem for P(t) involving the function R(t) = \(\int_0^t r(s) ds\) and initial condition \(P(0)=P_0\). Answer: \(P(t) = P_e - P_e e^{R(0)-R(t)}\left(1-\frac{P_{0}}{P_e}\right)\)

Step by step solution

01

Rewrite the differential equation as a separable equation

We are given the differential equation \(\frac{dP}{dt} = r(t) \left( 1- \frac{P}{P_e} \right) P\). In order to rewrite it, we can divide both sides by \(P \left( 1- \frac{P}{P_e} \right)\) and rearrange terms accordingly: $$\frac{1}{P\left(1-\frac{P}{P_e}\right)}\frac{dP}{dt}=r(t)$$
02

Integrate both sides with respect to t

Now we can integrate both sides with respect to t: $$\int \frac{1}{P\left(1-\frac{P}{P_e}\right)} \frac{dP}{dt} dt = \int r(t) dt$$ On the left side, dt will cancel out with dt in the integral, and we get: $$\int \frac{1}{P\left(1-\frac{P}{P_e}\right)} dP = \int r(t) dt$$
03

Perform the integration

On the left side, we perform a substitution: \(u=1-\frac{P}{P_e} \Rightarrow -\frac{1}{P_e}du = dP\). Also, we know that \(\int r(t) dt = R(t) + C\) where C is a constant. Thus, the equation becomes, $$-\frac{1}{P_e}\int \frac{1}{P(1-u)}(-P_edu) = R(t) + C$$ Now we can simplify the left side: $$\int \frac{1}{(1-u)} du = R(t) + C$$
04

Integrate the left side and apply initial conditions

Now, the left side of the equation can be integrated as: $$\int \frac{1}{(1-u)} du = -\ln |1-u|$$ So now, we have the following equation: $$-\ln |1-u| = R(t) + C$$ We know \(u=1-\frac{P}{P_e}\) and \(\frac{P}{P_e}=1-u\), so we can rewrite our equation as: $$-\ln\left|1-\frac{P}{P_e}\right| = R(t) + C$$ Now, we apply our initial condition \(P(0) = P_0\). Substituting this, we get: $$-\ln\left|1-\frac{P_{0}}{P_e}\right| = R(0) + C$$ So, $$C = -\ln\left|1-\frac{P_{0}}{P_e}\right| - R(0)$$
05

Solve for P(t)

Now that we have the value of C, we can write our equation as: $$-\ln\left|1-\frac{P}{P_e}\right|=R(t)-\ln\left|1-\frac{P_{0}}{P_e}\right| + R(0)$$ Raise both sides as powers of e: $$1-\frac{P}{P_e} = e^{R(0)-R(t)}\left(1-\frac{P_{0}}{P_e}\right)$$ Now, solve for P(t): $$P(t) = P_e - P_e e^{R(0)-R(t)}\left(1-\frac{P_{0}}{P_e}\right)$$ Thus, we have obtained the solution for P(t) involving the function R(t).

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Key Concepts

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

Initial Value Problem
An Initial Value Problem (IVP) in differential equations is a common scenario where you have to find a function that satisfies a differential equation along with an initial condition. In the exercise given, the differential equation describes how something changes over time, while the initial condition, \( P(0) = P_0 \), gives us the starting value of the function. By solving the IVP, we find the function \( P(t) \) that not only solves the differential equation but also matches the initial condition at \( t = 0 \). **Why Initial Conditions Matter**
  • Initial conditions specify the unique solution among the infinite possibilities the differential equation might admit.
  • They reflect real-world constraints or starting points, such as the initial population of a species or the initial volume of liquid in a tank.
Separable Equations
Separable Equations are a special type of differential equation that can be factored into two separate parts, one involving only the dependent variable and the other only the independent variable. This makes them particularly easy to solve using integration. **Steps to separate and solve such equations:**
  • Reorganize the equation by moving all terms involving the dependent variable, \( P \), to one side.
  • Move all terms involving the independent variable, \( t \), to the other side.
For the given problem, the differential equation is structured in a way that allows separation:\[\frac{1}{P\left(1-\frac{P}{P_e}\right)}\frac{dP}{dt}=r(t)\]By arranging terms, we isolate the variables to integrate independently. This is a crucial step, as it lets us integrate both sides, leading toward finding a solution.
Integration
Integration is the mathematical process used to find the original function from its rate of change, often represented by a differential equation. In the exercise, integration is utilized to solve the separated differential equation.Performing integration involves:
  • Applying the integration operator to both sides of the equation after separation.
  • Handling each integral independently, keeping track of constants of integration as they arise.
In the problem, integration transforms:\[\int \frac{1}{P\left(1-\frac{P}{P_e}\right)} dP = \int r(t) dt\]The right side of the equation becomes \( R(t) + C \), where \( C \) is a constant. Integrating the left helps us build a function that solutions the differential equation over time. Each solution represents how the system evolves from its initial condition.
Population Dynamics
Population Dynamics studies the changes in population sizes and densities over time, often through mathematical models like differential equations. The exercise showcases a classic model used to describe how a population \( P \) changes over time with growth rate \( r(t) \) and carrying capacity \( P_e \).**Key Components of the Model:**
  • Growth Rate \( r(t) \): This function represents the growth speed at which the population increases over time at different rates.
  • Carrying Capacity \( P_e \): The maximum population size that the environment can sustain indefinitely.
  • Logistic Model: The equation in the exercise represents a logistic growth scenario, commonly seen in population studies, where population growth is rapid initially, then slows as it approaches the carrying capacity.
Understanding these features helps in interpreting the solution \( P(t) \), providing insights into how a population adapts given initial conditions and environmental limitations.

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Most popular questions from this chapter

First order linear differential equations possess important superposition properties. Show the following: (a) If \(y_{1}(t)\) and \(y_{2}(t)\) are any two solutions of the homogeneous equation \(y^{\prime}+p(t) y=0\) and if \(c_{1}\) and \(c_{2}\) are any two constants, then the sum \(c_{1} y_{1}(t)+c_{2} y_{2}(t)\) is also a solution of the homogeneous equation. (b) If \(y_{1}(t)\) is a solution of the homogeneous equation \(y^{\prime}+p(t) y=0\) and \(y_{2}(t)\) is a solution of the nonhomogeneous equation \(y^{\prime}+p(t) y=g(t)\) and \(c\) is any constant, then the sum \(c y_{1}(t)+y_{2}(t)\) is also a solution of the nonhomogeneous equation. (c) If \(y_{1}(t)\) and \(y_{2}(t)\) are any two solutions of the nonhomogeneous equation \(y^{\prime}+p(t) y=g(t)\), then the sum \(y_{1}(t)+y_{2}(t)\) is not a solution of the nonhomogeneous equation.

A 5000 -gal aquarium is maintained with a pumping system that passes 100 gal of water per minute through the tank. To treat a certain fish malady, a soluble antibiotic is introduced into the inflow system. Assume that the inflow concentration of medicine is \(10 t e^{-t / 50} \mathrm{mg} / \mathrm{gal}\), where \(t\) is measured in minutes. The well-stirred mixture flows out of the aquarium at the same rate. (a) Solve for the amount of medicine in the tank as a function of time. (b) What is the maximum concentration of medicine achieved by this dosing and when does it occur? (c) For the antibiotic to be effective, its concentration must exceed \(100 \mathrm{mg} / \mathrm{gal}\) for a minimum of \(60 \mathrm{~min}\). Was the dosing effective?

Oscillating Inflow Concentration A tank initially contains \(10 \mathrm{lb}\) of salt dissolved in 200 gal of water. Assume that a salt solution flows into the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out at the same rate. Assume that the inflow concentration oscillates in time, however, and is given by \(c_{i}(t)=0.2(1+\sin t) \mathrm{lb}\) of salt per gallon. Thus, as time evolves, the concentration oscillates back and forth between 0 and \(0.4 \mathrm{lb}\) of salt per gallon. (a) Make a conjecture, on the basis of physical reasoning, as to whether or not you expect the amount of salt in the tank to reach a constant equilibrium value as time increases. In other words, will \(\lim _{t \rightarrow \infty} Q(t)\) exist? (b) Formulate the corresponding initial value problem. (c) Solve the initial value problem. (d) Plot \(Q(t)\) versus \(t\). How does the amount of salt in the tank vary as time becomes increasingly large? Is this behavior consistent with your intuition?

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ (2 y-\sin y) y^{\prime}+t=\sin t, \quad y(0)=0 $$

Oscillating Flow Rate A tank initially contains \(10 \mathrm{lb}\) of solvent in \(200 \mathrm{gal}\) of water. At time \(t=0\), a pulsating or oscillating flow begins. To model this flow, we assume that the input and output flow rates are both equal to \(3+\sin t \mathrm{gal} / \mathrm{min}\). Thus, the flow rate oscillates between a maximum of \(4 \mathrm{gal} / \mathrm{min}\) and a \(\mathrm{minimum}\) of \(2 \mathrm{gal} / \mathrm{min}\); it repeats its pattern every \(2 \pi \approx 6.28 \mathrm{~min}\). Assume that the inflow concentration remains constant at \(0.5 \mathrm{lb}\) of solvent per gallon. (a) Does the amount of solution in the tank, \(V\), remain constant or not? Explain. (b) Let \(Q(t)\) denote the amount of solvent (in pounds) in the tank at time \(t\) (in minutes). Explain, on the basis of physical reasoning, whether you expect the amount of solvent in the tank to approach an equilibrium value or not. In other words, do you expect \(\lim _{t \rightarrow \infty} Q(t)\) to exist and, if so, what is this limit? (c) Formulate the initial value problem to be solved. (d) Solve the initial value problem. Determine \(\lim _{t \rightarrow \infty} Q(t)\) if it exists.
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