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Chapter 2: Problem 21

Solve the initial value problem $$ \frac{d P}{d t}=r\left(1-\frac{P}{P_{e}}\right) P, \quad P(0)=P_{0} $$ by viewing the differential equation as a Bernoulli equation.

Short Answer

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Question: Solve the initial value problem involving a Bernoulli differential equation, given by: $$ \frac{d P}{d t}=r\left(1-\frac{P}{P_{e}}\right) P, \quad P(0) = P_0 $$ Answer: The solution to the given initial value problem is: $$ P(t) = \frac{1}{-rt + \frac{1}{P_0}} $$

Step by step solution

01

Identify the Bernoulli equation

The given differential equation is: $$ \frac{d P}{d t}=r\left(1-\frac{P}{P_{e}}\right) P $$ This is a Bernoulli equation with n=-1.
02

Apply substitution method

Let: $$ u=\frac{1}{P} $$ Now we compute \(\frac{du}{dt}\): $$ \frac{d u}{d t} = -\frac{1}{P^2}\frac{d P}{d t} $$ Substitute the given differential equation into this expression: $$ \frac{d u}{d t} = -\frac{1}{P^2}(r\left(1-\frac{P}{P_{e}}\right) P) = -r \left(1 - \frac{P}{P_e}\right) $$ Notice that we now have a linear differential equation in \(u\).
03

Solve the linear differential equation

The linear differential equation we need to solve is: $$ \frac{d u}{d t} = -r \left(1 - \frac{P}{P_e}\right) $$ Now substitute back using \(u=\frac{1}{P}\) and solve for P: $$ \frac{d\left(\frac{1}{P}\right)}{d t}=-r\left(1-\frac{P}{P_e}\right) \Rightarrow \frac{-d P}{P^2} = -r\left(1 - \frac{P}{P_e}\right) dt $$ Now integrate both sides: $$ \int \frac{-d P}{P^2} = \int -r\left(1 - \frac{P}{P_e}\right) dt $$ This leads to: $$ \frac{1}{P} = -rt + C $$
04

Use initial condition

Now we use the initial condition \(P(0) = P_0\) to find the constant C: $$ \frac{1}{P_0} = -r(0) + C \Rightarrow C = \frac{1}{P_0} $$ So we have: $$ \frac{1}{P} = -rt + \frac{1}{P_0} $$
05

Solve for P(t)

Finally, we solve for \(P(t)\): $$ P(t) = \frac{1}{-rt + \frac{1}{P_0}} $$ This is the solution to the initial value problem.

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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 is a type of differential equation paired with a specific initial condition. This means, at the start, or at "time zero" (like when t = 0), the system begins with a known value, given as \( P(0) = P_0 \) in our equation. It's like setting the starting line for an experiment.
  • The differential equation describes how variables change, like how the population grows.
  • The initial condition provides the starting point, telling us the initial population, \( P_0 \).
  • This helps us predict future values because knowing where you start helps you figure out where you鈥檒l go.
Overall, solving such problems helps model real-world phenomena by making calculations relevant to a specific context.
Differential Equations
Differential equations like \( \frac{dP}{dt} = r\left(1-\frac{P}{P_e}\right)P \) describe relationships involving rates of change. They play a crucial role in physics, engineering, and economics.
  • They express how a quantity changes over time, such as a population or speed.
  • The left side, \( \frac{dP}{dt} \), signifies how a variable, P, changes as time (t) passes.
  • The right side defines the relationship causing this change, such as environmental factors or growth rates, here modified by one less the fraction of P over carrying capacity \(P_e\).
Solving these equations helps to predict behavior under various conditions, providing insights into the dynamics of the system being studied.
Substitution Method
The substitution method is a handy technique for simplifying complex differential equations, like changing variables to make equations easier to handle. In this problem, we set \( u = \frac{1}{P} \) to transform our Bernoulli equation into a more manageable form.
  • This helps rewrite the equation, turning something cumbersome into something linear and straightforward.
  • By substituting, we convert difficult non-linear forms, like when working with \( \frac{dP}{dt} = r\left(1-\frac{P}{P_e}\right)P \), into linear equations in terms of another variable (u).
  • The process simplifies solving once we have an equation we can manage more directly.
This approach is like changing the lens's focus to better analyze and solve the equation.
Linear Differential Equation
Linear differential equations are the simpler siblings of more complex types because they define a direct proportionality relationship involving derivatives. Once we applied the substitution \( u = \frac{1}{P} \), the Bernoulli equation transformed.
  • In our case, the linear equation is \( \frac{du}{dt} = -r \left(1 - \frac{P}{P_e}\right) \).
  • This type of equation is easier to solve because everything is "in a straight line," or simply proportional without powers or complicated terms.
  • Linear equations allow straightforward integration or other solving methods, helping to achieve the final solution for \(P(t)\).
Understanding these properties makes tackling these equations less daunting, ensuring you can handle them with ease.

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Most popular questions from this chapter

(a) For what value of the constant \(C\) and exponent \(r\) is \(y=C t^{r}\) the solution of the initial value problem $$ 2 t y^{\prime}-6 y=0, \quad y(-2)=8 ? $$ (b) Determine the largest interval of the form \((a, b)\) on which Theorem \(2.1\) guarantees the existence of a unique solution. (c) What is the actual interval of existence for the solution found in part (a)?

Suppose that at some initial time the pendulum is located at angle \(\theta_{0}\) with an angular velocity \(d \theta / d t=\omega_{0}\) radians/sec. (a) Equation (15) is a second order differential equation. Rewrite it as a first order separable equation by adopting angle \(\theta\) as the independent variable, using the fact that $$ \theta^{\prime \prime}=\frac{d}{d t}\left(\frac{d \theta}{d t}\right)=\frac{d \omega}{d t}=\frac{d \omega}{d \theta} \frac{d \theta}{d t}=\omega \frac{d \omega}{d \theta} . $$ Complete the specification of the initial value problem by specifying an appropriate initial condition. (b) Obtain the implicit solution $$ m l^{2} \frac{\omega^{2}}{2}-m g l \cos \theta=m l^{2} \frac{\omega_{0}^{2}}{2}-m g l \cos \theta_{0^{-}} $$ The pendulum is a conservative system; that is, energy is neither created nor destroyed. Equation (16) is a statement of conservation of energy. At a position defined by the angle \(\theta\), the quantity \(m l^{2} \omega^{2} / 2\) is the kinetic energy of the pendulum while the term \(-m g l \cos \theta\) is the potential energy, referenced to the horizontal position \(\theta=\pi / 2\). The constant right-hand side is the total initial energy. (c) Determine the angular velocity at the instant the pendulum reaches the vertically downward position, \(\theta=0\). Express your answer in terms of the constants \(\omega_{0}\) and \(\theta_{0}\).

The motion of a body of mass \(m\), gravitationally attracted to Earth in the presence of a resisting drag force proportional to the square of its velocity, is given by $$ m \frac{d v}{d t}=-\frac{G m M_{e}}{r^{2}}+\kappa v^{2} $$ [recall equation (13)]. In this equation, \(r\) is the radial distance of the body from the center of Earth, \(G\) is the universal gravitational constant, \(M_{e}\) is the mass of Earth, and \(v=d r / d t\). Note that the drag force is positive, since it acts in the positive \(r\) direction. (a) Assume that the body is released from rest at an altitude \(h\) above the surface of Earth. Recast the differential equation so that distance \(r\) is the independent variable. State an appropriate initial condition for the new problem. (b) Show that the impact velocity can be expressed as $$ v_{\text {impact }}=-\left[2 G M_{e} \int_{0}^{h} \frac{e^{-2(\kappa / m) s}}{\left(R_{e}+s\right)^{2}} d s\right]^{1 / 2}, $$ where \(R_{e}\) represents the radius of Earth. (The minus sign reflects the fact that \(v=d r / d t<0 .)\)

A tank initially holds 500 gal of a brine solution having a concentration of \(0.1 \mathrm{lb}\) of salt per gallon. At some instant, fresh water begins to enter the tank at a rate of 10 \(\mathrm{gal} / \mathrm{min}\) and the well- stirred mixture leaves at the same rate. How long will it take before the concentration of salt is reduced to \(0.01 \mathrm{lb} / \mathrm{gal}\) ?

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?
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