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Chapter 2: Problem 5

(a) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1

Short Answer

Expert verified
The equilibrium solutions are \( P = 1 \) and \( P = 4 \). Population becomes extinct at \( t = -\frac{1}{3} \ln \frac{4(P_0-1)}{P_0-4} \).

Step by step solution

01

Identifying Equilibrium Solutions

First, solve the equation \( P(5-P) - 4 = 0 \) to find the equilibrium solutions. Factoring yields \( (P-1)(P-4) = 0 \), so the solutions are \( P = 1 \) and \( P = 4 \). These are the equilibrium solutions where the population remains constant.
02

Deriving the Differential Equation

The given differential equation is \( \frac{dP}{dt} = P(5-P) - 4 = -(P-4)(P-1) \). This shows how the rate of change of \( P \) is related to \( P \) itself and its distance from the equilibrium points.
03

Separation of Variables

To solve the differential equation, separate variables: \( \frac{dP}{(P-4)(P-1)} = -dt \). This allows us to integrate each side independently.
04

Integrating Both Sides

Integrate the left side using partial fractions \( \frac{1/3}{P-4} - \frac{1/3}{P-1} \) and the right side with respect to \( t \). This results in \( \frac{1}{3} \ln \left| \frac{P-4}{P-1} \right| = -t + c \).
05

Solving for the Constant and Function of \( P \)

Set \( t = 0 \) and \( P = P_0 \) to find that the constant \( c_1 = \frac{P_0 - 4}{P_0 - 1} \). Solve for \( P(t) \) to obtain the solution: \( P(t) = \frac{4(P_0 - 1) - (P_0 - 4) e^{-3t}}{(P_0 - 1) - (P_0 - 4) e^{-3t}} \).
06

Finding Time of Extinction

To determine when the population becomes extinct (\( P = 0 \)), set \( P = 0 \) in the expression \( \frac{P-4}{P-1} = \frac{P_0-4}{P_0-1} e^{-3t} \). Solving for \( t \) gives \( t = -\frac{1}{3} \ln \frac{4(P_0-1)}{P_0-4} \). This gives the time at which extinction occurs, provided \( P_0 < 1 \).

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Key Concepts

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

Equilibrium Solutions
Equilibrium solutions are points where the rate of change in a system, such as a differential equation, becomes zero. In simple terms, they represent the values at which the system remains constant over time.
In our problem, the differential equation is given as \( \frac{dP}{dt} = P(5-P) - 4 \). We solve the equation \( P(5-P) - 4 = 0 \) to find the equilibrium solutions. Factoring the expression gives \((P-1)(P-4) = 0\), leading to the solutions \( P = 1 \) and \( P = 4 \).
These points are called equilibrium points because when \( P \) is equal to 1 or 4, the population \( P \) neither increases nor decreases; it stays the same over time. Hence:
  • For \( P = 1 \): This represents a stable equilibrium, as any deviation from this point results in a system that returns back to equilibrium.
  • For \( P = 4 \): This is another equilibrium solution where the system neither grows nor diminishes.
Separation of Variables
Separation of Variables is a powerful method used to solve differential equations. It involves rearranging an equation so that each variable appears on a different side of the equation, allowing for independent integration.
In the differential equation \( \frac{dP}{dt} = -(P-4)(P-1) \), our goal is to isolate \( P \) on one side and \( t \) on the other. We rearrange it to \( \frac{dP}{(P-4)(P-1)} = -dt \).
Now, both sides can be integrated:
  • The left side involves partial fraction decomposition: \( \frac{1/3}{P-4} - \frac{1/3}{P-1} \)
  • The right side is simpler, as it integrates to \(-t + c\)
By integrating both sides, we find the relationship between \( P \) and \( t \), which is essential to understand how \( P \) changes over time given an initial condition.
Phase Portrait
A phase portrait is a visual representation that helps us understand the behavior of differential equations. It consists of arrows that indicate the direction of change in a given system.
For our differential equation \( \frac{dP}{dt} = -(P-4)(P-1) \), the phase portrait visually depicts how the population \( P \) will change over time for different initial conditions. By analyzing the arrows' directions:
  • For \( P_0 > 4 \), the population decreases towards 4, illustrating stability.
  • For \( 1 < P_0 < 4 \), the population increases towards 4, also illustrating stability.
  • For \( 0 < P_0 < 1 \), the population decreases to zero, showing extinction.
The phase portrait provides an intuitive understanding of possible future states of the system based on current conditions, helping predict long-term behavior and stability points.
Integration of Differential Equations
Integration of differential equations is a method used to find a solution to equations that involve derivatives. In our context, it helps to determine the relationship between the variables \( P \) and \( t \) over time.
After separating variables in the equation \( \frac{dP}{(P-4)(P-1)} = -dt \), we integrate both sides:
  • The left-hand side, after decomposition, becomes \( \int \left(\frac{1/3}{P-4} - \frac{1/3}{P-1}\right) dP \).
  • The right-hand side simplifies to integration over time \( - \int dt \), resulting in \(-t + c\).
After integration, solving for \( P \) yields a function that helps express \( P \) in terms of \( t \).
This solution function, \( P(t) = \frac{4(P_0-1) - (P_0-4)e^{-3t}}{(P_0-1) - (P_0-4)e^{-3t}} \), describes the population behavior over time, revealing how the initial population \( P_0 \) and time \( t \) influence \( P \). This mathematical integration enables predicting how populations evolve based on initial starting points.

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Most popular questions from this chapter

(a) If we know \(s(t)\) and \(i(t)\) then we can determine \(r(t)\) from \(s+i+r=n\) (b) In this case the system is \\[\begin{array}{l}\frac{d s}{d t}=-0.2 s i \\\\\frac{d i}{d t}=-0.7 i+0.2 s i\end{array}\\] We also note that when \(i(0)=i_{0}, s(0)=10-i_{0}\) since \(r(0)=0\) and \(i(t)+s(t)+r(t)=0\) for all values of \(t .\) Now \(k_{2} / k_{1}=0.7 / 0.2=3.5,\) so we consider initial conditions \(s(0)=2, i(0)=8 ; s(0)=3.4, i(0)=6.6\) \(s(0)=7, i(0)=3 ;\) and \(s(0)=9, i(0)=1\). We see that an initial susceptible population greater than \(k_{2} / k_{1}\) results in an epidemic in the sense that the number of infected persons increases to a maximum before decreasing to \(0 .\) On the other hand, when \(s(0)< k_{2} / k_{1},\) the number of infected persons decreases from the start and there is no epidemic.

For \(y^{\prime}+y=f(x)\) an integrating factor is \(e^{x}\) so that $$y e^{x}=\left\\{\begin{array}{ll} e^{x}+c_{1}, & 0 \leq x \leq 1 \\ -e^{x}+c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=1\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=2 e\) so that $$y=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq 1 \\ 2 e^{1-x}-1, & x>1 \end{array}\right.$$

(a) Let \(\rho\) be the weight density of the water and \(V\) the volume of the object. Archimedes' principle states that the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive direction to be down, the differential equation is \\[ m \frac{d v}{d t}=m g-k v^{2}-\rho V \\] (b) Using separation of variables we have \\[ \begin{aligned} \frac{m d v}{(m g-\rho V)-k v^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{\sqrt{k} d v}{(\sqrt{m g-\rho V})^{2}-(\sqrt{k} v)^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{1}{\sqrt{m g-\rho V}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g-\rho V}} &=t+c \end{aligned} \\] Thus \\[ v(t)=\sqrt{\frac{m g-\rho V}{k}} \tanh \left(\frac{\sqrt{k m g-k \rho V}}{m} t+c_{1}\right) \\] (c) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) the terminal velocity is \(\sqrt{(m g-\rho V) / k}.\)

The linear equation \(d x / d t=-\lambda_{1} x\) can be solved by either separation of variables or by an integrating factor. Integrating both sides of \(d x / x=-\lambda_{1} d t\) we obtain \(\ln |x|=-\lambda_{1} t+c\) from which we get \(x=c_{1} e^{-\lambda_{1} t} .\) Using \(x(0)=x_{0}\) we find \(c_{1}=x_{0}\) so that \(x=x_{0} e^{-\lambda_{1} t} .\) Substituting this result into the second differential equation we have $$\frac{d y}{d t}+\lambda_{2} y=\lambda_{1} x_{0} e^{-\lambda_{1} t}$$ which is linear. An integrating factor is \(e^{\lambda_{2} t}\) so that $$\frac{d}{d t}\left[e^{\lambda_{2} t} y\right]=\lambda_{1} x_{0} e^{\left(\lambda_{2}-\lambda_{1}\right) t}+c_{2}$$ $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{\left(\lambda_{2}-\lambda_{1}\right) t} e^{-\lambda_{2} t}+c_{2} e^{-\lambda_{2} t}=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+c_{2} e^{-\lambda_{2} t}$$ Using \(y(0)=0\) we find \(c_{2}=-\lambda_{1} x_{0} /\left(\lambda_{2}-\lambda_{1}\right) .\) Thus $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Substituting this result into the third differential equation we have $$\frac{d z}{d t}=\frac{\lambda_{1} \lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Integrating we find $$z=-\frac{\lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}+c_{3}$$ Using \(z(0)=0\) we find \(c_{3}=x_{0} .\) Thus $$z=x_{0}\left(1-\frac{\lambda_{2}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}\right)$$.

Separating variables we have \(d y /\left(\sqrt{1+y^{2}} \sin ^{2} y\right)=d x\) which is not readily integrated (even by a CAS). We note that \(d y / d x \geq 0\) for all values of \(x\) and \(y\) and that \(d y / d x=0\) when \(y=0\) and \(y=\pi,\) which are equilibrium solutions.
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