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Chapter 2: Problem 17

From \(\frac{1}{P-P^{2}} d P=\left(\frac{1}{P}+\frac{1}{1-P}\right) d P=d t\) we obtain \(\ln |P|-\ln |1-P|=t+c\) so that \(\ln \left|\frac{P}{1-P}\right|=t+c\) \(\frac{P}{1-P}=c_{1} e^{t} .\) Solving for \(P\) we have \(P=\frac{c_{1} e^{t}}{1+c_{1} e^{t}}\).

Short Answer

Expert verified
\( P = \frac{c_1 e^t}{1+c_1 e^t} \)

Step by step solution

01

Understand the Given Equation

We start with the equation \( \frac{1}{P-P^2} dP = \left(\frac{1}{P} + \frac{1}{1-P}\right) dP = dt \). This equation suggests we have a differential form that needs to be simplified or integrated.
02

Simplify the Expression

The left-hand side \( \frac{1}{P-P^2} \) can be factored as \( \frac{1}{P(1-P)} \). So, the equation becomes \( \left( \frac{1}{P} + \frac{1}{1-P} \right) dP = dt \). It shows the integrable form has already been simplified.
03

Integrate Both Sides

Integrating both sides gives us \( \int \left( \frac{1}{P} + \frac{1}{1-P} \right) dP = \int dt \). This leads to \( \ln |P| - \ln |1-P| = t + c \) where \( c \) is the constant of integration.
04

Combine Logarithmic Terms

The expression \( \ln |P| - \ln |1-P| \) can be combined using properties of logarithms: \( \ln \left| \frac{P}{1-P} \right| = t + c \). This step condenses the expression to a single logarithmic form.
05

Exponentiate to Solve for \( \frac{P}{1-P} \)

To remove the logarithm, exponentiate both sides: \( \left| \frac{P}{1-P} \right| = e^{t+c} = c_1 e^t \) where \( c_1 = e^c \) is another constant.
06

Solve for \( P \)

We express \( \frac{P}{1-P} = c_1 e^t \). Solving for \( P \), multiply both sides by \( 1-P \): \( P = c_1 e^t (1-P) \).
07

Express \( P \) Explicitly

Rearranging gives \( P = c_1 e^t - c_1 e^t P \) or \( P + c_1 e^t P = c_1 e^t \). Factor \( P \) out: \( P(1+c_1 e^t) = c_1 e^t \). Divide both sides by \( 1+c_1 e^t \) to solve for \( P \): \( P = \frac{c_1 e^t}{1+c_1 e^t} \).

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus that helps in finding the antiderivative or the area under a curve. In the context of differential equations, integration is crucial for solving equations where derivatives are involved. In this exercise, the equation given is \[\left(\frac{1}{P} + \frac{1}{1-P}\right) dP = dt\] Here, each side of the equation needs to be integrated separately to progress towards a solution.
- On the left side, we integrate terms involving the variable \(P\). - On the right side, we integrate the term involving \(t\).
After integration, we combine these results to find an expression that relates \(P\) and \(t\). At this step, it is also crucial to add the constant \(c\) to account for the indefinite nature of integration, which represents the family of solutions.
Exponential Functions
Exponential functions appear often in solutions to differential equations, especially when dealing with growth and decay problems. Here, the relationship \[\ln \left|\frac{P}{1-P}\right| = t + c\]can be converted to an exponential form by exponentiating both sides.
- Exponentiating helps remove the logarithm, transforming the equation to a clearer exponential expression.
The exponential form derived is\[\left| \frac{P}{1-P} \right| = e^{t+c} = c_1 e^t\]where \(c_1 = e^c \). This simplification using the exponential function assists in isolating and eventually solving for the variable \(P\), making the function exponentially dependent on \(t\). Exponential functions are characterised by their constant rate of growth, which is reflected in how \(P\) changes with respect to \(t\).
Logarithmic Equations
Logarithmic equations are vital when manipulating expressions involving exponential terms or solving for a variable trapped inside a logarithm. They simplify multiplication into addition or division into subtraction, a useful property in the context of this problem. In this step-by-step solution, a crucial point was the consolidation of the logarithmic terms:\[\ln |P| - \ln |1-P| = t + c\]Applying logarithmic rules, specifically the quotient rule, these can be combined into a single term:\[\ln \left| \frac{P}{1-P} \right|\]Logarithms help in transitioning between multiplicative and additive forms, simplifying the manipulation of complex equations. This is why understanding logarithmic properties is essential for solving differential equations that involve exponential functions.
Separation of Variables
The separation of variables is a technique used to solve ordinary differential equations (ODEs), where you literally separate the variables into different sides of the equation. In this exercise, the separation process is subtly already in place by presenting a differential form \[\frac{1}{P-P^2} dP = dt\]We separate the variables by ensuring that all terms containing \(P\) are on one side, while dividing the differential elements accordingly, and all terms containing \(t\) are on the other.
- Once separated, each side is independently integrated. This method simplifies the process of solving ODEs by transforming them into manageable algebraic forms.
It also allows the integration process to be applied effectively, leading towards a solution where the variable \(P\), the population in this context, can be expressed explicitly in terms of \(t\), time. This separation makes it easier to see how \(P\) evolves over time based on the given differential equation.

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Most popular questions from this chapter

Let \(I=I(t)\) be the intensity, \(t\) the thickness, and \(I(0)=I_{0} .\) If \(d I / d t=k I\) and \(I(3)=0.25 I_{0},\) then \(I=I_{0} e^{k t}\) \(k=\frac{1}{3} \ln 0.25,\) and \(I(15)=0.00098 I_{0}\).

(a) \((i)\) If \(s(t)\) is distance measured down the plane from the highest point, then \(d s / d t=v .\) Integrating \(d s / d t=16 t\) gives \(s(t)=8 t^{2}+c_{2} .\) Using \(s(0)=0\) then gives \(c_{2}=0 .\) Now the length \(L\) of the plane is \(L=50 / \sin 30^{\circ}=100 \mathrm{ft} .\) The time it takes the box to slide completely down the plane is the solution of \(s(t)=100\) or \(t^{2}=25 / 2,\) so \(t \approx 3.54 \mathrm{s}\). \((i i)\) Integrating \(d s / d t=4 t\) gives \(s(t)=2 t^{2}+c_{2} .\) Using \(s(0)=0\) gives \(c_{2}=0,\) so \(s(t)=2 t^{2}\) and the solution of \(s(t)=100\) is now \(t \approx 7.07 \mathrm{s}\). \((\text {iii})\) Integrating \(d s / d t=48-48 e^{-t / 12}\) and using \(s(0)=0\) to determine the constant of integration, we \(\operatorname{obtain} s(t)=48 t+576 e^{-t / 12}-576 .\) With the aid of a CAS we find that the solution of \(s(t)=100,\) or $$\begin{aligned} &100=48 t+576 e^{-t / 12}-576 &\text { or } \quad 0=48 t+576 e^{-t / 12}-676 \end{aligned}$$ is now \(t \approx 7.84 \mathrm{s}\). (b) The differential equation \(m d v / d t=m g \sin \theta-\mu m g \cos \theta\) can be written $$m \frac{d v}{d t}=m g \cos \theta(\tan \theta-\mu)$$. If \(\tan \theta=\mu, d v / d t=0\) and \(v(0)=0\) implies that \(v(t)=0 .\) If \(\tan \theta< \mu\) and \(v(0)=0,\) then integration implies \(v(t)=g \cos \theta(\tan \theta-\mu) t < 0\) for all time \(t\). (c) since \(\tan 23^{\circ}=0.4245\) and \(\mu=\sqrt{3} / 4=0.4330,\) we see that \(\tan 23^{\circ}<0.4330 .\) The differential equation is \(d v / d t=32 \cos 23^{\circ}\left(\tan 23^{\circ}-\sqrt{3} / 4\right)=-0.251493 .\) Integration and the use of the initial condition gives \(v(t)=-0.251493 t+1 .\) When the box stops, \(v(t)=0\) or \(0=-0.251493 t+1\) or \(t=3.976254 \mathrm{s}\). From \(s(t)=-0.125747 t^{2}+t\) we find \(s(3.976254)=1.988119 \mathrm{ft}\). (d) With \(v_{0} >0, v(t)=-0.251493 t+v_{0}\) and \(s(t)=-0.125747 t^{2}+v_{0} t .\) Because two real positive solutions of the equation \(s(t)=100,\) or \(0=-0.125747 t^{2}+v_{0} t-100,\) would be physically meaningless, we use the quadratic formula and require that \(b^{2}-4 a c=0\) or \(v_{0}^{2}-50.2987=0 .\) From this last equality we find \(v_{0} \approx 7.092164 \mathrm{ft} / \mathrm{s} .\) For the time it takes the box to traverse the entire inclined plane, we must have \(0=-0.125747 t^{2}+7.092164 t-100 .\) Mathematica gives complex roots for the last equation: \(t=\) \(28.2001 \pm 0.0124458 i .\) But, for $$0=-0.125747 t^{2}+7.092164691 t-100$$ the roots are \(t=28.1999 \mathrm{s}\) and \(t=28.2004 \mathrm{s} .\) So if \(v_{0} >7.092164,\) we are guaranteed that the box will slide completely down the plane.

Rewrite \(\left(5 x^{2}-2 y^{2}\right) d x-x y d y=0\) as $$x y \frac{d y}{d x}=5 x^{2}-2 y^{2}$$ and divide by \(x y,\) so that $$\frac{d y}{d x}=5 \frac{x}{y}-2 \frac{y}{x}.$$ We then identify $$F\left(\frac{y}{x}\right)=5\left(\frac{y}{x}\right)^{-1}-2\left(\frac{y}{x}\right).$$

(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.

The left-hand derivative of the function at \(x=1\) is \(1 / e\) and the right- hand derivative at \(x=1\) is \(1-1 / e\). Thus, \(y\) is not differentiable at \(x=1\).
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