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Explain how a stirred tank reaction works.
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Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.5.33
Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem
P'(t) = rP (1-P/K), P(0) = P₀
is P(t) = K/((K/P₀ − 1)e⁻ʳᵗ + 1)
Guida verificata passo dopo passo1
Start with the logistic differential equation given: \(P'(t) = rP \left(1 - \frac{P}{K}\right)\), where \(r\) and \(K\) are constants, and \(P(0) = P_0\) is the initial condition.
Rewrite the differential equation in separable form by dividing both sides by \(P \left(1 - \frac{P}{K}\right)\) and multiplying both sides by \(dt\):
\(\frac{dP}{P \left(1 - \frac{P}{K}\right)} = r \, dt\).
Simplify the left side by expressing the denominator as a single fraction:
\(P \left(1 - \frac{P}{K}\right) = P \left(\frac{K - P}{K}\right) = \frac{P(K - P)}{K}\), so the integral becomes
\(\int \frac{K}{P(K - P)} \, dP = \int r \, dt\).
Use partial fraction decomposition to rewrite \(\frac{K}{P(K - P)}\) as \(\frac{A}{P} + \frac{B}{K - P}\), then solve for constants \(A\) and \(B\). After finding \(A\) and \(B\), integrate both sides:
\(\int \left(\frac{A}{P} + \frac{B}{K - P}\right) dP = \int r \, dt\).
After integrating, apply the initial condition \(P(0) = P_0\) to solve for the constant of integration. Finally, solve the resulting equation for \(P(t)\) to obtain the explicit solution:
\(P(t) = \frac{K}{\left(\frac{K}{P_0} - 1\right) e^{-rt} + 1}\).
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Logistic Differential Equation
The logistic differential equation models population growth with a carrying capacity, expressed as P'(t) = rP(1 - P/K). Here, r is the growth rate, K is the maximum population, and P(t) is the population at time t. It describes growth that slows as the population approaches K.
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07:39
Classifying Differential Equations
Separation of Variables
Separation of variables is a method to solve differential equations by rewriting them so that each variable and its differential are on opposite sides. For the logistic equation, this involves isolating terms with P on one side and t on the other before integrating both sides.
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07:15Separation of Variables
Initial Value Problem and Integration Constants
An initial value problem specifies the value of the solution at a particular point, here P(0) = P₀. After integrating, the constant of integration is determined using this initial condition to find the particular solution that fits the problem.
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Percorso guidato
05:03Initial Value Problems
Pratica correlata
Domanda del libro di testo
9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.
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Domanda del libro di testo
23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.
B′(t) = 0.005B − 500, B(0) = 50,000
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Domanda del libro di testo
15–16. {Use of Tech} Solving logistic equations Write a logistic equation with the following parameter values. Then solve the initial value problem and graph the solution. Let r be the natural growth rate, K the carrying capacity, and P₀ the initial population.
r=0.2, K=300, P₀=50
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Domanda del libro di testo
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
u'(x) = e²ˣ⁻ᵘ
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Domanda del libro di testo
33–42. Solving initial value problems Solve the following initial value problems.
y'(x) = 4 sec² 2x, y(0) = 8
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