Textbook Question
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
y'(t) = eʸᐟ²sin t
Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- 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: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 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)
Verified step by step guidance1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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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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Related Practice
Textbook Question
33–38. {Use of Tech} Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem.
yy'(x) = 2x/(2 + y)², y(1) = −1
Textbook Question
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
Textbook Question
21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
u''(x) = 55x⁹ + 36x⁷ - 21x⁵ + 10x⁻³
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Textbook Question
20–22. {Use of Tech} Solving the Gompertz equation Solve the Gompertz equation in Exercise 19 with the given values of r, K, and M₀. Then graph the solution to be sure that M(0) and lim(t→∞) M(t) are correct.
r = 0.05, K = 1200, M₀ = 90
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Textbook Question
39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form
a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t).
Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems.
t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6
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