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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.5.15
Chapter 9, Problem 9.5.15

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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Start by writing the general form of the logistic differential equation: \[\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)\] where \(P(t)\) is the population at time \(t\), \(r\) is the natural growth rate, and \(K\) is the carrying capacity.
Substitute the given parameter values into the logistic equation: \[\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{300}\right)\].
Set up the initial value problem (IVP) by including the initial condition: \[P(0) = 50\].
To solve the IVP, separate variables and integrate: rewrite the equation as \[\frac{dP}{P(1 - P/300)} = 0.2\,dt\] and then perform partial fraction decomposition on the left side before integrating both sides.
After integrating, solve for \(P(t)\) explicitly in terms of \(t\) and the constants, then use the initial condition \(P(0) = 50\) to find the constant of integration. This will give you the logistic growth function describing the population over time.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Logistic Growth Model

The logistic growth model describes population growth that starts exponentially but slows as the population approaches a maximum limit called the carrying capacity (K). It is represented by the differential equation dP/dt = rP(1 - P/K), where r is the natural growth rate and P is the population at time t.
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Exponential Growth & Decay

Solving Initial Value Problems (IVPs)

An initial value problem involves solving a differential equation with a given initial condition, such as P(0) = P₀. Solving the logistic equation IVP means finding a function P(t) that satisfies both the logistic differential equation and the initial population value.
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Initial Value Problems

Graphing Solutions of Differential Equations

Graphing the solution to a logistic equation shows how the population changes over time, typically starting at P₀, growing rapidly, and leveling off near the carrying capacity K. This visual helps interpret the behavior and stability of the population model.
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Related Practice
Textbook Question

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.

p'(x) = 16/x⁹ - 5 + 14x⁶

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Textbook Question

27–30. Newton’s Law of Cooling Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following cases. Then answer any additional questions.


A pot of boiling soup (100°C) is put in a cellar with a temperature of 10°C. After 30 minutes, the soup has cooled to 80°C. When will the temperature of the soup reach 30°C 

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.

(t² + 1)³yy'(t) = t(y² + 4)

Textbook Question

Explain how to solve a separable differential equation of the form

g(t)y'(y) = h(t)

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Textbook Question

27–30. Newton’s Law of Cooling Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following cases. Then answer any additional questions.


An iron rod is removed from a blacksmith’s forge at a temperature of 900°C . Assume k=0.02 and the rod cools in a room with a temperature of 30°C When does the temperature of the rod reach 100°C? 

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Textbook Question

7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.

u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³