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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.2.41a

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
a. Find the equilibrium solutions. 


y′(t) = y(y - 3)

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Identify the given autonomous differential equation: \(y'(t) = y(y - 3)\).
Recall that equilibrium solutions occur where the derivative \(y'(t)\) is zero for all \(t\), meaning \(f(y) = 0\).
Set the right-hand side equal to zero: \(y(y - 3) = 0\).
Solve the equation \(y(y - 3) = 0\) by finding values of \(y\) that satisfy it. This means solving \(y = 0\) or \(y - 3 = 0\).
Conclude that the equilibrium solutions are the constant functions \(y(t) = 0\) and \(y(t) = 3\).

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Autonomous Differential Equations

An autonomous differential equation is one where the derivative y' depends only on the variable y, not explicitly on the independent variable t. This means the rate of change of y depends solely on y itself, simplifying analysis and allowing the direction field to be independent of t.
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Classifying Differential Equations

Equilibrium Solutions

Equilibrium solutions occur when y'(t) = 0 for some constant y = y0, meaning the function f(y0) = 0. These solutions represent constant functions where the system remains steady over time, corresponding to horizontal lines in the direction field.
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Solutions to Basic Differential Equations

Finding Equilibrium Points by Solving f(y) = 0

To find equilibrium solutions, set the right-hand side function f(y) equal to zero and solve for y. For example, given y' = y(y - 3), solve y(y - 3) = 0 to find y = 0 and y = 3 as equilibrium points.
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Critical Points Example 2
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A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.


a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C₁eᵗ + C₂e⁻ᵗ. You may assume this function is the general solution.

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Cooling time Suppose an object with an initial temperature of T₀ > 0 is put in surroundings with an ambient temperature of A, where A < T₀/2. Let t₁/₂ be the time required for the object to cool to T₀/2.


a. Show that t₁/₂ = −1/k ln((T₀ − 2A)/(2(T₀ − A))).

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42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.

a. Find the general solution of the equation.


y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1


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{Use of Tech} Endowment model An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem B′(t)=rB−m, for t≥0, with B(0)=B0. The constant r>0 reflects the annual interest rate, m>0 is the annual rate of withdrawal, B0 is the initial balance in the account, and t is measured in years.


a. Solve the initial value problem with r=0.05, m=\$1000/year, and B0=\$15,000 Does the balance in the account increase or decrease?

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52-56. In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations.


where P(t) is the population, for t ≥ 0, and r > 0 and K > 0 are given constants.


a. Verify by substitution that the general solution of the equation is P(t) = K/(1 + Ce⁻ʳᵗ), where C is an arbitrary constant.

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Solving Bernoulli equations Use the method outlined in Exercise 43 to solve the following Bernoulli equations.


a. y′(t) + y = 2y²

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