Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. The general solution of the differential equation y'(t) = 1 is y(t) = 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 9.2.44a
Direction field analysis Consider the first-order initial value problem y'(t)=ay+b,y(0)=A for t≥0 where a, b, and A are real numbers.
a. Explain why y=−b/a is an equilibrium solution and corresponds to a horizontal line in the direction field.
Verified step by step guidance1
Recall that an equilibrium solution to a differential equation is a constant solution where the derivative is zero for all values of \( t \). This means \( y'(t) = 0 \) along that solution.
Given the differential equation \( y'(t) = a y + b \), set \( y'(t) = 0 \) to find the equilibrium solution: \( 0 = a y + b \).
Solve for \( y \) to get \( y = -\frac{b}{a} \). This value of \( y \) makes the derivative zero, so the solution is constant and does not change with \( t \).
Since \( y = -\frac{b}{a} \) is constant, its graph is a horizontal line in the \( t y \)-plane, which corresponds to a horizontal line in the direction field.
In the direction field, the slope at every point on this line is zero, confirming that \( y = -\frac{b}{a} \) is an equilibrium solution.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Solution
An equilibrium solution to a differential equation is a constant solution where the derivative is zero. For y'(t) = ay + b, setting y' = 0 gives y = -b/a, meaning the solution does not change over time and remains constant.
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Solutions to Basic Differential Equations
Direction Field
A direction field is a graphical representation showing the slope of solutions to a differential equation at various points. At an equilibrium solution, the slope is zero, so the direction field shows horizontal line segments, indicating no change in y.
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05:45Understanding Slope Fields
Initial Value Problem (IVP)
An IVP specifies a differential equation along with an initial condition, such as y(0) = A. This condition determines a unique solution curve passing through the point (0, A) in the direction field.
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05:03Initial Value Problems
Related Practice
Textbook Question
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)
Textbook Question
33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.
a. Approximate the value of y(T) using Euler’s method with the given time step on the interval [0,T].
y′(t) = -2y, y(0) = 1; Δt = 0.2, T = 2; y(t) = e⁻²ᵗ
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Textbook Question
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(2 - y)
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Textbook Question
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) = 2y + 4
Textbook Question
46–48. Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case, carry out the indicated analysis using direction fields.
Drug infusion The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation m′(t)+km(t)=I, where m(t) is the mass of the drug in the blood at time t≥0, K is a constant that describes the rate at which the drug is absorbed, and I is the infusion rate. Let I=10mg/hr and k=0.05 hr^−1.
a. Draw the direction field, for 0≤t≤100, 0≤y≤600.
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