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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
Tutti i libri di testoBriggs 3rd EditionCh. 9 - Differential EquationsProblema 9.R.20b
Capitolo 9, Problema 9.R.20b

Direction fields The direction field for the equation y′(t)=t−y, for |t|≤4 and |y|≤4, is shown in the figure.
b. Use the direction field to sketch the solution curve that passes through the point (0,−1/2).
Direction field plot showing slope vectors for y′(t) = t minus y over t and y from -4 to 4.

Guida verificata passo dopo passo
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Identify the initial point given in the problem, which is (0, -\(\frac{1}{2}\)). This is where your solution curve will start on the direction field.
At the initial point, observe the slope of the direction field. The slope is given by the differential equation y\'(t) = t - y. Substitute t = 0 and y = -\(\frac{1}{2}\) into the equation to find the slope at this point: y\'(0) = 0 - (-\(\frac{1}{2}\)) = \(\frac{1}{2}\).
Using the slope \(\frac{1}{2}\) at the initial point, sketch a small line segment starting at (0, -\(\frac{1}{2}\)) that rises gently, reflecting this positive slope.
Follow the direction field arrows from the initial point, moving step-by-step along the slopes indicated by the field. At each new point, estimate the slope from the nearby direction field vectors and continue drawing the curve smoothly, ensuring it aligns with the slope directions.
Continue this process both forward and backward in t, making sure the curve remains consistent with the slope directions shown in the direction field, thus sketching the solution curve that passes through (0, -\(\frac{1}{2}\)).

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Direction Fields (Slope Fields)

A direction field is a graphical representation of a first-order differential equation showing small line segments with slopes given by the differential equation at various points. It helps visualize the behavior of solutions without solving the equation analytically. Each segment indicates the slope of the solution curve passing through that point.
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Understanding Slope Fields

Initial Value Problem and Solution Curves

An initial value problem specifies a differential equation along with a point through which the solution curve must pass. Using the direction field, one can sketch the solution curve starting at the initial point by following the slope directions, illustrating how the solution evolves over the domain.
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05:03
Initial Value Problems

Differential Equation y′(t) = t − y

This equation defines the slope of the solution curve at any point (t, y) as the difference between t and y. Understanding how the slope depends on both variables is crucial for interpreting the direction field and predicting the shape of solution curves, especially near the initial condition.
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Classifying Differential Equations
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Domanda del libro di testo

Euler’s metho d Consider the initial value problem y′(t)=1/2y,y(0)=1. 

a. Use Euler’s method with Δt=0.1 to compute approximations to y(0.1) and y(0.2). 

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Logistic growth in India The population of India was 435 million in 1960 (t=0) and 487 million in 1965 (t=5). The projected population for 2050 is 1.57 billion.

e. Discuss some possible shortcomings of this model. Why might the carrying capacity be either greater than or less than the value predicted by the model?

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A predator-prey model Consider the predator-prey model

x′(t) = −4x + 2xy, y′(t) = 5y − xy

c. Find the equilibrium points for the system.

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Euler’s method Consider the initial value problem y′(t)=1/2y,y(0)=1. 

b. Use Euler’s method with Δt=0.05 to compute approximations to y(0.1) and y(0.2). 

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Direction fields The direction field for the equation y′(t)=t−y, for |t|≤4 and |y|≤4, is shown in the figure.

d. Complete the following sentence. The solution of the differential equation with the initial condition y(0)=A, where A is a real number, approaches the line _____ as t→∞.

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11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(t) = -3y + 9, y(0) = 4

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