Textbook Question
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = y³sin t, y(0) = 1
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 9.2.13
12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.
y'(t) = 4−y, y(0) = −1
Verified step by step guidance1
Identify the differential equation given: \(y'(t) = 4 - y\). This means the slope of the solution curve at any point \((t, y)\) depends on the value of \(y\) at that point.
Set up a grid of points in the window \([-2, 2] \times [-2, 2]\) for \(t\) and \(y\). At each point \((t, y)\), calculate the slope \(y' = 4 - y\) to determine the direction of the solution curve.
At each grid point, draw a small line segment with slope equal to \(4 - y\). This collection of line segments forms the direction field, visually representing the behavior of solutions to the differential equation.
Use the initial condition \(y(0) = -1\) to locate the starting point on the \(t\)-\(y\) plane at \((0, -1)\). From this point, sketch a curve that follows the direction field, moving in the direction indicated by the slope segments.
Observe the general behavior of the solution curve: since \(y' = 4 - y\), when \(y\) is less than 4, the slope is positive, and when \(y\) is greater than 4, the slope is negative. This suggests the solution will approach the horizontal line \(y = 4\) as \(t\) increases.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Direction Fields
A direction field is a graphical representation of a first-order differential equation showing the slope of the solution curve at various points. Each small line segment indicates the slope y' at that point (t, y), helping visualize the behavior of solutions without solving the equation explicitly.
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Understanding Slope Fields
Initial Value Problems
An initial value problem specifies a differential equation along with a starting point, such as y(0) = -1. This condition determines a unique solution curve passing through the given point, allowing us to sketch or find the particular solution that fits the initial data.
Recommended video:
05:03Initial Value Problems
Solving First-Order Differential Equations
Understanding how to solve or analyze first-order differential equations like y' = 4 - y is essential. This often involves recognizing the equation type (e.g., separable or linear), which helps predict solution behavior and sketch solution curves consistent with the direction field.
Recommended video:
06:06Solving Separable Differential Equations
Related Practice
Textbook Question
Explain how the growth rate function determines the solution of a population model.
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) = C₁eᵗ + C₂teᵗ; u''(t) - 2u'(t) + u(t) = 0
Textbook Question
The general solution of a first-order linear differential equation is y(t) = Ce⁻¹⁰ᵗ − 13. What solution satisfies the initial condition y(0) = 4?
Textbook Question
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P − 0.001P²; P(0) = 10, P(0) = 40, P(0) = 80
Textbook Question
5–10. First-order linear equations Find the general solution of the following equations.
y'(x) = −y + 2
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