Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.2.16

Chapter 9, Problem 9.2.16

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(x) = sin y, y(−2) = 1/2

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Identify the differential equation given: \(\frac{dy}{dx} = \sin y\). This means the slope of the solution curve at any point \((x, y)\) depends only on the value of \(y\) at that point.

Set up the window for the direction field as \(x \in [-2, 2]\) and \(y \in [-2, 2]\). This defines the range of \(x\) and \(y\) values where you will plot the slope segments.

At various points \((x, y)\) in the window, calculate the slope \(\frac{dy}{dx} = \sin y\). Since the slope depends only on \(y\), for each horizontal line at a fixed \(y\), the slope will be constant.

Draw small line segments at each point with the slope \(\sin y\). For example, at \(y=0\), \(\sin 0 = 0\), so the slope is zero and the segments are horizontal. At \(y=\frac{\pi}{2}\) (approximately 1.57), \(\sin y = 1\), so the slope is 1, and so on.

To sketch the solution curve passing through the initial condition \(y(-2) = \frac{1}{2}\), start at the point \((-2, 0.5)\) on the graph and follow the direction indicated by the slope segments, moving from left to right, tracing a curve that aligns with the slopes.

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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' = f(x, y) at that point, helping visualize the behavior of solutions without solving the equation explicitly.

Differential Equation y' = sin(y)

The given differential equation y' = sin(y) relates the slope of the solution curve to the sine of the dependent variable y. Understanding how sin(y) varies helps predict the slope patterns in the direction field, such as zero slopes at multiples of π and positive or negative slopes elsewhere.

Initial Condition and Solution Curves

An initial condition like y(−2) = 1/2 specifies a starting point on the direction field, allowing the sketching of a unique solution curve passing through that point. This curve follows the slope directions indicated by the field, illustrating how the solution evolves from the initial value.

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Initial Value Problems

Related Practice

Textbook Question

25–28. Two steps of Euler’s method For the following initial value problems, compute the first two approximations u1 and u2 given by Euler’s method using the given time step.

y′(t) = 2−y, y(0) = 1; Δt = 0.1

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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(1−P/800); P(0) = 100, P(0) = 400, P(0) = 700

Textbook Question

45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes

p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).

Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.

y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 4

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Textbook Question

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

y'(t) = t lnt + 1

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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.

u'(x) = e²ˣ⁻ᵘ

Textbook Question

33–42. Solving initial value problems Solve the following initial value problems.

y'(x) = 4 sec² 2x, y(0) = 8

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(x) = sin y, y(−2) = 1/2

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Key Concepts

Direction Fields

Differential Equation y' = sin(y)

Initial Condition and Solution Curves

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(x) = sin y, y(−2) = 1/2