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 9.2.38c

Chapter 9, Problem 9.2.38c

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.
c. Sketch the solution curve that corresponds to the initial condition y0=1.

y′(t) = 2y + 4

Verified step by step guidance

Identify the given differential equation: \(y'(t) = 2y + 4\). This is an autonomous equation because the right-hand side depends only on \(y\), not explicitly on \(t\).

Find the equilibrium solutions by setting \(y'(t) = 0\), which means solving \(2y + 4 = 0\) for \(y\). This will give the constant values of \(y\) where the solution does not change over time.

Solve the equation \(2y + 4 = 0\) to find the equilibrium solution(s) \(y_0\). This equilibrium solution corresponds to a horizontal line in the direction field.

Since the initial condition is \(y(0) = 1\), analyze the behavior of the solution near \(y=1\) by considering the sign of \(y'(t)\) when \(y=1\). This will tell you whether the solution is increasing or decreasing at that point.

Sketch the solution curve starting at the point \((0,1)\) on the \(t\)-\(y\) plane. Use the information about the equilibrium solution and the slope \(y'(t)\) at \(y=1\) to draw the curve showing how \(y\) changes with \(t\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Autonomous Differential Equations

An autonomous differential equation has the form y' = f(y), where the rate of change depends only on y, not explicitly on t. This means the behavior of solutions depends solely on the current value of y, making the direction field time-invariant and simplifying analysis of solution curves.

Equilibrium Solutions

Equilibrium solutions occur when y' = f(y) = 0, meaning the solution y(t) remains constant over time. These correspond to horizontal lines in the direction field and represent steady states where the system does not change, providing key reference points for sketching solution behavior.

Sketching Solution Curves with Initial Conditions

To sketch a solution curve for a given initial condition y(0) = y0, identify equilibrium points and analyze the slope y' = f(y) at y0. The sign and magnitude of y' determine whether the solution increases or decreases, guiding the shape of the curve over time.

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Summary of Curve Sketching

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.

c. Sketch the solution curve that corresponds to the initial condition y0=1.

y′(t) = 6 - 2y

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Textbook Question

{Use of Tech} Free fall Using th e background given in Exercise 47, assume the resistance is given by f(v)=−Rv, for t≥0, where R>0 is a drag coefficient (an assumption often made for a heavy medium such as water or oil).

c. Find the solution of this separable equation assuming v(0)=0 and 0<v<g/b.

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Textbook Question

29–32. {Use of Tech} Errors in Euler’s method Consider the following initial value problems.

c. Which time step results in the more accurate approximation? Explain your observations.

y′(t) = 4−y, y(0) = 3; y(t) = 4−e⁻ᵗ

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Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

c. Find the equilibrium points for the system.

x′(t) = −3x + xy, y′(t) = 2y − xy

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Textbook Question

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.

(Use of Tech) Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form y'(t) = -kyⁿ(t), where y(t) is the concentration of the compound, for t ≥ 0, k > 0 is a constant that determines the speed of the reaction, and n is a positive integer called the order of the reaction. Assume the initial concentration of the compound is y(0) = y₀ > 0.

c. Let y₀ = 1 and k = 0.1. Graph the first-order and second-order solutions found in parts (a) and (b). Compare the two reactions.

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Textbook Question

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.

c. Why is the condition A < T₀/2 needed?

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