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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
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

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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.
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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.
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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

33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

c. Repeating parts (a) and (b) using half the time step used in those calculations, again find an approximation to y(T).


y′(t) = 6 - 2y, y(0) = -1; Δt = 0.2, T = 3; y(t) = 3 - 4e⁻²ᵗ

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Textbook Question

{Use of Tech} Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation

 dP/dt=kP(1−P/A),P0=P_0, 

where K is a positive infection rate, A is the number of people in the community, and P0 is the number of infected people at t=0. The model also assumes no recovery. 


c. For a fixed value of K and A, describe the long-term behavior of the solutions, for any P0 with 0<P0<A. 

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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.

c. What is the equilibrium solution?

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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