Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. Find an explicit solution of the given initial-value problem. dy = (y - 1)2 + 0.04, y(0) = 1 dx y = Use a graphing utility to plot the graph of the solution in a neighborhood of (0, 1). Compare the graph with those obtained in Problems 39, 40, and 42 of Section 2.2. y y y 1.0015 1.0015 1.015 1.0010 1.0010 1.010 1.0005 1.0005 1.005 -0.015-0.010-0.005 0.005 0.0100.015 -0.015 -0.010-0.005 0.9995 0.005 0.010 0.015 -0.015-0.010-0.005 0.9995 0.005 0.010 0.015 0.995 0.9990 0.9990 0.990 0.9985 0.9985 0.985 y 1.015 1.010 1.005 -0.015 -0.010-0.005 0.995 0.005 0.010 0.015 0.990 0.985
Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. Find an explicit solution of the given initial-value problem. dy = (y - 1)2 + 0.04, y(0) = 1 dx y = Use a graphing utility to plot the graph of the solution in a neighborhood of (0, 1). Compare the graph with those obtained in Problems 39, 40, and 42 of Section 2.2. y y y 1.0015 1.0015 1.015 1.0010 1.0010 1.010 1.0005 1.0005 1.005 -0.015-0.010-0.005 0.005 0.0100.015 -0.015 -0.010-0.005 0.9995 0.005 0.010 0.015 -0.015-0.010-0.005 0.9995 0.005 0.010 0.015 0.995 0.9990 0.9990 0.990 0.9985 0.9985 0.985 y 1.015 1.010 1.005 -0.015 -0.010-0.005 0.995 0.005 0.010 0.015 0.990 0.985
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. Find an explicit solution of the given initial-value problem.
dy
= (y – 1)2 + 0.04, y(0) = 1
dx
y =
Use a graphing utility to plot the graph of the solution in a neighborhood of (0, 1). Compare the graph with those obtained in Problems 39, 40, and 42 of Section 2.2.
y
y
y
1.0015
1.0015
1.015
1.0010
1.0010
1.010
1.0005
1.0005
1.005
X
X
-0.015 -0.010-0.005
0.9995
-0.015-0.010–0.005
0.995
0.005 0.010 0.015
-0.015 -0.010-0.005
0.005 0.010 0.015
0.005 0.010 0.015
0.9995
0.9990
0.9990
0.990
0.9985
0.9985
0.985
y
1.015
1.010
1.005
-0.015 –0.010–0.005
0.005 0.010 0.015
0.9
0.990
0.985
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First of integrate the given equation as follows
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