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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 + 0.01, y(0) = 1 ax 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 1.004 1.0004 1.002 1.0002 x x -0.004 0.002 -0.004 0.002 -0.002 0.004 -0.002 0.004

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 + 0.01, y(0) = 1 ax 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 1.004 1.0004 1.002 1.0002 x x -0.004 0.002 -0.004 0.002 -0.002 0.004 -0.002 0.004

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 33CR
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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 +0.01, y(0) = 1 dx 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 1.004 1.0004 1.002 1.0002 x -0.004 0.002 -0.004 -0.002 0.002 0.004 -0.002 0.004
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 +0.01, y(0) = 1 dx 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 1.004 1.0004 1.002 1.0002 x -0.004 0.002 -0.004 -0.002 0.002 0.004 -0.002 0.004
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