Explanation Consider the function. y = { 1 − e − x , 0 ≤ x ≤ 1 ( e − 1 ) e − x , x > 1 } So, d y d x = { e − x , 0 ≤ x ≤ 1 − ( e − 1 ) e − x , x > 1 } Find the both side limit of the above differential equation

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 2.3, Problem 53E

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To discuss: The reason that, it is technically incorrect to say that the function y={1−e−x,0≤x≤1(e−1)e−x,x>1} is a solution of the IVP on the interval [0,∞).

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Chapter 2.3, Problem 52E

Chapter 2.3, Problem 54E

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Chapter 2 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - dydx=1xy (a) y(0) = 0 (b) y(1) = 0 (c) y(2) = 2...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

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The reason that, it is technically incorrect to say that the function y = { 1 − e − x , 0 ≤ x ≤ 1 ( e − 1 ) e − x , x > 1 } is a solution of the IVP on the interval [ 0 , ∞ ) .

Solution Summary: The author explains that it is technically incorrect to say that the function y=leftl1-e-x is a solution of the IVP.

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To discuss: The reason that, it is technically incorrect to say that the function y={1−e−x,0≤x≤1(e−1)e−x,x>1} is a solution of the IVP on the interval [0,∞).

Want to see the full answer?

Chapter 2 Solutions

The reason that, it is technically incorrect to say that the function y = { 1 − e − x , 0 ≤ x ≤ 1 ( e − 1 ) e − x , x &gt; 1 } is a solution of the IVP on the interval [ 0 , ∞ ) .

Solution Summary: The author explains that it is technically incorrect to say that the function y=leftl1-e-x is a solution of the IVP.

Concept explainers

Videos

To discuss: The reason that, it is technically incorrect to say that the function y={1−e−x,0≤x≤1(e−1)e−x,x>1} is a solution of the IVP on the interval [0,∞).

Want to see the full answer?

Chapter 2 Solutions

The reason that, it is technically incorrect to say that the function y = { 1 − e − x , 0 ≤ x ≤ 1 ( e − 1 ) e − x , x > 1 } is a solution of the IVP on the interval [ 0 , ∞ ) .