Program Explanation Given information: The differential equation is d y d x = ( x − 2 ) 2 and initial condition is y ( 2 ) = 1 . Concept used: The differential equation d y d x = f ( x ) g ( y ) can be calculated with the help of separating the variables as, ∫ d y g ( y ) = ∫ f ( x ) d x The integral of the function x n can be calculated as, ∫ x n d x = x n + 1 n + 1 + c Here, c is called the constant of integration. Calculation: The given differential equation is d y d x = ( x − 2 ) 2 . Separate the variables of the above differential equation as, d y = ( x − 2 ) 2 d x Now integrate the equation on both the sides. ∫ d y = ∫ ( x − 2 ) 2 d x Now evaluate the above integral as, y = ( x − 2 ) 3 3 + C

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

Question

Chapter 1.2, Problem 2P

Program Plan Intro

To find: A particular function y=f(x) that satisfy the differential equation dydx=(x−2)2 and the initial condition y(2)=1 .

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Chapter 1.2, Problem 1P

Chapter 1.2, Problem 3P

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Chapter 1 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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Prob. 10P Ch. 1.3 - Prob. 11P Ch. 1.3 - Prob. 12P Ch. 1.3 - Prob. 13P Ch. 1.3 - Prob. 14P Ch. 1.3 - Prob. 15P Ch. 1.3 - Prob. 16P Ch. 1.3 - Prob. 17P Ch. 1.3 - Prob. 18P Ch. 1.3 - Prob. 19P Ch. 1.3 - Prob. 20P Ch. 1.3 - Prob. 21P Ch. 1.3 - Prob. 22P Ch. 1.3 - Prob. 23P Ch. 1.3 - Prob. 24P Ch. 1.3 - Prob. 25P Ch. 1.3 - Prob. 26P Ch. 1.3 - Prob. 27P Ch. 1.3 - Prob. 28P Ch. 1.3 - Verify that if c is a constant, then the function...Ch. 1.3 - Prob. 30P Ch. 1.3 - Prob. 31P Ch. 1.3 - Prob. 32P Ch. 1.3 - Prob. 33P Ch. 1.3 - (a) Use the direction field of Problem 5 to...Ch. 1.3 - Prob. 35P Ch. 1.4 - Prob. 1P Ch. 1.4 - Prob. 2P Ch. 1.4 - Prob. 3P Ch. 1.4 - Prob. 4P Ch. 1.4 - Prob. 5P Ch. 1.4 - Prob. 6P Ch. 1.4 - Prob. 7P Ch. 1.4 - Prob. 8P Ch. 1.4 - Prob. 9P Ch. 1.4 - Prob. 10P Ch. 1.4 - Prob. 11P Ch. 1.4 - Prob. 12P Ch. 1.4 - Prob. 13P Ch. 1.4 - Prob. 14P Ch. 1.4 - Prob. 15P Ch. 1.4 - Prob. 16P Ch. 1.4 - Prob. 17P Ch. 1.4 - Prob. 18P Ch. 1.4 - Prob. 19P Ch. 1.4 - Prob. 20P Ch. 1.4 - Prob. 21P Ch. 1.4 - 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A particular function y = f ( x ) that satisfy the differential equation d y d x = ( x − 2 ) 2 and the initial condition y ( 2 ) = 1 .

Solution Summary: The author explains how the differential equation dy-dx=(x-2)2 can be calculated with the help of separating the variables.

Concept explainers

To find: A particular function y=f(x) that satisfy the differential equation dydx=(x−2)2 and the initial condition y(2)=1 .

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Chapter 1 Solutions