(a) Program Explanation Given information: The differential equation is d x d t = k x 2 . Concept used: The differential equation d y d x = f ( x ) g ( y ) can be calculated with the help of separating the variables, which is done as ∫ d y g ( y ) = ∫ f ( x ) d x Calculation: The given differential equation is d x d t = k x 2 . Separate the variables of the above differential equation as d x x 2 = k d t Now, integrate the equation on both the sides (b) Program Plan Intro To find: A solution of the initial valueproblem x ′ = k x 2 , x ( 0 ) = 0 .

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 1.1, Problem 43P

(a)

Program Plan Intro

To show: Thegeneral solution of the differential equation dxdt=kx2 is x(t)=1(C−kt) .

(b)

Program Plan Intro

To find: A solution of the initial valueproblem x′=kx2 , x(0)=0 .

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Chapter 1.1, Problem 42P

Chapter 1.1, Problem 44P

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Chapter 1 Solutions

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

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Ch. 1.1 - Prob. 11P Ch. 1.1 - Prob. 12P Ch. 1.1 - Prob. 13P Ch. 1.1 - Prob. 14P Ch. 1.1 - Prob. 15P Ch. 1.1 - Prob. 16P Ch. 1.1 - Prob. 17P Ch. 1.1 - Prob. 18P Ch. 1.1 - Prob. 19P Ch. 1.1 - Prob. 20P Ch. 1.1 - Prob. 21P Ch. 1.1 - Prob. 22P Ch. 1.1 - Prob. 23P Ch. 1.1 - Prob. 24P Ch. 1.1 - Prob. 25P Ch. 1.1 - Prob. 26P Ch. 1.1 - Prob. 27P Ch. 1.1 - Prob. 28P Ch. 1.1 - Prob. 29P Ch. 1.1 - Prob. 30P Ch. 1.1 - Prob. 31P Ch. 1.1 - Prob. 32P Ch. 1.1 - Prob. 33P Ch. 1.1 - Prob. 34P Ch. 1.1 - Prob. 35P Ch. 1.1 - Prob. 36P Ch. 1.1 - Prob. 37P Ch. 1.1 - Prob. 38P Ch. 1.1 - Prob. 39P Ch. 1.1 - Prob. 40P Ch. 1.1 - Prob. 41P Ch. 1.1 - Prob. 42P Ch. 1.1 - Prob. 43P Ch. 1.1 - Prob. 44P Ch. 1.1 - Prob. 45P Ch. 1.1 - Prob. 46P Ch. 1.1 - Prob. 47P Ch. 1.1 - Prob. 48P Ch. 1.2 - Prob. 1P Ch. 1.2 - Prob. 2P Ch. 1.2 - Prob. 3P Ch. 1.2 - Prob. 4P Ch. 1.2 - In Problems 1 through 10, find a function y=f(x)...Ch. 1.2 - Prob. 6P Ch. 1.2 - Prob. 7P Ch. 1.2 - Prob. 8P Ch. 1.2 - Prob. 9P Ch. 1.2 - Prob. 10P Ch. 1.2 - Prob. 11P Ch. 1.2 - Prob. 12P Ch. 1.2 - Prob. 13P Ch. 1.2 - Prob. 14P Ch. 1.2 - Prob. 15P Ch. 1.2 - Prob. 16P Ch. 1.2 - Prob. 17P Ch. 1.2 - Prob. 18P Ch. 1.2 - Prob. 19P Ch. 1.2 - Prob. 20P Ch. 1.2 - Prob. 21P Ch. 1.2 - Prob. 22P Ch. 1.2 - Prob. 23P Ch. 1.2 - A ball is dropped from the top of a building 400...Ch. 1.2 - Prob. 25P Ch. 1.2 - Prob. 26P Ch. 1.2 - Prob. 27P Ch. 1.2 - Prob. 28P Ch. 1.2 - A diesel car gradually speeds up so that for the...Ch. 1.2 - Prob. 30P Ch. 1.2 - Prob. 31P Ch. 1.2 - Prob. 32P Ch. 1.2 - On the planet Gzyx, a ball dropped from a height...Ch. 1.2 - Prob. 34P Ch. 1.2 - Prob. 35P Ch. 1.2 - Prob. 36P Ch. 1.2 - Prob. 37P Ch. 1.2 - Prob. 38P Ch. 1.2 - If a=0.5mi and v0=9mi/h as in Example 4, what must...Ch. 1.2 - Prob. 40P Ch. 1.2 - Prob. 41P Ch. 1.2 - Prob. 42P Ch. 1.2 - Prob. 43P Ch. 1.2 - Prob. 44P Ch. 1.3 - Prob. 1P Ch. 1.3 - Prob. 2P Ch. 1.3 - Prob. 3P Ch. 1.3 - Prob. 4P Ch. 1.3 - Prob. 5P Ch. 1.3 - Prob. 6P Ch. 1.3 - Prob. 7P Ch. 1.3 - Prob. 8P Ch. 1.3 - Prob. 9P Ch. 1.3 - 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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Thegeneral solution of the differential equation d x d t = k x 2 is x ( t ) = 1 ( C − k t ) .

Solution Summary: The author explains that the general solution of the differential equation dxt=kx2 can be calculated by separating the variables.

Concept explainers

To show: Thegeneral solution of the differential equation dxdt=kx2 is x(t)=1(C−kt) .

To find: A solution of the initial valueproblem x′=kx2 , x(0)=0 .

Want to see the full answer?

Chapter 1 Solutions