Program Explanation Given information: The differential equation is d y d x + 1 = 2 y and initial conditionis y ( 1 ) = 1 . Concept used: The differential equation is 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 Calculation: The given differential equation is d y d x + 1 = 2 y . Separate the variables of the above differential equation as, d y d x = 2 y − 1 d y 2 y − 1 = d x Now integrate the equation on both the sides as, ∫ d y 2 y − 1 = ∫ d x 1 2 ln ( 2 y − 1 ) = x + C ln ( 2 y − 1 ) = 2 x + 2 C 2 y − 1 = e 2 C e 2 x The above expression can be written as, 2 y − 1 = e 2 C e 2 x y = e 2 C e 2 x + 1 2 Here, C is the constant of integration

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

5th Edition

ISBN: 9780134872971

Author: Edwards, C., 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 1.4, Problem 23P

Program Plan Intro

To find: The particular solution of the initial value problem dydx+1=2y and the initial conditionis y(1)=1 .

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Chapter 1.4, Problem 22P

Chapter 1.4, Problem 24P

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Chapter 1 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- For Differential Equations and Boundary Value Problems: Computing and Modeling Tech Update

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The particular solution of the initial value problem d y d x + 1 = 2 y and the initial conditionis y ( 1 ) = 1 .

Solution Summary: The author explains how the differential equation can be calculated with the help of separating the variables as, c

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To find: The particular solution of the initial value problem dydx+1=2y and the initial conditionis y(1)=1 .

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Chapter 1 Solutions