Program Explanation Given information: The differential equation is d y d x = 6 e 2 x − y and initial condition is 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 = 6 e 2 x − y . Separate the variables of the above differential equation as, d y d x = 6 e 2 x e − y e y d y = 6 e 2 x d x Now integrate the equation on both the sides as, ∫ e y d y = ∫ 6 e 2 x d x e y = 6 e 2 x 2 + C e y = 3 e 2 x + C y = ln ( 3 e 2 x + C ) Here, C is the constant of integration. Therefore, the general equation of the differential equation d y d x = 6 e 2 x − y is y = ln ( 3 e 2 x + C )

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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Question

Chapter 1.4, Problem 27P

Program Plan Intro

To find: The particular solution of the initial value problem dydx=6e2x−y and the initial condition is y(0)=0 .

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Chapter 1.4, Problem 26P

Chapter 1.4, Problem 28P

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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 = 6 e 2 x − y and the initial condition is y ( 0 ) = 0 .

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

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To find: The particular solution of the initial value problem dydx=6e2x−y and the initial condition is y(0)=0 .

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Chapter 1 Solutions