Program Explanation Given information: The differential equation is a d v d x = 1 + v 2 such that v = d y d x and d v d x = d 2 y d x 2 and the initial condition is y ′ ( 0 ) = 0 . Calculation: The given differential equation is as follows, a d v d x = 1 + v 2 Here, v = d y d x and d v d x = d 2 y d x 2 . Now use the method of separation of the variable to solve the above differential equation. a 1 + v 2 d v = d x Now integrate both sides. ∫ a 1 + v 2 d v = ∫ d x a ∫ d v 1 + v 2 = ∫ d x a sinh − 1 v = x + c Therefore, the equation is as follows, a sinh − 1 v = x + c ........ (1) It is given that y ′ ( 0 ) = 0 then v ( 0 ) = 0

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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Chapter 1.4, Problem 69P

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Program Description: Purpose of the problem is to show that v(x)=sinh(xa) to solve the differential equation advdx=1+v2 and then show that y(x)=acosh(xa)+C by integrate the equation v(x)=sinh(xa) .

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Chapter 1.4, Problem 68P

Chapter 1.5, Problem 1P

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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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Program Description: Purpose of the problem is to show that v ( x ) = sinh ( x a ) to solve the differential equation a d v d x = 1 + v 2 and then show that y ( x ) = a cosh ( x a ) + C by integrate the equation v ( x ) = sinh ( x a ) .

Solution Summary: The author explains how to solve the differential equation by using the method of separation of the variable.

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Program Description: Purpose of the problem is to show that v(x)=sinh(xa) to solve the differential equation advdx=1+v2 and then show that y(x)=acosh(xa)+C by integrate the equation v(x)=sinh(xa) .

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Chapter 1 Solutions