Explanation The given differential equation is: y ″ + ( y ' ) 2 = 2 e − y Let y ' = v ( t ) and differentiate the equation d y d t = v with respect to t : y ' = v d d t ( d y d t ) = d d t ( v ) d 2 y d t 2 = d v d t This is a second order differential equation of the form y ″ = f ( y , v ) . d v d t = d v d y d y d t = d v d t v Therefore, the value of second derivative is y ″ = v d v d y . Substitute y ' = v and y ″ = v d v d y in y ″ + ( y ' ) 2 = 2 e − y . v d v d y + ( v ) 2 = 2 e − y v d v d y + v 2 − 2 e − y = 0 ( v 2 − 2 e − y ) + v d v d y = 0 Compare the above equation with M ( y , v ) + N ( y , v ) d v d y = 0 . Here M ( y , v ) = v 2 − 2 e − y and N ( y , v ) = v . Differentiate M ( y , v ) = v 2 − 2 e − y with respect to v: d d v M ( y , v ) = d d v ( v 2 − 2 e − y ) M v = 2 v Differentiate N ( y , v ) = v with respect to y: d d v N ( y , v ) = d d v ( v ) N y = 0 Therefore, M v ≠ N y . Thus, the differential equation ( v 2 − 2 e − y ) + v d v d y = 0 is not exact solution. To make it exact, multiply equation ( v 2 − 2 e − y ) + v d v d y = 0 throughout by an integrating factor. Consider M v − N y N = 2 v − 0 v = 2 then the required integrating factor is: μ ( y ) = exp ∫ 2 d y = e 2 y Multiply equation ( v 2 − 2 e − y ) + v d v d y = 0 with μ ( y ) . ( v 2 e 2 y − 2 e y ) + v e 2 y d v d y = 0 Compare the above equation with M ( y , v ) + N ( y , v ) d v d y = 0 . Here M ( y , v ) = v 2 e 2 y − 2 e y and N ( y , v ) = v e 2 y . Differentiate M ( y , v ) = v 2 e 2 y − 2 e y with respect to v: d d v M ( y , v ) = d d v ( v 2 e 2 y − 2 e y ) M v = 2 v e 2 v Differentiate N ( y , v ) = v e 2 y with respect to y: d d v N ( y , v ) = d d v ( v e 2 y ) N y = 2 v e 2 y Since M v = N y , the equation ( v 2 e 2 y − 2 e y ) + v e 2 y d v d y = 0 has become exact solution. As the differential equation ( v 2 e 2 y − 2 e y ) + v e 2 y d v d y = 0 is exact, there is a function ψ ( y , v ) such that ψ y ( y , v ) = M ( y , v ) and ψ v ( y , v ) = N ( y , v )

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

11th Edition

ISBN: 9781119336617

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley (WileyPLUS Products)

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Chapter 2, Problem 35MP

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To solve: The given differential equation y″+(y')2=2e−y.

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Chapter 2, Problem 34MP

Chapter 2, Problem 36MP

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, 11e WileyPLUS Registration Card + Loose-leaf Print Companion

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The given differential equation y ″ + ( y ' ) 2 = 2 e − y .

To solve: The given differential equation y″+(y')2=2e−y.

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Chapter 2 Solutions