Explanation Given: The differential equation is ( 1 t + 1 t 2 − y t 2 + y 2 ) d t + ( y e y + t t 2 + y 2 ) d y = 0 . Procedure used: Solution to a differential equation : Begin with considering the case of a differential equation M d x + N d y = 0 , where M and N are functions of ( x , y ) . Compare the given equation to obtain values of M and N . Differentiate M with respect to y by applying partial derivate and differentiate N with respect to x by applying partial derivate. Case I: Differentiation of M with respect to y is not equal to differentiation of N with respect to x . The differential equation is not exact and the differential equation can be solved by the integrating factor. Case II: Differentiation of M with respect to y is equal to differentiation of N with respect to x . The differential equation is exact and the solution of differential equation can be found by integrating M ( x , y ) with respect to d x considering y as a constant and N ( x , y ) with respect to d y considering x as a constant. Calculation: Consider the differential equation. ( 1 t + 1 t 2 − y t 2 + y 2 ) d t + ( y e y + t t 2 + y 2 ) d y = 0 Here, M = ( 1 t + 1 t 2 − y t 2 + y 2 ) and N = ( y e y + t t 2 + y 2 ) Differentiate M with respect to y by applying partial derivate. ∂ M ∂ y = ∂ ( 1 t + 1 t 2 − y t 2 + y 2 ) ∂ y = − ( t 2 + y 2 ) + y ( 2 y ) ( t 2 + y 2 ) 2 = y 2 − t 2 ( t 2 + y 2 ) 2 Differentiate of N with respect to t by applying partial derivate. ∂ N ∂ x = ∂ ( y e y + t t 2 + y 2 ) ∂ x = ( t 2 + y 2 ) − t ( 2 t ) ( t 2 + y 2 ) 2 = y 2 − t 2 ( t 2 + y 2 ) 2 Thus, ∂ N ∂ x = ∂ M ∂ y . Therefore, the given differential equation is exact. Consider the equations given below. ∂ f ∂ x = ( 1 t + 1 t 2 − y t 2 + y 2 ) … … ( 1 ) ∂ f ∂ y = ( y e y + t t 2 + y 2 ) … … ( 2 ) Integrate equation(1)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 2.4, Problem 20E

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Chapter 2.4, Problem 19E

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Chapter 2 Solutions

Advanced Engineering Mathematics

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Whether the given differential equation is exact or not. If it is exact, then find the solution of differential equation.

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Chapter 2 Solutions