Explanation Approach: a) An equation is an exact equation when M ( x , y ) d x + N ( x , y ) d y = 0 and ∂ M ∂ y ( x , y ) = ∂ N ∂ x ( x , y ) holds for all ( x , y ) in R and if the compatibility condition is applicable. b) If M d x + N d y = 0 is exact, the partial derivative ∂ F ∂ y = N . Integrate the equation with respect to y . F ( x , y ) = ∫ N ( x , y ) d y + h ( x ) c) To determine h ( x ) , take the partial derivative with respect to x on both the sides of the equation and substitute ∂ F ∂ x for M . d) Integrate h ' ( x ) to obtain h ( x ) up to a numerical value and substitute in the equation mentioned in point (b) to find the value of F ( x , y ) . e) The solution of the equation M d x + N d y = 0 is given by F ( x , y ) = C . Calculation: Consider the given differential equation. e t ( y − t ) d t + ( 1 + e t ) d y = 0 Compare the equation e t ( y − t ) d t + ( 1 + e t ) d y = 0 with the standard equation for the exact equation. M d t + N d y = 0 From the above comparison, M = e t ( y − t ) N = ( 1 + e t ) Check for the compatibility condition for the equation to be the exact equation. Take the partial differentiation of M with respect to y . ∂ M ∂ y = ∂ ∂ y e t y − e t t = e t Take the partial differentiation of N with respect to t . ∂ N ∂ t = ∂ ∂ t ( 1 + e t ) = e t From the above, it is clear that the value of ∂ N ∂ t and ∂ M ∂ y are equal. ∂ M ∂ y = ∂ N ∂ t e t = e t Therefore, the differential equation e t ( y − t ) d t + ( 1 + e t ) d y = 0 is exact equation. Consider the equation for F ( t , y ) . ∂ F ∂ t = M F ( t , y ) = ∫ N ( t , y ) d y + h ( t ) Substitute ( 1 + e t ) for N in the above equation and integrate with respect to y

7th Edition

ISBN: 9780135902738

Author: Nagle

Publisher: PEARSON

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Chapter 2.4, Problem 13E

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The equation et(y−t)dt+(1+et)dy=0 is exact or not, and the solution of the equation if it is exact.

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Chapter 2.4, Problem 12E

Chapter 2.4, Problem 14E

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Ch. 2.2 - Prob. 1E Ch. 2.2 - In Problems 1-6, determine whether the given...Ch. 2.2 - Prob. 3E Ch. 2.2 - Prob. 4E Ch. 2.2 - Prob. 5E Ch. 2.2 - Prob. 6E Ch. 2.2 - Prob. 7E Ch. 2.2 - Prob. 8E Ch. 2.2 - Prob. 9E Ch. 2.2 - Prob. 10E

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The equation e t ( y − t ) d t + ( 1 + e t ) d y = 0 is exact or not, and the solution of the equation if it is exact.

The equation et(y−t)dt+(1+et)dy=0 is exact or not, and the solution of the equation if it is exact.

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