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 x + 1 ) d t + ( e t − 1 ) d x = 0 Compare the equation ( e t x + 1 ) d t + ( e t − 1 ) d x = 0 with the standard equation for the exact equation. M d x + N d y = 0 From the above comparison, M = ( e t x + 1 ) N = ( e t − 1 ) Check for the compatibility condition for the equation to be the exact equation. Take the partial differentiation of M with respect to x . ∂ M ∂ x = ∂ ∂ x ( e t x + 1 ) = e t Take the partial differentiation of N with respect to t . ∂ N ∂ t = ∂ ∂ t ( e t − 1 ) = 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 equation ( e t x + 1 ) d t + ( e t − 1 ) d x = 0 is exact

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ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Question

Chapter 2.4, Problem 24E

To determine

The solution of the equation (etx+1)dt+(et−1)dx=0 if the initial value is x(1)=1.

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Chapter 2.4, Problem 23E

Chapter 2.4, Problem 25E

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The solution of the equation ( e t x + 1 ) d t + ( e t − 1 ) d x = 0 if the initial value is x ( 1 ) = 1 .

The solution of the equation (etx+1)dt+(et−1)dx=0 if the initial value is x(1)=1.

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