undefined Solution To test: Whether the differential equation e − y d x + e − x ( − e − y + 1 ) d y = 0 is exact or not and find the solution of the given equation. The solution of the given ODE is C = e x + e y − y _ . The given differential equation is e − y d x + e − x ( − e − y + 1 ) d y = 0 . Here the given factor is f = e x + y . Multiply the equation e − y d x + e − x ( − e − y + 1 ) d y = 0 by e x + y . e − y ⋅ e x + y d x + e − x ( − e − y + 1 ) ⋅ e x + y d y = 0 e x + y − y d x + ( − e − x − y + e − x ) ⋅ e x + y d y = 0 e x d x + ( e y − 1 ) d y = 0 Step 1 : Test for exactness : Take M = e x N = e y − 1 . ∂ M ∂ y = ∂ ∂ y ( e x ) = 0 ∂ N ∂ x = ∂ ∂ x ( e y − 1 ) = 0 Hence ∂ M ∂ y = ∂ N ∂ x . Therefore, it is exact. Step 2: Implicit differential solution : Let u = ∫ M d x + k ( y ) . u = ∫ M d x + k ( y ) = ∫ e x d x + k ( y ) = e x + k ( y ) ...... ( 1 ) Differentiate the above equation with respect to y as follows. ∂ u ∂ y = ∂ ∂ y ( e x + k ( y ) ) = ∂ ∂ y ( e x ) + ∂ k ∂ y = 0 + ∂ k ∂ y Take ∂ u ∂ y = N 1 then, 0 + ∂ k ∂ y = e y − 1 ∫ ∂ k ∂ y d y = ∫ ( e y − 1 ) d y k = e y − y + C Substitute the value of k in the equation (1). Therefore, u ( x , y ) = e x + e y − y + C C = e x + e y − y Step 3: Checking an implicit solution : Check the implicit solution by differentiating implicitly as follows. d u = ∂ u ∂ x d x + ∂ u ∂ y d y = ( e x + e y − y ) d x + ( e x + e y − y ) d y = e x d x + ( e y − 1 ) d y = 0 Hence, the result checked. Thus, the solution of the given ODE is C = e x + e y − y _ .

10th Edition

ISBN: 2819770198774

Author: Kreyszig

Publisher: WILEY CONS

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