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

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