To test: Whether the differential equation 2 cosh x cos y d x = sinh x sin y d y is exact or not and find the solution of the given equation. The given differential equation 2 cosh x cos y d x − sinh x sin y d y = 0 is exact _ and the solution of the given ODE is C = 2 cos y sinh x _ . The given differential equation is 2 cosh x cos y d x − sinh x sin y d y = 0 . Step 1 : Test for exactness : Take M = 2 cosh x cos y N = − sinh x sin y . ∂ M ∂ y = ∂ ∂ y ( 2 cosh x cos y ) = 2 cosh x sin y ∂ N ∂ x = ∂ ∂ x ( − sinh x sin y ) = − cos x sin y Hence ∂ M ∂ y ≠ ∂ N ∂ x . Therefore, the given ODE is not exact. It is known that if the differential equation M ( x , y ) d x + N ( x , y ) d y = 0 is not exact, then multiply the equation by a integrating factor u ( x , y ) as follows. Here the given ODE is not exact so, multiply the given equation by integrating factor which reduces to exact equation as follows. Obtain the integrating factor as follows. M y ( x , y ) − N x ( x , y ) N ( x , y ) = − 2 cosh x sin y − ( − cosh x sin y ) 2 cosh x cos y = − cosh x sin y 2 cosh x cos y = tan y 2 Further proceed as follows. e ∫ tan y d y 2 = e − 1 2 ∫ sin y d y cos y = e − 1 2 ∫ d t t = e − ln | cos y | + C = e − 1 2 ln | cos y | ⋅ e c = C | cos y | ( ∵ C = e c ) So, the integrating factor will be F ( y ) = 1 cos y . Therefore, multiply the equation by 1 cos y . 2 cosh x cos y d x − sinh x sin y 1 cos y d y = 0 Now check for the exactness of the equation as follows. Take M 1 = 2 cosh x cos y and N 1 = − sinh x sin y 1 cos y then, ∂ M 1 ∂ y = ∂ ∂ y ( 2 cosh x cos y ) = − cosh x cos y ⋅ ( − sin y ) = cosh x sin y cos y ∂ N 1 ∂ x = ∂ ∂ x ( sinh x sin y 1 cos y ) = sin y cosh x cos y Hence ∂ M 1 ∂ y = ∂ N 1 ∂ x . Therefore, it is exact. Step 2: Implicit differential solution : Let u = ∫ M d x + k ( y ) . u = ∫ M d x + k ( y ) = ∫ 2 cosh x cos y d x + k ( y ) = 2 cos y sinh x + k ( y ) ...... ( 1 ) Differentiate the above equation with respect to y as follows. ∂ u ∂ y = ∂ ∂ y ( 2 cos y sinh x + k ( y ) ) = − sin y 2 cos y ⋅ 2 sinh x + ∂ k ∂ y = − sinh sin y cos y + ∂ k ∂ y Take ∂ u ∂ y = N 1 then, − sinh sin y cos y + ∂ k ∂ y = sinh sin y cos y ∂ k ∂ y = 0 k = C Substitute the value of k in the equation (1). Therefore, u ( x , y ) = 2 cos y sinh x + C C = 2 cos y sinh x 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 = ∂ ∂ x 2 cos y sinh x d x + ∂ ∂ x 2 cos y sinh x d y = 2 cosh x cos y d x − sinh x sin y d y = 0 Hence, the result checked. Thus, the given differential equation 2 cosh x cos y d x − sinh x sin y d y = 0 is Exact _ and the solution of the given ODE is C = 2 cos y sinh x _ .

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