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                      Boyce/DiPrima 9 ed, Ch 2.6: 
                      Exact Equations & Integrating Factors
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                      Elementary Differential Equations and Boundary Value Problems, 9 edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
                      Consider a first order ODE of the form
                                               M(x,y)+N(x,y)y′=0
                      Suppose there is a function ψ such that
                                   ψx(x,y)=M(x,y), ψy(x,y)= N(x,y)
                      and such that ψ(x,y) = c defines y = φ(x) implicitly. Then
                          M(x,y)+N(x,y)y′= ∂ψ +∂ψ dy = d ψ[x,φ(x)]
                                                           ∂x      ∂y dx        dx
                      and hence the original ODE becomes 
                                               d     [          ]
                                                  ψ x,φ(x) =0
                                               dx
                      Thus ψ(x,y) = c defines a solution implicitly.  
                      In this case, the ODE is said to be exact. 
              Example 1:  Exact Equation
                 Consider the equation:
                                 2x+y2+2xyy′=0
                 It is neither linear nor separable, but there is a function φ
                 such that     ∂ϕ =2x+y2 and ∂ϕ =2xy
                                ∂y                ∂x
                 The function that works is      ϕ(x,y)= x2 + xy2
                 Thinking of y as a function of x and calling upon the chain 
                 rule, the differential equation and its solution become 
                     dϕ = d (x2 + xy2) =0⇒ϕ(x,y)= x2 +xy2 =c
                     dx    dx
            Theorem 2.6.1
            Suppose an ODE can be written in the form
                     M(x,y)+N(x,y)y′=0      (1)
            where the functions M, N, M and N are all continuous in the 
                                    y     x
            rectangular region R: (x, y) ∈ (α, β ) x (γ, δ ). Then Eq. (1) is 
            an exact differential equation iff 
                  My(x,y)= Nx(x,y), ∀(x,y)∈R   (2)
            That is, there exists a function ψ satisfying the conditions
                 ψx(x,y)=M(x,y), ψy(x,y)= N(x,y)   (3)
            iff M and N satisfy Equation (2).  
             Example 2: Exact Equation   (1 of 3)
              Consider the following differential equation. 
                (ycosx+2xey)+(sinx+x2ey −1)y′=0
              Then 
                M(x,y)= ycosx+2xey, N(x,y)=sinx+x2ey −1
              and hence
                M (x,y)=cosx+2xey = N (x,y) ⇒ ODE is exact
                  y                     x
              From Theorem 2.6.1, 
               ψ (x,y)=M = ycosx+2xey, ψ (x,y)= N =sinx+x2ey −1
                 x                          y
              Thus
                                    (          y )           2 y
              ψ(x,y)=∫ψx(x,y)dx=∫ ycosx+2xe dx= ysinx+x e +C(y)