and hence

X( ∂Z− ∂Y) + Y( ∂X− ∂Z) + Z( ∂Y− ∂X ) = 0;
∂y ∂z∂z ∂x∂x ∂y

conversely it can be proved that this is sufficient in order that μ may exist to render μ(Xdx + Ydy + Zdz) a perfect differential; in particular it may be satisfied in virtue of the three equations such as

∂Z − ∂Y= 0;
∂y ∂z

in which case we may take μ = 1. Assuming the condition in its general form, take in the given differential equation a plane section of the surface φ = C parallel to the plane z, viz. put z constant, and consider the resulting differential equation in the two variables x, y, namely Xdx + Ydy = 0; let ψ(x, y, z) = constant, be its integral, the constant z entering, as a rule, in ψ because it enters in X and Y. Now differentiate the relation ψ(x, y, z) = ƒ(z), where ƒ is a function to be determined, so obtaining

∂ψ dx + ∂ψdy + ( ∂ψ− dƒ ) dz = 0;
∂x ∂y∂z dz

there exists a function σ of x, y, z such that

∂ψ = σX, ∂ψ= σY,
∂x ∂y

because ψ = constant, is the integral of Xdx + Ydy = 0; we desire to prove that ƒ can be chosen so that also, in virtue of ψ(x, y, z) = ƒ(z), we have

∂ψ − dƒ= σZ, namely dƒ= ∂ψ − σZ;
∂z dzdz ∂z