if this can be proved the relation ψ(x, y, z) − ƒ(z) = constant, will be the integral of the given differential equation. To prove this it is enough to show that, in virtue of ψ(x, y, z) = ƒ(z), the function ∂ψ/∂x − σZ can be expressed in terms of z only. Now in consequence of the originally assumed relations,

∂ψ = μX, ∂φ= μY, ∂φ = μZ,
∂x ∂y ∂z

we have

∂ψ / ∂φ= σ= ∂ψ/ ∂φ ,
∂x ∂xμ ∂y ∂y

and hence

∂ψ ∂φ ∂ψ ∂φ = 0;
∂x ∂y∂y ∂x

this shows that, as functions of x and y, ψ is a function of φ (see the note at the end of part i. of this article, on Jacobian determinants), so that we may write ψ = F(z, φ), from which

σ = ∂F; then ∂ψ= ∂F+ ∂F ∂φ= ∂F+ σ· μZ = ∂F+ σZ or ∂ψ− σZ = ∂F ;
μ ∂φ∂z ∂z∂φ ∂z∂z μ∂z ∂z ∂z

in virtue of ψ(x, y, z) = ƒ(z), and ψ = F(z, φ), the function φ can be written in terms of z only, thus ∂F/∂z can be written in terms of z only, and what we required to prove is proved.

Consider lastly a simple type of differential equation containing two independent variables, say x and y, and one dependent variable z, namely the equation