----- = [mu]Y, ------ = [mu]Z;
dPx vy vz

these equations require

dP dP
---([mu]Y) = ---([mu]Z), &c.,
dPz dPy

and hence

/dPZ dPY\ /dPX dPZ\ /dPY dPX\
X( --- - --- ) + Y( --- - --- ) + Z( --- - --- ) = 0;
\dPy dPz/ \dPz dPx/ \dPx dPy/

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

dPZ dPY
--- - --- = 0;
dPy dPz

in which case we may take [mu] = 1. Assuming the condition in its
general form, take in the given differential equation a plane section
of the surface [phi] = 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 [psi](x, y, z) = constant, be its
integral, the constant z entering, as a rule, in [psi] because it
enters in X and Y. Now differentiate the relation [psi](x, y, z) =
[f](z), where [f] is a function to be determined, so obtaining

dP[psi] dP[psi] /dP[psi] df\
-------dx + -------dy + ( ------- - -- )dz = 0;
dPx dPy \ dPz dz/

there exists a function [sigma] of x, y, z such that

dP[psi] dP[psi]
-------- = [sigma]X, ------- = [sigma]Y,
dPx dPy

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

dP[psi] df df dP[psi]
------- - -- = [sigma]Z, namely -- = ------- - [sigma]Z;
dPz dz dz dPz

if this can be proved the relation [psi](x, y, z) - f(z) = constant,
will be the integral of the given differential equation. To prove this
it is enough to show that, in virtue of [psi](x, y, z) = [f](z), the
function dP[psi]/dPx - [sigma]Z can be expressed in terms of z only.
Now in consequence of the originally assumed relations,

dP[psi] dP[phi] dP[phi]
------- = [mu]X, ------- = [mu]Y, ------- = [mu]Z,
dPx dPy dPz

we have

dP[psi] /dP[phi] [sigma] dP[psi] /dP[phi]
------- / ------- = ------- = ------- / -------,
dPx / d