Page 206 - Foundations Of Differential Calculus

9. On Differential Equations 189
This same condition is required if Pdx+Qdy is to be a determined differen-
tial, that is, one that arises from the differentiation of some finite function
of x and y. This brings us back to what we have previously observed in
paragraph 312, that the equation dZ = Pdx + Qdy with Z a function of z
alone, while P and Q are functions of x and y, can be real only if

∂P ∂Q
= .
∂y ∂x
Both of these cases are interrelated, since if R is a function of z alone we
can substitute dZ for Rdz, where Z is a function of z.
322. In order to illustrate this criterion let us consider the following equa-
tion:
2 3 2 3 2 2 2
6xy z − 5yz dx + 5x yz − 4xz dy + 4x y − 6xyz dz =0.
When this is compared to the general form, we obtain

2 3 ∂P 3 ∂P 2 2
P =6xy z − 5yz , =12xyz − 5z , =6xy − 15yz ,
∂y ∂z

2 3 ∂Q 3 ∂Q 2 2
Q =5x yz − 4xz , =10xyz − 4z , =5x y − 12xz ,
∂x ∂z
2 2 2 ∂R 2 2 ∂R 2 2
R =4x y − 6xyz , =8xy − 6yz , =8x y − 6xz .
∂x ∂y

With these values discovered, the equation giving the solution is

2 3 2 2 2 3 2 2
6xy z − 5yz −3x y − 6xz + 5x yz − 4xz 2xy +9yz
2 2 2 3
+ 4x y − 6xyz 2xyz − z =0.
But when this expression is simplified, each term is negated by another, so
that 0 = 0, which indicates that the given equation is real.
323. When the expression obtained in this way from the criterion fails to
vanish, this is a sign that the given equation is imaginary. However, when
a finite equation is found in this way from the criterion, provided that
it is consistent with the differential equation, it indicates the relationship
that the variables have to each other. Furthermore, this is the way in which
those cases arise that we recall from paragraph 310. Suppose that the given
equation is

(z − x) dx +(y − z) dy =0.