Page 197 - Foundations Of Differential Calculus

180 9. On Differential Equations
That is, we substitute
2
2 dy d x
d y −
dx
2
for d y, and
2 2 2 2 3
3 3d xd y 3dy d x dy d x
d y − + −
dx dx 2 dx
3
for d y, etc. When this is done it becomes clear whether the resulting
equation is the same as the given equation. If this is the case, the given
equation gives a determined relationship between x and y, just as we have
shown at length.
303. In order that this become perfectly clear let us take a given equation
in which no constant differential seems to be given:
2 2 2 2
Pd x + Qd y + Rdx + Sdx dy + Tdy =0.

When we let dx be constant, the equation becomes
2 2 2
Qd y + Rdx + Sdx dy + Td =0.
From this now, the consideration of a constant differential is removed in
the previously prescribed manner, to obtain
2
Qdy d x 2 2 2
− + Qd y + Rdx + Sdx dy + Tdy =0.
dx
Since this equation differs from the original only in the first term, we must
see whether P = −Q dy/dx. If this is the case, we conclude that the given
equation exhibits a fixed relationship between x and y, which can be found
by the rules of integral calculus, whichever differential is taken to be con-
stant. However, if it is not true that P = −Q dy/dx, then the given equation
is impossible.
304. It follows that unless the given equation
2 2 2 2
Pd x + Qd y + Rdx + Sdx dy + Tdy =0

is meaningless, it is necessary that Pdx + Qdy =0. This can happen in
two ways. First, the equation

Qdy
P = − , or Pdx + Qdy =0,
dx
is an identical equation. Second, the equation Pdx + Qdy = 0 is itself a
first-order differential equation whose differentiation gave rise to the given
equation. In the second case, the equation Pdx + Qdy = 0 corresponds to
the given equation and contains the same relationship between x and y.For