Page 185 - Foundations Of Differential Calculus

168 9. On Differential Equations
Since P and Q are functions of x, their differentials have the form dP = pdx
and dQ = qdx so that
2ydy + Pdy + yp dx + qdx =0,

with the result that
dy yp + q
= − .
dx 2y + P

2
283. Now, just as the finite equation y +Py+Q = 0 gives the relationship
between y and x, so the differential equation expresses the relationship, or
the ratio of dy to dx. However since
dy −yp + q
= ,
dx 2y + P
we cannot know this ratio dy : dx unless we know the function y. Things
could not possibly be otherwise, since from the finite equation y has two
values. Either of these two values has its own proper differential, and either
differential will appear depending on which value is substituted for y.Ina
similar way, the function y can be defined by a cubic equation. In this case,
dy/dx will have three values, depending on which of the three values of y
is substituted. If in a given finite equation y has four or more values, then
of necessity, dy/dx has just as many significations.
284. Nevertheless, the function y can be eliminated, since we have two
equations containing y, namely the finite and the differential. In that case,
however, the differential dy will take on as many values as y had in the
original finite equation. Thus there are this many ratios of dy to dx. Let
2
us take the preceding example, y + Py + Q = 0, whose differential is
2ydy + Pdy + ydP + dQ = 0, from which we obtain
Pdy + dQ
y = − .
2dy + dP
When this value for y is substituted in the previous equation, we have
2 2 2 2 2
4Q − P dy + 4Q − P dP dy + QdP − PdPdQ + dQ =0,
whose roots are
1
1 2 PdP − dQ
2
dy = − dP ± .
2 P − 4Q
These two differentials correspond to the two values of y from the original
finite equation
1 1
2
y = − P + P − 4Q.
2 2