Page 198 - Foundations Of Differential Calculus

9. On Differential Equations 181
this reason the solution can be found without the aid of integral calculus.
Indeed, when Pdx + Qdy = 0 is differentiated, we obtain
2 2
Pd x + Qd y + dP dx + dQ dy =0,
and when this is subtracted from the given equation, there remains
2
2
Rdx + Sdx dy + Tdy = dP dx + dQ dy.
Since dy = −P dx/Q, the differentials can be completely eliminated to
indicate the relationship between x and y.
305. Let us suppose that in the solution of some problem we arrive at the
equation
3 2 2 2 2 2 2 2 2 2
x d x + x yd y − y dx + x dy + a dx =0
and that there is no assumption about a constant differential. Since it is
3 2
clear that the equation is not absurd, it follows that x dx + x ydy =0 or
xdx + ydy = 0, whose differential is
3 2 2 2 2 2 2 2
x d x + x yd y +3x dx +2xy dx dy + x dy =0.
When this equation is subtracted from the given equation, there remains
2 2 2 2 2 2
a dx − y dx − 3x dx − 2xy dx dy =0,
or

2 2 2
a dx − y dx − 3x dx − 2xy dy =0.
Since xdx + ydy = 0,wehave
2
2xy dy = −2x dx,
2 2 2 2 2 2
so that a dx−y dx−x dx =0, or y +x = a . Now this equation expresses
a true relationship between x and y, and it agrees with the differential
xdx + ydy = 0, which we found before. This agreement follows unless
it were manifestly clear that the given equation were impossible. Since in
2 2 2
this case that is not true, it is valid to find x + y = a without integral
calculus.
306. For the sake of an example of an impossible equation, let us consider
2 2 2 2 2 2
y d x − x d y + ydx − xdy + adxdy =0,
2
in which no constant differential is assumed. Then we would have y dx −
2
x dy = 0. When this is differentiated, we have
2 2 2 2
y d x − x d y +2ydx dy − 2xdxdy =0.