123 
330] ON DIFFERENTIAL EQUATIONS AND UMBILICI. 
Substituting these values, and observing that the exponent of b + cu is 
(-2 + 1-A + 1-B + 1-C, = 1-A-B-C) = 0, 
the integral equation is 
const. = x (f+ gu + V U)~ l x 
(f +gu + a(b + cu) + \Zlfy~ A (f+ gu + (3 (b + cu) + fbJ) 1-B (f+gu + <y(b + cu) + f IT) 1 * 0 ; 
or, observing that the exponent 1 of x is 
= -l + (l-A) + {l -B)+(l-G), 
and putting for shortness □ = (fx + gy) 2 + (bx + eg) 2 , the integral equation finally is 
const. = (fx + gy + VO ) -1 x 
{fx + gy + a (bx + cy) + V □ ) 1- ^ (fx + gy + /3 (bx + cy) + VD ) 1_b (fx+gy+y (bx + cy)+VD Y~ c ,. 
where the quantities a, /3, y, A, B, C are given by 
(b + cv) (v 2 - 1) + 2 (/+ gv) — c(v — cl)(v—^)(v — y), 
c(v 2 —1) + Zgv _ A B ^ G 
c (v — a.) (v — /3) (v — y) v — a v — $ v — y ' 
Consider the curve 
0 = (fx+gy+a (bx+cy)+V □) l ~ A (fx+gy + /3 (bx+ cy) + V □) ^ l ~ B (fx + gy + y (bx + cy) + V □) l ~ c t 
which corresponds to the value = 0 of the constant. If, for instance, 
fx + gy + a (bx + cy) + V □ =0, 
this equation gives 
(bx + cy) {(bx + cy) (a 2 - 1) + 2 (fx + gy) a] = 0 ; 
or say 
(bx + cy) (a 2 - 1) + 2 (fx+gy) a = 0. 
But we have 
(b + ca) (a 2 — 1) + 2 (/ + ga) a = 0, 
and the equation therefore is 
(bx + cy) (f+ goi) - (fx + gty) (b + ca) = 0 ; 
that is 
Ccf-bg) (y-ax) = 0; 
or simply y — ax = 0; that is, the directions of the curve at the origin, or point x = 0, 
y= 0, are given by the equations y—ax = 0, y — /3x = 0, y — yx = 0. This is right, since 
from the differential equation we obtain at the origin 
(b + cp)(p 2 -l)+2(f+gp)p, =c(p-a)( 2 )-/3)(p-y), =0. 
16—2