538 PROFESSOR A. C. DIXON ON THE SINGULAR SOLUTIONS 
The differential equations are then reduced* to 


3 (dt)? = 
(u—v)t(_+f#) (a+ tA) (6+ tB) +10)” * 
and 
(dt)? x 
> om 
“(uw — »v) (1 + #) (@ + tA) (© + EB) (ce + 10) 
If Aw’ + By,*? + Cy? — 1 = 0, then ¢, uv, or v vanishes, say ¢; the first equation 
is satisfied, and the second is reduced to 
u (du)? a v (dv)? 
(1 + w) (a + wA) (6 + wB) (¢ + wl) ~ (1 + v) (a + vA) (0 + WB) (¢ + 00) * 


The variables are here separated, so that we have one first singular solution. 
The other is given by taking 
ax + by? + cy. —1= 0, that ist = o, 
and 
(du)? (dv)? 
u(1 + u) (a + wA) (b+ uB) (ce + uC) v(1 + v) (a + vA)(b + WB) (ce + vO) 

Either of these two solutions gives as the second singular solution 
ax* + by, + cy? —1=0, 
Aa? + By,’ + Cy.” — 1 = 0. 
* Tn carrying out the reduction we may use the formula 
_Aax? Bby,? Coys? Mite ts (a — A) (b — B) (e —C) (un — 4) (uw — wu) (uh — v) 
Gp wA b+ eB et paC lta @+ nA) Ob ABC aC) iG) icra Cer 


and differentiate it, considering « as constant and ¢, w, v, 7), y2 as functions of a. 
The expression 
Aa - Bb ie Ce 
nk (be)? + 5g En? + ae Cw) 
may then be expressed in terms of dt, du, dv by means of the formule for 2, y;, y2, in terms of f, u, v. 
Tt will thus be found that the equation 

Bas MPs As) Mw Ser) Op AG. DoE Bay Chen O meer 
( Aa Bbp,? Cep,” ) 
FenN We Teas = Ose 70 
( Aaz Bboy, p, Ceyss\’ — ( Aax* Bby,? Coys? 1 ) 

reduces to 
d? 
=(u—») ((—p) 1 +8 (a + Ad) (© + BE) (CC + CH) — ¢ 
