154 Proceedings of the Royal Society of Edinburgh. [Sess. 
If the coefficients in this equation are functions of p and q only, so that 
x, y , and 0 are quite absent, the complete integral is given by 
z — lx + my + n, 
l and m being connected by equation (4), where p and q have been replaced 
by these quantities respectively. 
This condition is fulfilled when 
l = a, m = /3, 
and hence the singular onefold is a particular case of the complete integral. 
It is important to remark that in many cases the equation might be 
reduced to the form (4) by a substitution, and in such cases also the 
singular onefold is included in the complete integral. 
Example 3. — In illustration suppose X, Y, Z to be functions of x, y, and 
z respectively, then the differential equation 
where 
or 
or 
2(pZ + X)’%Z + Yp = 0 
(1) m and n > 1 
(2) m — 0 n > 1 
(5) 
(3) m > 1 n = 0 
has a onefold of singular solutions which is included in the complete 
integral. 
For example, 
( p - x) 2 (q - y) 2 - (p - x) 2 - (q - y) 2 = 0 
when reduced by the substitution 
gives x 2, + y 2 — 2z = c as a onefold of singular solutions, and 
/v»2 1 
z = — + ax + /3y + c 
where a 2 /3 2 — a 2 — /3 2 = 0 as complete integral. 
On considering equation (3), it seems that when p and q appear to 
integral powers, the lowest degree possible with a given onefold of singular 
solution is the second, the coefficients being functions of x, y, z only. It can 
easily be shown, however, that every such equation is reducible as regards 
p and q, that is to say, it is impossible for an irreducible equation of the 
