156 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Hence z = 0 is a singular solution, and z — ax 2 is a singular onefold 
giving z — 0 for a = 0. 
But the original equation may be written in the form 
(px - 2 z) 2 + ( px - 2 z)qz + q 2 yz = 0, 
which is reducible as regards p and q to 
2 xp + (z + Jz 2 - 4yz)q — 4z. 
Lagrange's Subsidiary System is 
dx dy i dz 
'2x z + Jz 2 -4yz 4 z 
two integrals of which are 
2 = cpc 2 and Jz - 4y 4- Jz = c 2 . 
Hence the solution is 
z = x 2 f( Jz - 4y + Jz), 
and the singular onefold is given by f— arbitrary constant. 
It has already been shown that f(xyz) = c 1 is one integral of the 
equation 
*Al fzp + (J 2 + >/ *As 2 -'f'Mte g - ‘Al/z - (^2 + ^ ^2 2 - 'f'Mfy • ( 7 ) 
Suppose the total expression 
^2 2 “ 
be given, but the separate functions \js v \[s 2 , and \[s 8 be not given, then we 
can easily find expressions for \js v \fr 2 and \js s , so that 
l 1 ^3 =0 ( 8 ) 
should be a solution of equation (7), for we need merely calculate p and q 
from (8), insert their values in (7), and then choose \js x and \js 2 , so that the 
resulting equation should be an identity. 
Hence equation (7) is a general form for a linear equation of the first order 
having any given integral f(xyz) — c v and a given surface — — O 
as a solution which is a locus of branch points. In the last example dis- 
cussed, in fact, it can easily be seen that z = 4>y is a solution of the equation, 
yet it is not contained in the final solution. 
Example 5. — f(xyz) = c = z — 2y is to be an integral, 
/;=o, fy= - 2, f z = 1, 
then (7) becomes 
or 
l Ali ? + («/'2+ = ^2 2 “Ws)> 
