( 762 ) 
we find the divided equation 
dv uttl 
— ==, du. 
v wt 
EXAMPLE V. 
The equation 
(#? —y) y = ay —1 
has a singular point in «=1, y=41, which is the vertex of a 
critical pencil with the polar line x + y + 1 = 0. 
By the substitution 
Su 
OE ee = 5 i 
this pencil passes into w == const., whilst the polar line is transformed 
into v= 0. We then find 
dv u—2 
v os w—u-+l 
du. 
EXAMPLE VI]. 
The equation 
ayy =a + y¥? 
has the critical pencil y= me with the polar line z — 0 passing 
through the vertex. In connection with this the substitution 
furnishes the divided equation 
udu +- dv = 0. 
The integral curves 
yt oe == Cz? 
are conics having in QO a contact of four points with z=0 as 
tangent. 
d 
ExAMPLE VII. a ae, — 
da 
uty Ty’. 
This equation of Bernouni has a critical pencil y= me with the 
polar line «= 0. By the substitution z—1:v, y=u:v it is trans- 
formed into w-?du--dv=0. Out of this we find #’—y+Cry=0. 
