( 848 ) 
1. Let «=0, y=0 be a singular point, so that we have 
7 (0,0) = 0 and g (0,0) = 0. : 
The tangent to the integral curve in the point (2, mz) is represented by 
Y — mx DELA (Ac), 
Jz 
where Jr =f (@, Mz), Jz — 9 (#, m2). 
If this tangent is to pass for each value of z through a fixed 
point, it must be possible to determine 2, and y, in such a way, 
that for all values of z 
@) fu — Yo Ir + (mg — fr) 4 = 0 
is satisfied. 
The left member of this equation must therefore after division by 
a power of w be linear in a. So we can put 
Jr = Aar! 4+ Aar, 
Gz = Bar + Bar, 
and we have then still the condition that (mg, — f) may contain 
Only aa 
As fr and gx originate from f(x,y) and g (z, y) by the substitution 
y= maz, we can put 
J=A ys + A, gris J.H A, oe + Be) (y, 2), 
g= By" + By a+... Brat + Ay, a), 
where H, and H, represent homogeneous polynomia of order (n—41). 
Then mgr — fr becomes equal to 
(Bymrt! + Bm +... + Bm) — 
— (Aym" + Aymt—1 +... + A,)} a + (mB— Aar. 
As (mg,— fr) must be, independent of m, of order (n —1) we 
have the conditions 
B= 0 Bis ALO <k <a); A, =e 
So En 
S= (Ayr! + Ayyt 2 a H.H Ari tT) y + AC) (9,2), 
g= (Ayr + Ayr Pa +... + Ani #"—!) 2 + HED (y,2). 
So we can put 
dy  yH,—Y(y,2)+ 4,0) (y,2) | 
de « Hy’—)(y,#) HHD (y,2) ( 
Now that the general form of the differential equation has been 
found, we can easily give the substitution indicated by Prof. pr Vaiss. 
We can replace (1) by 
Hed Hod 
Xv 
a Hel) En Hel) Z 
