( 465 ) 
In both cases this expression vanishes. Therefore both conditions 
are found by writing u, =O in the equation (10). In this way the 
conditions we looked for, are the following 
where the last row 
4. When n= 3, 
1 Orr Or Ops 0 
b, 
— Bee cle 70 
b, 
b, 
0 — 1 5, 0 
b, ih den Ed ret ab) 
b, 
0 0 En Pega 
0 0 0 
@Duz=oy Td e § 
is given by the relations (11). 
the general integral 
NL RE | 
= nd a = sale Pec agit, 
yv Huy +wy+4, 
shows, that the differential equation must be of the form 
dy ay Hay Hay Hay’ + ay’? + ay + a, 
= 14 
da b,y* + 4b,y° + 6b,y° + 4b,y + b, VE 
with the following relations between the coefficients a, 6, A, u: 
Ga, =d, \ 
Oa, = ud, + 4,' — Al, 
Oa, = ud, Hud, +4 — Au, — As 
Ga, = ud, EE ud, ae 4,2, ' = Jo ns hu, FE do, eS Ao 
Oa, = ud, Hud, Hud, — Agu,’ — Al — Alo 
6a, = Hod, oF ud. nie Aobty’ aes Alo (15) 
Ga, = Udo i Au. 
6b, =À, — Ay, 
40b, = 22, — 
24,u, 
60d, = 34, +5 Au, En Ast, Ee 3À, U, 
40b, = 2d,u, — 2A,u, 
8b, DE Jo, ig du, 
