49 
ORDRE SAT; Ree 
which after reduction gives finally the condition 
12 (a + €) (a — 28 — c) (B a—28 4 5c) = 0. 
This condition breaks up into three others from which the first 
a+c=0O has already been examined in Art. 2. 
9. Introducing the second, we must examine the case where 
2a' = —- i (8a — 28 + 3c) 
2c = (ba — 28 + 5e) 
2b = 38a+ 5e 
2p =de 
or, remembering that b == ig 
a’ = — (a + 2c) 
ers (Le 1-0) 
26 == = 38a + BC 
26°—== t(a ==). 
This case has already been met with in Art. 7. 
10. Finally we have the relations 
2a’ = — i (Ba — 28 + 3e) 
2e == t(5a — 28 + 5e) 
2b; Za + de 
23 = 3a-+ 5e 
which are identical with 
ae 
ec = ta 
2b = 2ib' — i (Za + 5e). 
The differential equation reduces in this case to 
dy —e Hier? + (Ba + de) ey + ray" 
dx y + av’? +i (8a + Be) ay + cy? 
whose general integral may be constructed from the two particular 
integrals 
j 1 
(a + 3e) (ew — iy)? + Uw + zw) + ——-=0 
ate 
and 
(a + 3c) (w — ty)? + Bila? +4) =0 
which are easily found. 
This general integral 
{ (a + 30) (we — dy)? + Bila? + y)P 
= cons. 
1 3 
(a + 30) (w — iy)? + Bi @ + iy) + rs 
Proceedings Royal Acad. Amsterdam. Vol. XV. 
