( 463 ) 
B bj 
B == — a a, + boa, — b,a, + bya, co ay. 
1 1 
Hence u, satisfies the following equation of Riccati 
a 1 
Ho ET A (6,>—, by) BT = (6,’a,—b,b,a, + b,?a,— 6, bya, tT bra) — 
1 1 
a, : 
Sets nn ee 8) 
1 
We now proceed to find the substitution of PAINLEvÉ. 
From the general integral 
erste ad 1s 
| y+ hy + Uo 
it is evident that u, is that particular solution of the equation (9) 
which satisfies the equation 
by + bake 
ner rd nt 
1 
if we attribute to y that particular integral of (6) which corresponds 
to the value C= ox. 
Therefore 
a by’ + boy 
f boy zl b, 
is the substitution which reduces the differential equation (6) to (9). 
3. From the preceding we may also deduce the conditions which 
must be satisfied by the given differential equation. For the three 
last equations (7) give 
d bz nee cms: Eed b, _ d Wido Hod 
de \ b, de 2,-—U,d, dx \b, de 2,—U,Az 
6 (5,b,'— 6, b,') mT Doda + b,a,'—b9A,'— 6, Azu, Horde, 
or 
and 
6 (6,5,'—4,6,') = bokod2 — bm yA,’ 55 bood ER bilt, —baAoldo- 
Combining each of these with the five first equations (7) and 
eliminating 2, 4,’ 2, u, u, we may write the conditions 
a, i, 9 0 0 0 | 
ds pt 0 —A, 0 | 
a, B 1 —à, —A, fee! 
a, wn, > Be, A, —A, | a 
| 4% 0 0 u, 0 —A, | 
| (b,5,,,—6, b, —b, —b,A, ba, | 
31* 
