( 1251 ) 
from which always the coefficients ¢ may be determined, and 
u, = (10a + 5e) t, + bt, 
2u, — 6u, = 10bt, + (12a + 10c)¢, + 20t, 
du, — Su, = det, + Tht, + (14a + 15c)t, + 308, 
Au, — 4u, = Act, + 46t, + (16a + 20c)t, + Abt, 
ou, — du, = det, + bt, + (18a + 25c)#, + 5b¢, 
6u, — 2u, = 2ct, — bt, + (20a + 30e) t, 
— U,=— ct, — dbt.. 
From these th2 coefficients « may be found only when 
(50a + 30c)t, + 12bt, + (14a + 18c)t, + Abt, + (18a + 3Cc)t, — 20bt, = 0 
or when 
20 | 4 
gd + (la + 6e) (26, — 54) + > 0 (Bt,—4e,) + 
ue = (34a + 30c) (4t, —8¢,) + 40 (5t, — 2t,) + = (190a +.210c)(—4,) = 0. 
Substituting now the values s, we have ihe condition 
(ate) [26s, + (Ila + 9c) s, + 6s, + (17a + 27c) 5, —146s,] = 9 
or 
Glare ee bo Bei 0 vana si. PS) 
If 6=0, tbe differential equation reduces to 
dy —a + (3a-+ 5c) ay 
da ya? + cy? 
which has been considered in Art. 5. 
If a+c=O0 we have 
dy — a J be? — 2ary — by? 
da rf y Haa + 2bay — ay? 
which has been treated in Art. 7. 
When however ac = — 6? — 2c’ the differential equation takes a 
new form 
dy — ox + bex* — (3b? + c*) ey — bey? 
dz cy — (B? + 2c?) x? + be ay + cPy? © 
To solve this we will try to find particular integrals. If the conic 
we + 2Hay + By’ + 2Gy + 2Fy + C=0 
satisfies the equation, 
a+ Hy + G __—ewx + boet — (3b? + C°) ey — bey? 
Hat By+F cy —(6? + 2c*) a? + 2be ey + cy? 
82* 
