PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 209 

A. a® Baa*-! vat—} a” xX a’ X (@ +1) 
2 En a ED Pee 
a-y a-Yy a—y (a-—‘)? ait a—y aoe 
xX 
“Ea a (° oe AEH. , 
In precisely the same manner we may integrate the general equation with 
constant coefficients. 
Let us apply the formula of Prop. 6 to the following example. 
Ex. 2. Uy—3 (@+1) Uy 414+ 2(@t1) (+2) uw, 4 0=0. 
Calling | (1 +5) 1 we get 


4 1 6 4 |(a+2) 
my —B (eo +1) (14s) % +2 +1) (C8 +2)A4+4,) * wy =0 
Now since ef? (1+ Ag) zeW" (1 + ay)he” ug 
we have 
rae aa 
w— 3(L+e7*)e7 "(1+ Ay) ef ay +2 (e? +1) (4 26-*ye-!4 fal atlas : ef 4, =0. 
l 4 [a+2) , 
or Uy —3 (1+ Ay)’ e% uy, +2(e° +1) (14+4,)' 2’ ef uw, =0. 
Pot 8 +a,) + +a,)) for + ages) 
where a, in the former operates on the z in the latter. 
uy — 8 (1+ Ay) eu +2(1 te “Jed (L+a,). (L+4,) ef wy =0 
or ty —B (+ Ay) ety + 2 (1+4,) € . (1404) eu, =0 
or ty —3(L+ Ay)! ey +2 (14 Ay)! ef P uy =0 
which can be resolved into the two 
l= Gn), e hay =0 and {1-—2(1+ Ay)! e! } uy =0 



or U,—(*#+1) 4,41=9 and u,—2 (#@ +1) u,,,=0 
U,=- a U,= B 
where (ia ja + 1 or U, ~ 9x jeri 
and therefore generally, ~, ae which is the complete solution of the equa- 
e+ 
tion. 
It is evident that the process employed in Example 1, applies equally in this 
Example, when a function of 2 appears on the right-hand side of the equation. — 
Hence 
Ex. 3. Uy —3 (@41) Wey +2 (441) (742) Upp o=X gives 
