MR. BOOLE ON A GENERAL METHOD IN ANALYSIS. 
243 
^+f+/i(4^(/iW+/ 2 W)|+/ 3 (0^0. 
Multiply by s, and put s—i 6 , we have 
+ ’ 
whence, by the fundamental theorem, 
u = y Zu m i m ^ — 1u m x m , 
Hence, if for u m we write F m (£), the solution will assume the following form, 
w = F o(0+ F i(0^+F 2 (0^ 2 , (19.) 
where F 0 (£) is arbitrary, and in general 
d_^ ,4\ i i (^ + (/i^+/2(0)^+/3(0) f — i(0 
'^ F 40 4/i(0 F »(*) — ^ 5 
whence we have, as the law of derivation, 
F *(o= + </i(<) +/ 2 «)|+/,m) 
To this we may add a precisely similar solution in ascending powers of y, the two 
together constituting the complete integral. 
We cannot, from the above, deduce Euler’s solution, because that solution is 
expressed in ascending powers of t, and not of x or y. If, however, for f 2 (t), 
f ?t {t) we substitute Euler’s forms of those functions, and make t=z 6 , we shall obtain 
the result in question. 
The general solution (19.) has been given in order to illustrate the fact before 
adverted to, that when T 0 of (17.) is not simply a function of D, the derivation of the 
coefficients of the final series may involve operations which it is difficult to perform. 
We shall now show how, by a linear transformation, the difficulty may be evaded. 
Assume x—y—s , x+y — t, then 
and 
d dds d dt d d 
dx ds dx' di dx ds ' dt ’ 
d d ds d dt d d 
dy dsdy'dtdy dt ds ’ 
and transforming the original equation, we find 
- </.(0 -/. «)*- (A (0 +M<) = o- 
Multiply by s 2 and make s—z 6 , we have, on reduction, 
D(D-l) M -(/ 1 ( l f)-/«))(D-l) £ '«-(^ + (/ 1 W+/ 2 W)|+/ 3 (0)s 2 '«=O, 
whence, by the fundamental theorem, 
