1900. No. Be ON PARTIAL DIFFERENTIAL EQUATIONS, ETC. I I 
The actual result will be: 
ludo [da + B3 + Cy + Då + H|=0, 
which, while we suppose D + 0, reduces to the given equation. 
If we now equate to zero the second factor of the first equation and 
the first factor of the second equation, we shall have 
Ou Ou 
I ou , ou 
age spar — 
Qu\ du ou\ ou Ou ou g 
EE x —° 
whence we derive the following system of linear partial differential 
equations : 
DT AE =0 EE) 
Again equating to zero the first factor of the first equation of (11) and 
the second factor of the second equation, we have, in the same manner, 
Pl = i 25 en Ht = 0 
° 
Da Ve 
2 Ou ‘ou Ou 
WS the +e 
That these systems are relevant to the solution of the problem 
under consideration, may be shown in the manner that has been used 
for system (12). 
It is however easy to see that the two linear systems (12‘) and 
(12“) give no new linear systems, but are both equivalent to the linear 
system (12), which we obtain by giving to å its different values deter- 
mined by the cubic (A). 
