404] 
ON THE ROTATION OF A SOLID BODY. 
139 
Euler remarks that as the right-hand side of the equation contains only the 
variable u, the solution will be effected if we can find a function of u, a multiplier 
of the left-hand side; he had elsewhere explained the method of finding such 
multipliers, and applying it to the equation in hand, the multiplier of the left-hand 
side, and therefore of the equation itself, is found to be * 
, x ,. V(G) 
the same thing K _ 2L MNGu 
K - 2LMNGu ’ 
or what is 
Multiplying by this quantity, the right-hand side may for shortness be represented 
by dU, so that 
dU _ (B- 2LMNFu) V (G) du 
(K - 2LMNGu) V {(2Lit + 51) (2Mu + 33) (2Nu + 1)} ’ 
and U may be considered as a given function of u, or what is the same thing of t, 
As regards the left-hand side, attending to the equation K = EG — F 2 , the radical 
multiplied into \f (G) may be presented under the form 
V[{(£ - 2) 2 ) (K- 2LMNGu) -(Gv- (Z)F) 2 }]; 
and consequently the left-hand side becomes 
(K - 2LMNGu) Gdv + LMNG (Gv - <£>F) du 
(K - LMNGu) V{(£ - 2) 2 ) (K - 2LMNGu) - (Gv - (£)F) 2 } 5 
which putting for the moment K — 2LMNGu=p 2 , Gv — ( 5)F=q, G — 2) 2 =/ 2 , becomes 
_pdq qdp the integral of which i 
p V (f 2 p - q) 
the integral is 
fp 
is sin -1 -5r ; hence restoring the values of p, q, f, 
Gv-SF 
sin" 
V (G - $> 2 ) v (K - 2LMNGu)' 
Hence considering the constant of integration as included in U, or writing 
(H- 2LMNFu) ^(G) du 
U 
= ® + /(T3 
(K - 2LMNGu) V {(2Lu + 51) (2Mu + 33) (2Nu + <£)} ’ 
we have for the required integral of the differential equation 
'V J? 
v (tr - 'Z)‘) v It A — zauhiv cru)\ 
whence also 
sm ^(G-^)^{(K-2LMNGu) j 
Gv-^F 
\J (G — 2) 2 ) V \(K — 2LMNGu)\ 
= sin U, 
and 
-s/ri(G - 2> 2 ) (K - 2LMNGu) - (Gv - 2)i0 2 }] _ _ TT 
(G — !D 2 ) V {(K — 2LMFGu)\ 
18—2