92 Proceedings of the Royal Society of Edinburgh. [Sess. 
-i 
or 
v = log + - sin TCOST + —— sin 2 r _ ( T + 1) ) 
l n 4 n L ) In 
a — clq exp v = a 0 exp ( - J | 1 + -h sin r cos t + -h_ sin 2 r 
4^ 
. (116) 
the equation (11a), (116), together with 
— , tan r -p — = — tan (nt + C) 
n 2 n u 
• (He) 
express the solution, which is immediately identified with the known 
form, since 
1 + — sin r cos r + sin 2 r = cos 2 r sec 2 (nt + C) 
?Z 4 // 2 
'2 
sin r = — cos t f — tan (nt + C) — 
n l n 2 n) 
so that 
exp v . sin r = exp ( - \ xt ) sin ( n't + C — tan 1 Ph j 
. (lid) 
The formulse just given are exact, but are less suited for use than 
approximations which are easily carried as far as x 2 /^ 2 - I now give 
these, still for the case R = 0. 
Starting from equations (10), 
% = - O-ifS i - 1 ) cos 2t + ^ sil1 2t 
ridt \n ' 2 
n 
with 
r = nt + e 
we have a first approximation 
n 2 /V 2 = 1, 
Substitute for e, and we have for the second approximation 
/< 
e = 6a , COS Zr. 
0 4 ri 
de 
ndt 
i(p - 1 - §3 - i(^ - l) cos (2»'«+ 2c„) + ZC sin (2n'< + 2 e 0 ) 
-^y c ° s (^+H) • • • 
giving 
ft' 2 _ 
e = e 0 - cos (2ft'£ + 2e 0 ) - T h ~ 2 sill (2nt + 2e 0 ) - sin (4ft'i( + 4e 0 ) . (12a) 
It is clear that to this order we may replace n' by n in the equation for e, 
since x 3 /n 3 is insensible. 
