112 
DR. G. R. GOLDSBROUGH ON THE INFLUENCE OF 
Again, in the special case where r — 2 n, we find in place of equations (12), the 
following 
2c h2n nA 0 cos —r) + A" h2n — 2KC h2n X 0 cos (n0 — r) —2 kX.\ %2 „ ') 
+ cq 0 A, ,„ + cq o„A 0 sin (n0 —t)+|A 0 {sin (3n0 —t) — sin (n0 —t) cos 2t j 
, x • o i n I ( 15 ) 
— cos ( ?i(p —t ) sm 2 t ! =0, j 
2 c 1> 2 n '/iiX 0 sm {n<p r) + A 1 , 2 ?! + 2 /cc 1j 2 K Aq sm (n<p r) + 2kA i,o„ + 05 ,oXi, 2 n = 0. J 
In order to avoid the explicit appearance of 0, we must have 
a l 2n = If cos 2t. 
Since we have already stipulated that A must not contain any term in cos (n 0 —r), 
c, 2n must be so chosen as to make quantities involving cos(n 0 — r) annul. Hence we 
must have 
{ 2 ci,o»nA 0 — 2 /cCi, 2 n X 0 —|A 0 sin 2 t} cos (n</>— t) — 2 /X'i, 2n = 0, 
i. 2c 1j2b wX 0 + 2/cCj, 2? , A uj - sm (n<p t) +X i,2« + 05,oXi,2h = 0. 
Whence 
n (e 5 ,„-vr) sin 2 t 
1,a * 4KA, 0 -n‘) ’ 
v _ x (w 2 -«i.o) (0 5 ,o + w 2 ) sin 2tA 0 sin (n 0 - T ) 
-^ 1,2 n — 8 / 4 \ • 
KU (« 1 , 005.0 -n) 
To the value for X li2 „ must be added the further particular solution arising from 
the term |-A 0 sin (3 n<f> — r) in (l5). It is 
a = ( 05 ,o — 9n 2 ) A 0 sin (3n0-r) 
1 ’ 2 " 16K o 0 5 .o-9n 4 ) 
Y _ 3ukA 0 cos (3n0 — t) 
8 (cq.o05.o-9n 4 ) 
Proceeding in this way, we have the following results 
Terms not involving argument 0 : 
In A 
In X 
In c 
Also 
A 0 sm {n(j >— t). 
X 0 cos (n<j >— t). 
None. 
2 KnX 0 = - (a uo —n 2 ) A 0 , 
cq.o = n 2 + 4K 2 n 2 /(0 5 , o —n 2 ). 
when 
