790 Sir Oliver Lodge on 



In the same way the differential equation for y can be 

 obtained; it only differs in having -\-nrib substituted for — n'bs. 



Therefore x and y are both functions of s. 



Hence we may assume that they can be expanded in powers of 

 sin 6 ; an expansion which readily takes the form 



x = a -\- a l slri6-)-a 2 cos2d-\-a :i sin3d -f- (iii.) 



y = h o + b l sinO + b 2 cos20 + h 2 smSd-\~ (iv.) 



Substituting in the original equations, (i.) and (ii.), we get two 

 equations like 



n(a -f- a x sin 6 + « cos 20 + a 3 sin 3t) + . . . .) 



= -(^ + j6 1 )8infl + (ft I +J,)ooB26-|(6 9 +6 4 )Biiiaar+ 



Equating coefficients in this pair of identities, we find 



«.=<>, CO *.*-* (»> 



«*i=-(&o+£& a ), to » ,ft i s= -(«ft+i«A W 



n« 2 = 6 t + ^, (5) n'6 2 = « l + a 3 , (e) 



na 2r+ i= —±(2r+l)(b 2 r+fer+2), w'&ar+i— - j(2r+l)(a 2r -f #&•+»). 



Taking rt l5 5 l5 as the arbitrary constants, and using these equa- 

 tions in numerical order, we find in succession 5„ a 2 , b 3 , a 3 . . . . 

 in terms ot* «,, b r and the known constants n, ri, b. This is the 

 General Solution. 



With 6 1 = as the special case, we find successively all the even 

 as and odd 6's zero. 



Hence (iii.) and (iv.) reduce to 



' 05= a a siii 0-f- a 3 sin 30 + # 5 sin 50-f (iii'.) 



y=6 o + 6 2 cos20 + 6 4 cos40+ (iv'.) 



where 



n« 3 = - f(& 2 -f 6 4 ), n'6 2 =aj+ tf 3 , 



»«.= -«&« + &.), n'b=2(a :i + a 5 ), 



fta-irfi = - i( 2/ ' + ] XV + V+2), n'6 2r = iia 2r -i + air+i). 



