356 Mr. J. Prescott on the 



and adding to the corresponding sides of (42), we get, on 

 neglecting /3, 



. d 2 z .-r, dz Wclf , IX g /x <r » mx ,, ON 

 A^ -iBco^- j^f (s-^) = JKA-iB«T). (48> 



Similarly, multiplying (45) by i and adding to (44), again 

 dropping £, 



f =(/-l)^ (49) 



We will write, for shortness, m for-^-r, and r for — r-; 



' 2 A gA 



then equation (48) gives 



J 2 -2imJ- J(,-t)=f (l-2mT). . (50) 



We have to solve equations (49) and (50) . 

 49. From equation (49) 



1 -dyfr 



Therefore, 



dT " dT ,+ /-l\ 1 dT 2 * dTj' 



iT2 - y_ -l ^ T2 + y _ x j- dT z • 



Substituting for 2 and its differential coefficients in (50) r 

 writing <£ for -^, and multiplying up by (/— 1), the result- 

 ing equation is 



irfg + (/+i-M) g - (£***»)# 



= (/-l)|(l-2mT). . . (51) 



This equation being linear, the complete value of <f> 

 consists of two parts, one of which is the particular inte- 

 gral corresponding to the terms on the right-hand side, and 

 the other is the complementary function, that is, the value of 

 (f> satisfying the equation obtained by dropping the terms on 

 the right-hand side. We will first deal with the comple- 

 mentary function. 



