522 Mr. A. 0. Allen on 



Thus T ,v, — ?/, z, — it-, and /are proportional to 



-R 



S -(^D+G+R+S) 

 Q+S S 

 KSD KSD 



•R 



-(r/D+G+R+S) S 



8 -S 



-L 



S 



-S 



(KSD+1) -KR 



-L LD+P+R 



KSD 



-(KSD+1) -KR 



-L 



S 



-S 



-(KSD+1) -KR 



-L LD+P+R 



R 







-KR 



LD+P+R 

 R -S 



Q + S 



KSD 



LD+P+R 

 R 

 

 



R 





 







— s 



Q+S 

 KSD 







g 



Q+S 

 KSD 



-R 



R 



-G/D+G+R+S) 

 S 

 KSD 







G/D+G+R+S) S 



g <5< 



KSD -(KSD+1 



Of these only the 3rd and 5th need now be considered; from 

 them we learn that the 5th operating on z equals the 3rd 

 operating on/, viz. {KQRS(P + R) -LR(Q + S)}/. As for 

 the 5th, which may be written 



LD+P+R 







-R 







R 







-(^D+G+R) 



s 







Q 







-S 







-1 



-1 



-(KSD+1) 



it is clearly a cubic in D, of which the roots may be called 

 a, /3, 7, so that the complementary function for z is 



*«*. 



St 



c 1 e" v + c 2 e H ° + c s e yt . 



A particular integral is found by dividing 

 (KQRS(P + R)-LR(Q + S)}/ by 



P+R 







-R 







R 







G+R 



S 







Q 







-s 







-l 



-1 



-1 



L e. by -(P + R)QS-(Q+S)(GP + GR + PR). If there 

 is to be a continuous balance, we must have c, = c 2 =Cg= 0, 



and L(Q + S) = KQS(P + R). The initial values of z, z, and 



