138 
PROF. H. F. BAKER ON CERTAIN LINEAR 
In these equations, as u r , v s are to be polynomials in £, £ -1 , the absolute terms, 
those involving £°, must vanish. For r — 1 this gives 
h 1 — q 1 = JcJS, h 1 + q 1 — ^B -1 . 
We thus write, using hyperbolic functions, 
/q = k lC ha, q 1 — kyshct, B = e -a . 
With these we find at once by integration the values of tq, v u save for the absolute 
terms in these, which we denote by P^ Q x respectively. The conditions for these are 
to be found by considering the absolute terms in the equations for r = 2 ; and so on 
continually. In general, when we have found 
Mj, Vi, U 2 , V 2 , •••, U r -li ^r-15 
and have found u r , v r , save for their absolute terms, P r , Q r , we find, on taking the 
absolute terms in the equations which involve du r+1 /dr and dv r+1 /dT, and adding and 
subtracting these terms, that the two quantities 
Jcysha (P r -Q r )-h r+1 , kychu (P r -Q r )-g r+ i 
are thereby expressed in terms of known quantities. It is at once seen that there 
would be no loss of generality in putting P 1} P 2 , P 3 , ... all zero. Carrying out the 
work, and writing M r for P r —Q r , we obtain 
q = k 1 sha.\ + (M. 1 k 1 cha.—k 2 sh2oi)\ 2 
+ {|-M 1 2 ^ 1 e a — 2M. 1 k 2 ch2a. + k 1 3 sha (k>ch 2 a—^) + k 1 k 2 shu. + k x chcx. (M 2 —MAi)} A 3 + ..., 
where 
/q = k x cha, 
h 2 — M. 1 k 1 shot—^k 1 2 ch2a., 
h 3 — ^M. 2 k x e a — M 1 k 2 sh2a + k x chcn (2 sh 2 a—\) + kjc 2 cha + kysha. (M 2 —MjPi)- 
Also 
—ixe l(1+2q)t = 1 — e"t+XWj + X 2 W 2 + ... 
in which 
Wi = _1 + P x — k x sha. + (—?! + Mi — kjsha.) e~ a £—^k 1 e~ a £ 2 , 
w 2 = K- 2 (^+R 2 )+r i [iPA-fe-‘+^ s (i«" , -*A«)] 
+ P 2 —P x k x sha —M -Jc^cha. + k 2 sha ( cha+e a ) 
— [P 2 — Ma + P^sAa + M^e -0 — k 2 sha. (c/ia + e _a )] 
+1 [ ■-l-PA + JMA + \k 2 e a - k 2 (sha + ie a )] 
-We- a (k 2 + W)' 
