SATISFYING LINEAR DIFFERENTIAL EQUATIONS. 7 



The remaining equations of this group give x 3 except for x 3 \ in terms of x^ and x, 1 . 



Proceeding in this way as far as the (p+l) th equation of X, x^...x p are all found 

 except for their first elements, while the first elements of these equations give 

 6 l pJrl ...0i, l p+r being not a priori independent of x^...x r l hitherto unknown. 



It has been shown above, however, that the determination of 6 l p ^ 1 does not require 

 a knowledge of x. 



In general, in fact, () l p - r is given independently of x^.-.x, 1 . 



To prove this, the way in which x^ enters into .x-* r+1 will be first considered. This 

 may be stated .as follows : 



The coefficient of x in x\ +l is equal to that part of x r k which is independent of 



For r = 1 this is at once seen by writing down the equations 



In general the equation for .r A r+1 is 



Assuming the statement above to be true for 1, 2...r, and that 9 V , ..., p - r+ i are 

 independent of x 1 l ...x l r - l , the above equation shows that the coefficient of x^ in 

 (pt pi)x k r + l is the part independent of a?! 1 ,..., ,r r l in 



r I 



, V /-,!* .''j- - r, 



li i \ a p-r-n+l^j T...-rt* p _ r+< + iX s j-, 



i.e., u 



so that under the above assumptions the statement holds for 1, 2..., r+1. 

 Also, under the same assumptions, from the equation giving l ; ,- r , viz., 



we deduce that the coefficient of x/ in ^p-,. is the part independent of x^ in 



This expression differs only from the left-hand member of the equation for 6 l p - r+ i 

 by multiples of x^^.x 1 ^^ and therefore, on the assumption that this equation gives 

 ^P-! independently of a?/..., the part independent of these quantities in the above 

 expression must, when 6 l p - r +i is determined, vanish, so that 6 l p - r is independent of x, 1 . 



