SATISFYING LINEAR DIFFERENTIAL EQUATIONS. 29 



o (tv+<r) = n (w) n [rr 1 (w) o-n (w)] 



resolves the required matrix into a product of two, of which the first, 11 (w), consists 

 simply of diagonal terms of which the r th is 



The second matrix has zero everywhere in and to the left of the diagonal ; and, 

 since in the matrices ^ the element in the r th row and s th column was zero unless 

 m = Q m ' '' (m = p+l, %), it contains no exponential expressions. It can therefore be 

 completely integrated in finite terms, just as was done in the case p = 1 (p. 23). 



15. A simple example may be appended of the application of the method to a particular system. 

 Consider the equation 



Putting iji - y, y, = y?y\ we have the linear system 



y'=t o, 



= 17 l }L+( Q 'U+/ \~L 



The characteristic equation is 



-0 or 



-P, -S-p 



giving equal roots - 1 for p. 



With p = ( j the transformed system becomes 



OM./O . 



= uy say. 

 Considering now the subsidiary equations 



Then the equations to be satisfied are 



II. (1) xj - (1 + #1) *i l = '2) (2 - 61) x -1=0, 



I. and II. (2) give (6!- 1) 2 = 0. 



We take 9\ 1 therefore, so that a^ 1 = 1 and x-2 = 2xi l . 



Again 



III. (1) xo? - &i zj 2 - 0, (2) (1 - 6>,) xj - a, 1 = 0, 



of which (2) gives Xi l = 0, so that x = 0. 



