TKANS ACTIONS OF SECTION A. 617 



7. On the Logarithmic Case in Linear Differential Equations, 

 By Professor A. C. Dixon, Sc.D., F.B.S. 



If X is a linear difterential operator whose coefficients are functions of the 

 independent variable x it is commonly possible to solve the equation 



Xy = (1) 



by finding a function (j){m, x) of x and a parameter m, such that 



X(f){m,x)=f(in).y]f{m,.v) (U) 



where >/r is a function that is finite in general, and then choosing m so that 

 f(m) = and ^(m, x) is not identically 0. 



When the equation /(m) =0 has a repeated root a solution may generally bo 

 derived by differentiating ^ with respect to in, and it is my present object to 

 justify this proceeding in a very simple way. 



Differentiate (2) with respect to in ; then, since m does not appear in X, 



SO that if f(m), f(m) are zero 



^ cpim, X) 



is a solution of (1), as well as (/)(»i, x). 



The simplest case is v/hen X has constant coefficients and may be written 

 F(D), Then 



F(D)e'"' 5:/(w2)e'«' 



and by differentiating as to m 



F(D).re"''- = /'(m) , c"''' +/(>«) . .re"'-^' 



80 that e^"', xe"'-^ are both solutions of r(D)y = if m is a repeq,ted root of 

 /•(m)=0. 



In the very general case when 



X=Fo(.i'D) + .i'Fj(jjD) + .r2'F„(,rl))+ . , , 

 the procedure is to form the series 



oo 

 71=0 



where A,, = 1, and 



A,,F„(m + A.T) + A/,_iFi(»j + Z;>--r)+ ... =0 (A; = l, 2, . . .) 



so that each coefficient is a definite algebraic function of m, and we then havo 



X^(m, .r) = F(,(>«) . A"' 



the solution being completed by solving the equation Fo(»?.) = 0. 

 Differentiating as to m we have 



^ 3 



X g^^ ^(»i, .r) = Fo'(»0 . .i'" + F,Xto) . .(."'log.r. 



And thus if m is a repeated root of Fo(«0 = 

 are both solutiops of (1), 



rf) and rr— 

 dm 



