Linear Differential Equations. 191 



or 



(ir— m,)(7r— »z 2 )w=0, and w = (7r — ??i l )~\'7r— m 2 ) _1 0; 



It will often happen that the conversion of symbols will very 

 greatly facilitate the solution of an equation. Thus, if 



x*D z u + kx' 2 T>u-2u=X, 



change D into x and x into — D, and we have 



DVm + AD 2 «w-2m=X; 

 or by reduction, 



{kx + x <2 )V 2 u+{2k + 4x)~Du = X. 



This may easily be put under the form 



D{ (kx+x 2 ) 2 J)u} = {kx + x*)X, 

 whence 



i^D-'ikx + x^-B-'ikx+x^X. 

 Now changing the symbols back again, we have 



u=x-*(W + k~D)- 2 x- 1 (D* + k'D)X 



for the solution required, the correctness of which is very easily 

 verified. If the last term of the given equation had been 

 — n(n + l)u instead of —2m, it would have been integrable after 

 the conversion of symbols and n — 1 differentiations. 

 Again, let 



x*Wu+{k*x*-2)u=X. 



By the same conversion of symbols as in the last example we 

 have 



DVw + (£ 2 D 2 -2)?<=X, 



and 

 whence 



(* 2 + ar ! )D 2 M + 4dD«=X, 

 D{(/t 2 + a 2 ) 2 Dw} = (* 2 + # 2 )Xj 



i< = D- 1 (A 2 + a' 2 )- 2 D _1 (A 2 + a: 2 )X. 



Change the symbols back again, and we have 



m = ^-'(D 2 + A 2 )~^- 1 (D 2 + F)X, 



which may be easily shown to satisfy the proposed. 



Some! lines the commutation of symbols may be employed with 

 effect in the middle of a solution, or in some of the steps of it; 

 but to give examples would too much enlarge this paper. If the 



last example had been 



.r 2 D 2 u+(^ 2 -w)« = X, 



