Linear Differential Equations. 479 



and therefore : " 



Du=.x~''^wa-'X~'^\[x)u. 



Change u into a?D + \{x) in the first member, and into ttu in the 

 second, and we have 



'DxJ)u-\-J)\{x)u=^x~'^Tr^u—x~^\[x)'Tru, 

 or 



xTi% + (\+\{x)^l)u—x-^'Tr%--x-'^\[x)'iru—^{x)u. 

 Whence by eliminating J)u and dividing by Xj there results 

 D^^x-Vw-x-^Q. + %\{x)')'jru-x-^(\[xY+\[x) H-a^X^)). 



In like manner, we should find D%, D^u, &c. We can thus 

 eliminate D, and find a resulting equation containing only ir 

 and Xy which by (A) can be put under the form required. 



Also if 



vsu—Dxu-\-\(J))u, 

 then 



Change u into J)ocu + X(D)w in the first member, and into 'ou in 

 the second, and we have 



a?D^ + ^X(D)w=D"Vw-D'~'X(D)tsrw, 

 or 



and 



A=D-Vm+D-'(1-X(D))^m+D-'(x(D)2-X(D)+DX'(D)), 



Similarly, we should find A, x^u, &c. ; and thus eliminating 

 X, we should have a resulting equation in -dt and D, which by (C) 

 may be put under the form 



^=f{'^)u+f,{'^)J)u+M'u^Wu+ (2) 



But if 



'mu=-xJ)u-\-\{J))Uf 

 change u into D~ V ; then 



'^'Y)~^u-xu + \{'D)'D-\ 

 and 



a?w='crD"^M~X(D)D"^M. 



By continuing to change u into xu^ and eliminating x from 

 the second member, we should have the values of x'^u, a^u, &c. 

 in terms of ct and D. Thus by eliminating x, we should have^ 

 as in the last case, a resulting equation in the form of (2). 



We might have proceeded in the second case with 



a7M = D"'ii7w-D"^X(D)w 

 2K2 



