LINEAR DIFFERENTIAL EQUATIONS. 
43 
and the solution is 
}, 
where y) means '^{x, y — 'k^x). 
Transforming- this equation to new independent valuables p and q, by the equations 
p-=y — 'kY^-\-TiX, —q=y — \x—hx, 
or 
we have, as a soluble form, 
d^u 1 /jo + g-x (du du\ , / 1 / 9 I 07 t m{m—l)\ 
pg + 2A^V^;- V^+5^; + \4k\^^ + ^^\d^'^d^) ) - J^+qf )^ = ^(P’ 9)’ 
the function X disappearing in the process. 
U (xx be of the form we obtain a well-known equation, solved by Euler in a 
series when there is no second term, 
d^ a /du du\ a{a—l)—m{m—l) 
dpdq'p + q\dp~'dpj' 
which is therefore soluble ; the solution being 
.r’”-''(D2— D'2)”*-‘{a?-’(D2— D'2)-”{^“-“+*9^(^, y)}}^ 
where is determined from <p by the equations 
pzz=x-\-y and q=x—y. 
lfa=0, we have the solution of 
d% m{m — 1 ) 
i ) — nvyuL — A I . . 
?)> 
=Hp, q)^ 
dpdq {p-\-qf 
and if a=m, we have the solution of 
d'^u m /du du\ 
^'^pTq\^'^'^) 5) ; 
from which the solution of 
d^ 
=:c2^ 2771-1 
d% 
If 
may be obtained by making 
<p=0 
s=[p-\-q)~^’^'^^ 
2m-l 
Returning to the general form and making X®=0, we have for the solution of 
d^U Tcyd’^U 
dx^ dip 
G 2 
