IN THE DIFFERENTIAL CALCULUS. 51 
2 ax 
where v is found from the solution of the equation of the first order, 
D+2n—2 2 DA = 
a a ee 
This solution is of the ordinary form when x—p is an integer. 
This last form appears to have been overlooked by Professor Boous. 
- n—p—1 
-1 
It may be remarked, that we have omitted (- 5) 0 in the first solution. 
The reason is, that equation (5) being of the second order, will contain two arbi- 
trary constants, and will thus render the solution of the given equation complete, 
without the introduction of any other terms. 
Ex. 5. (a- ar is —(2m+1) ose —(m?—g?)u=0. 
This equation has been solved by Professor Booz ; but it is requisite to shew 
that his solution is not confined to integral values of m. 
The symbolical form of the equation is 
(D+m—2)?—¢? 
DOL 
Let u=e—™ f(—D)v; then 
D—2)?-¢? 
(D—m) (D—m— th D+2)e?4 v=0. 
D(D-1) f Ns 
Dam Dama +9) 
or f(-D)= am 
u— 4u=0. 

f(-D)o- 
By making f(—D)= 
this equation is reduced to 

/-—D+m d m 
and =e7mé =(—1)-™ (4) 
uU=eE ae (-—1) a v 
d\m 
Fa (=) {a cos (¢ sin—!x) + B sin (g sin~! )} 
This equation is, however, susceptible of a solution in a different form from 
that which we have just exhibited. 

Let u=f (-3) v, then 
f (-3) > Oe tO 7 De )eo=0 
VOL. XX. PART I. oO 
