INTEGRATION OF DIFFERENTIAL EQUATIONS. Ill 

 Therefore as before, 



and a n = m~ p N 



tn-vi + l \ 

 \ m J 



Here we make f(k) Jc p , and <^> (Jc) 



m 



m-i 



no complementary function being added. 

 Hence, precisely as before, we find that 

 _ d~ p X 



V JT.-P m-l 



tf* A;- 



is the solution required. 



Let us now consider the more general equation, 



If 3^ = S . a n a? n , there will be 

 n... (n s + l)(nsj}m) (n s 1)... 



a n +^a n _ m = ...... (7). 



This equation is analogous to (2); but m-l is replaced 

 by s. Assume, therefore, 



5 - (^ - 1) m}...(n - s + m) Jc~ p b n _ mJ 

 and n...(n-s + l)(n-s)(n-s-l)...(n-m+ l)b n + kb n _ m = 0. 

 Hence, as before, 



2b n x n = (k) X, 



where X denotes the same function of x that it did in the former 



case. 



2 

 Consequently b n = ^V<j> (A;) k m , and 



