1054 HON. LORD M'LAREN ON SYSTEMS OF 



homogeneous part. The arbitrary term may either be treated as a coefficient or reduced 

 to unity by division, and in the auxiliary equation it is replaced by % n in order to form 

 a homogeneous function equated to zero. 



The method of this section is essentially the same as is implied in the following : — 



A solution of the equation 



x 5 + 6x*y + 3x i + 5xy z + 2x s + 4h/ 3 = 

 is 



_ a 

 b 

 where K is a root of the equation 



(a 5 + 6a 4 &) + (3a 4 + 5a6 3 )R + (2a 3 + 4& 3 )R 2 = . 



For on substituting we have the left-hand side 



_ a 5 6a 4 6 3a 4 5ab 3 2a 3 46 3 

 ~R 5+ R 5 + R 4+ ^ 4_ + R3 + R3 



_ a 5 + 6a 4 6 + (3a 4 +5a& 3 )R+ (2a 3 +4& 3 )R 2 

 R 5 



~R 5 



= 0, as it should be. 

 And generally — 

 If (p n (x, y) denote a homogeneous equation of the n th degree, the equation 



<t>n{x, y)+<p n -i(x, y)+ • • • +4>i{x, y)+<po(x,y)=0 



has for solution 



a 



x =n 



b 



where R is a root of the equation 



0„(a, b) + 'R ( p n _ 1 (a, b) + W<p n _ 2 (a, b)+ . . . +R"- 1 1 (a, b) + BT<p (a > b)=--0. 



In the case of a heterogeneous function of three variables, there is a choice of four 

 solutions. (1) A quantity P may be introduced which will make the equation homo- 

 geneous as a whole ; and the equation may then be solved for P by treating the arbi- 

 trarily assumed quantities, £ rj, £, as known quantities. Then, dividing by P", values 

 of x, y, z are found for P= 1. (2, 3, and 4) Any one of the quantities (say z) may be 

 assumed as constant during the operation, or z = a and a quantity P is then to be intro- 



