Resuming the equation (8), and substituting in it p 2 for 7 



Equations of the Second Order. 39 



is symmetrical about the maximum ordinate b. In short, the form 



of the curve resembles that of a cycloid at a certain distance from 



c 2 

 the axis of abscissae. If we put e 2 for I — 7^, and e 2 sin 2 cj> for 



c 2 

 1 - Hi the equation is transformable into 



(1— e 2 sin 2 </>) 2 



which may be integrated by elliptic- functions. Thus it is always 

 possible to describe the evolute. It remains to be shown that the 

 curve which solves the proposed problem is one of its involutes. 



dd 2 

 dij 



and for y' its value ~(2a—p), we obtain 



(*+*)£ (2.-p)«-V 



« 2 -£$ (2«-p) 2 



As this equation contains p, it is of the second order, and its in? 



tegral will involve two arbitrary constants in addition to k. That 



integral cannot, therefore, satisfy A=0, which is of the second 



order. But it is important to remark that the equation obtained 



by eliminating the three arbitrary constants is satisfied by the 



dA 

 values of q and r deduced respectively from A = and — ^- = 0. 



It is evident that the equation resulting from such elimination 

 is obtained by simply eliminating k by differentiation from the 

 foregoing equation ; which is to be done as follows. 



By solving the equation as a quadratic with respect to k, and 

 putting for the sake of shortness P 2 for 1 + p 2 , it will be seen that 



P 2 , / P 2 a 4 P|U 



*"" Y^\y\2a-pf + 4i , 



which, since P= , ^ . , may be put under the form 



k ~ 2y 2 (p-a) 2 V + \(2a-p) 2 +i J J 

 If this equation be differentiated to get rid of k, the result will 

 be a complicated equation containing y, p 3 p } and ~. Now I 



