FUNCTIONAL DIFFERENTIAL EQUATIONS. 139 



Let us suppose Pz and P^z constant ; 



x = I + a.z ; 



therefore y*=2ax + C. 

 On substitution we find 



Thus, in order to a real result, we must suppose a negative, 

 e.g. let a = & 2 ; then 



(7 ....................... (18), 



the equation to a parabola, which accordingly is a solution of 

 the problem, and the only simple one it admits of. 



When a = 0, it becomes two straight lines parallel to the 

 axis. 



The other factor 1 + ^"x = gives, on integration, 



..................... (19), 



the equation to a circle ; but on substitution, we find 



Ja 2 = Ja 2 - a, 

 which leads to no result, unless a = 0. 



A solution of this problem, by Poisson, is given at p. 591 

 of the last volume of Lacroix's great work. It is apparently 

 equivalent in point of generality to (17) ; and the author 

 points out its incompleteness in the case of a = 0. The pre- 

 ceding views show distinctly the nature and origin of the new ' 

 solution which then presents itself. Mr Babbage also has 

 considered this problem at the end of his second essay on the 

 Calculus of Functions (vide Phil. Trans. 1816, p. 253). But 

 I believe it will be found that his solution is erroneous. 



Notwithstanding the length this paper has already reached, 

 I must endeavour to point out, as briefly as possible, my reasons 

 for thinking so. 



Mr Babbage confines himself to the case of a = 0. He 

 begins by demonstrating the existence of a relation, equivalent, 

 excepting a difference of notation, to ty [x + ^ifr'x] = ty'x, but in 

 doing this, loses sight of the other factor 1 -f ffl'x 0. 



