814 PROFESSOR CHRYSTAL ON THE 



Now, by (12), since (15) has a singular solution we must have 



ac + c 2 -4Z> = 0. 



Hence 



2clu d£ _ 



The integral is therefore 



(u±af£=A, 

 or 



a%dz »/( — acx 2 — 4ay) = B , 

 the rational form of which is 



(ax-Bf+a(4y+cx 2 ) = Q (30), 



representing a family of parabolae the envelope of which is obviously the parabola 



4y+cx 2 =0. (31). 



To find the Condition that the Equation (15) may have an Algebraic Integral 



(«4:0, b^O, c= t=0). 

 Proceeding as before, we reduce the equation to the form 



2udu cl£ . 

 uP+au—f f 

 where 



(32), 



f=c 2 +ac- i (33). 



Let 



u 2 + cm -/= (u — X)(u — fx) . 

 Then 



2u 2 / X ^ \ 



u 2 +au—f A-ywU-X w-/J' 

 and the integral of (6) is 



x ^_log( % -X)- x ^ w log( tt - M ) + log^=C . . . (34). 

 If 



j8 = o? + 4/= a 2 + 4c 2 + 4ac - 1 66 , 



= (a+2c) 2 -16&, ....... (35), 



then 



2\=-a+JJ, 2v.= -a-jJ; 



and we get for the primitive 



(^+0W( M "- x )-(7|- 1 ) log(tt " M) ' Flog ^ =C; 



