Equations by Factors and Differentiation. 393 



F being substituted for y— p + -p^. Hence, putting for brevity 



p3 v p p4 



R for a + = ^r- and Q for — , we shall have 



2q 2 q 



h=R±(EQ / + B?)*. 



Hence, differentiating to get rid of A, 



Consequently, by squaring to remove the radical, 



This is the differential equation it was required to find; and 



dF 



since, as is readily seen, it is satisfied if both F = and -y— =0, 



we may infer, just as in the preceding problem, that the complete 

 integral of the equation 



a ayq 



= ?/- 



y/l+p* (1+jp 8 )* 



embraces all curves parallel to a certain curve described by the 

 focus of an hyperbola rolling on a straight line. It is plain that 

 the curve itself must be included among those embraced by the 

 integral. 



The same result may be obtained by conducting the argument 

 in a somewhat different manner, as follows. If instead of substi- 

 tuting F, R, and Q for the respective functions y— p + -p^f, 



P 3 vV P 4 



a + F, ^r, and — , we had operated with the functions them- 



2q 2 q r 



selves, in place of the equation (e) we should have obtained 

 4« 2 ((P P -^)? 2 -y : ) S =P^(P 2 + ?J/) 3 (3i>J 2 -PV). . ( V ) 



Let us now suppose that F = y— ~ + -^f. Then from this 

 equality and its first derived equation it will be found that 



P 3 F P 3 P 2 



q ~ ay a y' 



" ay dx ay 2 ay y 2 P 2 



By substituting these values of q and r in the equation (rj), and 



