612 Mr. DE morgan, on METHODS OF INTEGRATING ' 



or X = (x - b)(y - a), a very obvious solution. Hence we must eliminate a from 

 sr = {x-(l>a){y - a), = (x - (pa) + (y - a) <p'a. 



In my Differential Calculus, p. 717, I gave a method for the general case z = (b {p, q) ; but 

 the following, derived from the present method, is preferable. 



Let f(X, a) =f(p(X,aX). X-^ dX. 



Then Z = Av(.r,l] + ^FQ, 



with which proceed as before. 



Of the instances which I have tried by the other method, 



pq = px + qy gives p.r + qy = xy, from which 





In this case we may conveniently take the retransformed equations 



q = ap, px + qy — s: = -^ p(] + b, which with pq = px + qy, 



give 2a (z + b) = (x + ay)', say 2ax + b = (x + ay)'. 



Again, (p + q) {px + qy) = 1 transforms into 



1 



■y 



{x + y) (px + qy) = 1, or sr + ---- = f l^ 



X + y \x 



Treat this by the method, and assume fz = + b, which will show that the general 



solution can be obtained by eliminating a between 



X — V ay 



«= -+ — ^+\l/a, 



a 'f - y 



X ~ y y 



= ~ + — i — + xf, a. 



a- X — y ' 



The equation ^p + ^q - 2x transforms into 



.y/a; + ^y = 2p, or sr = 1 xi + 1 xyi +fy, 

 X = i Xi + i YK 

 y = iXV-i+fV, 

 z = ^Xi-lXYi + Yf'y-fV. 

 This is not an easy form. But if we take the retransformed equations • 

 q = a?, px + qy - z == ^ pi + :L pqi + ft^ 

 and join pi + qh = g.r with them, we find 



z = ^ (2 X — aY + a^ y + b, a primary solution, being the one already obtained. 



