526 Mr DE MORGAN, ON SOME POINTS 



\|^, = eo , or \J/ XX = oo , this is obvious. If \|/ and >|/> be finite, and \|/ a or \|/ 6 infinite, the pro- 

 ducing relation can only be (a, b) = 0. If \|/, be finite and \|/„ infinite, either \^ M = oo , or 

 (a, 6) = 0. If yjr aix be infinite and not <\r xx , either yp xxx =co , or (a, 6) = 0. So that Xy =0 ° 

 may either not show certain intraneous singular solutions, or may show intraneous solutions 

 which are not singular : but it must contain all extraneous solutions ; and every case of it 

 which makes ^ x + y y y '+ ^.^ finite or zero must be singular. 



M. Cauchy's theorem may be extended to biordinal equations, so far as regards a case of 

 the primitive, by a proof which may be copied from that of the case already given, and need 

 not be inserted here. If y = P be a solution of y" = ■% {oo, y, y), P being a function of w, the 

 solution is extraneous or intraneous according as f dz: ^ ^ (a?, P, P x + ») - ^ (x, P, P x )} begins 

 finite or infinite, z being the only variable. 



Returning to the criterion, let elimination of a and b from y = >J/, y = \|/>, and 

 ^ai'ta^ ^b^a* = ° g* ve V — w (•» y)t tne singular primordinal : we also obtain a and b each in 

 terms of oo and y. These values of a, b, y', substituted in y = \J/, y = df„ make them iden- 

 tical; whence 



If either or both Z? y , = oo , b = constant, give \|a = 0, we have, if \}/ a in this case undergo 

 substitution for oo, ■ur y = co: that is, attempts at solution made through >^ a =0, give either 

 singular primitives of the singular primordinal, or the failures, already noticed, in which y is 

 infinite. And \j, a =0, \^ 6 = 0, combined with y = \|/, give either the singular primitive of the 

 singular primordinal, or the failure adverted to. 



The following question may be asked, and at first presents some difficulty: — Since 

 dA : A y ,= (y"— ^dx is an absolute identity, how is it that some solutions of A y , = oo , the 

 ordinary ones, give y" — ^ = 0, and others, the singular ones, do not? What can it matter 

 to the deduction of y'=x fr° m dA.as = {y"-^)dx, whether the symbol oo arose in one 

 way, or another? The answer is, that independently of our not being sure that Ay, = os gives 

 y" = y, it may happen that a differential relation the ordinary solutions of which make A y , 

 infinite, may have singular solutions which do not make A , infinite. Let ^=00 be produced 

 by v(x, y, y) = 0, and by help of v = v {oc, y, y) throw A into the form V {oo, y, v). We 

 have then A y = Vjo^ . Now, v = giving V v = oo , gives A y , = cc , except only when accom- 

 panied by v y ,= 0. And v = is precisely the condition under which the singular solutions of 

 v = must be looked for. 



Again, if v = satisfy y" = ■%, it means that u = makes »„+ v y y'+ v y ,y "= identical 

 with y" = %. In the case of v y , = 0, this identity can no longer be affirmed ; here v = makes 

 v x + v y y' vanish, and y" can only be determined by another differentiation. 



We may now prove the assumption on which the ordinary view of the theory is founded. 

 The coexistence of y = ^, y = \f, x demands that a and 6 shall either be constants, or compen- 

 satory; as in \js a da + \j/ b db = 0. Now ^ a ^ hx - ^(,^« = 0, except when satisfied by ^ =0, 



