DIFFERENTIAL EQUATIONS OF THE SECOND ORDER. 27 



and removing factors B and C respectively, we find 



(Discrt. of A) (, - f^.) J = , 

 (Discrt. ofA)(^ y -^)3 = 0. 



Now it bus been seen that not more than one of the three quantities 



can vanish. Consequently 



the Discriminant of A = 0, 



in other words, the equation A = 0, that is 



A/ 3 + Hpq + B</ - Gq - F/> -1- C - 0, 



can be resolved into two equations linear in > and (/. This establishes the 

 proposition. 



But if <, instead of being a function of two arguments and TJ, were a function 

 of only a single argument u, then instead of three equations we infer only one 

 equation of the type 



AM, S + Hu.^ + BM/ + Gu,n, + Fyy'.- + C, c = 0, 



and its resolubility cannot be established. (It is, of course, not the case that it 

 is not resoluble in particular cases ; it is not resoluble in general.) Hence what, (It/ 1 

 equation A = cannot be resolved into two equations, linear in p and q, v infer 

 that the arbitrary functions which occur in the integral equivalent are functions of 

 only a single argument. 



14. In the case when A = is resoluble into two linear equations, and when the 

 other equations possess integrable combinations, a method can be constructed for 

 obtaining those combinations. Thus take the example considered in 11, where the 

 deduction of the combinations is fortuitous in the sense that no indication of the kind 

 of combination is given. Let 



8 (a, b, c,f, (j, h, I, m, n, v, x, >j, z) = 

 be an integrable combination ; that is, we must have 



r-. 'r = - 



ax ay 



Either (i) one of them, or (ii) a linear cross between them, or (iii) both of them, must 

 be satisfied in virtue of the set of equations. 



E 2 



