DIFFERENTIAL EQUATIONS OF THE SECOND ORDER. 21 



We therefore have two solutions. The first is 



p -q = 0, 

 leading to 



z = function of x + y ; 

 the second is 



P + 1 = 0, 

 leading to 



y function of x + z. 



The other three equations, particular to the equation under consideration, are 

 da^ dg_ dh_ dy <kl . d<l , 2a h g 



dx r dx dx ' *' dy "~ dx T ify + y + s = ". 



^ _ ^ - ' #. ._ (1 L , 4 , 2A - 5 - / _ 2/ - m - M 



die -^ rfo dx, r * dy dc ' ~ dy ' y + z (y + z)- ' 



dx dx d.v * dy dx di/ i/ + z (// + ; p ) 3 



that is, 



' * ^ ' ~^T^ =' 



2h. -I - f 21 - m ' 



dy) 



m 



, ^ tf = ' 



Taking first the case 



p + 1 - 0, 

 we have 



But 



2 _ h - ff - 2g +/+ c = ( j + p ^) (21 - m - n) = ^ (2l - m - n), 



and 



d 1 



dxy.+ z~ (y + 0) 3 (y + z)* ' 



so that 



d . , ,. d /2l m n\ 



-7- (a h g + /) + -T- I - - ) s= 0. 



fte v ^ \ y + z I 



We thus recover the differential equation, which is an integral of the system ; the 



