DIFFERENTIAL EQUATIONS OF THE SECOND ORDER, 

 and consequently there exists a function, say 4> (x -f y, z), such that 



15 



and therefore we have 



m n = <E> (cc + y, z), 



where * is an arbitrary function. We might proceed from this equation to the 

 primitive. 



Next, take the relation 



2 + 1 = 0, 



deduced from the characteristic ; this shows that z + y is a function of x. The three 

 deduced equations now are 



dh ! dh dy dg 



_L db _L ^ _L 4 



df df dc dc 



dx dy dx dy 



= 



J 



and inserting the value q = 1 in the three relations of identity, they become 



dh da dg dg ~\ 



dx dy dy dx 



. nA > 



dx dy " P dy ilr, '' 



Combining these, we find 



df dg dc dc 



dx dy dy dx 



dh da^ 

 dy dy 



db dh^ 

 Ty " 'dy 



dy dy 

 so that h a, b h, f g, are functions of x. Hence, as we must have 



f- g = b- h 

 by the original equation we take for an integral equation, as in the former case, 



a h G 2 (x, y + z), 



