240 Mr. Greatheed's ?iew Method of 



My method, then, for solving the proposed partial differ- 

 ential equation is this : Substitute -7— for n 9 or let n repre- 

 sent -7-, and proceed exactly as in the above solution of the 



equation between two variables, substituting for the arbitrary 

 constant an arbitrary function of y. The result then will be 



^ \ — 1 -bx d 



+ e d - a yf{ay). 



_ 4 (A) 



b \dy) 



d 



■bx 



By the proposition which was premised, s d ' ay f { a y) 

 is equivalent to f (a y— b x). Also, since integration is the re- 

 verse operation to differentiation, / — — J is equivalent to 

 f dy = y. Hence, finally, 



* = C ~f + f(ay-bx) 9 



which is the correct solution of the proposed equation. I shall 

 proceed to explain the meaning, and prove the legitimacy, of 

 the several steps in the solution. 



bx d 



In the first place, multiplying by the factor s a d y * is 

 equivalent to changing all functions of y to which it is prefixed 



bx 



into the same functions of it -\ • Now % is an unknown 



r a 



function of x and y, suppose it equal to $ (#,3/), then 



bx d j 7 b x d j < 



-— d z b — ■—- a z 



S a dy — — -U — g a dy _- — 



d x a ay 



is equivalent to 



d f bx~\ b d f bx \ 



— | \x 9 y + -j + - -^{x 9 y + _ j, 



which is the total differential coefficient of f < x 9 y H I 



with respect to x. 



As for the other side of the equation, namely, 



bx d 

 g a dy . _ 



a 



bs d 



•The more proper expression would be "prefixing the symbol e a d »'\ 

 but since it is convenient, and the operations are exactly analogous, I shall 

 use the term u multiplying." 



