PARTIAL DIFFERENTIAL EQUATIONS. 
187 
§ 39. II. As a second example take the equation 
F (r, s, t, p — siy, q — ty, z — qy + ^ ty 2 , x ) = 0. 
Here the third row of the array is 
d(s,t), 0,0, 
so that the functions s, t satisfy the auxiliary equations. Put, then, s = a, t = b ; 
thus 
q = ax + by + c 
z = axy + i by* + cy + X, 
the last term being a function of x only. The differential equation thus becomes 
F (d 2 X/c/x 2 , a, b, dX./dx, ax -f c, X, x) = 0, 
an ordinary equation of the second order giving X in terms of x and two more 
arbitrary constants ; hence the finding of a complete primitive is reduced to the solu¬ 
tion of the equation last written. 
§ 40. III. If the equation is of the particular form F(r, s, t, — rx — sy,q — sx — ty, 
z — px — qy^rx 1 sxy-\-^ty~) = 0, the first three rows of the array are 
d (r, s ) 0 0 
d (r, t) 0 0 
d (.?, t) 0 0. 
Hence any two of the three functions r, s, t will satisfy the auxiliary equations, 
and a complete primitive is given by putting 
r — a, s = h, t — b. 
Hence p = ax-\-hy-\-g, q = hx-\-by-\-f 
z = c + gx+fy + l ( ax 3 + 2 hxy + % 2 ), 
where a, b, c, f, g, h are constants satisfying the relation 
F (a, h, b, g, f c) = 0. 
This is a case in which other solutions are readily given by supposing the para¬ 
meters variable and functions of one variable only, say a. The variations must 
satisfy the conditions 
x 2 da -f- 2 xy dh + y-db + 2x dg + 2y df + 2 dc = 0, 
x da + y dh + dg = 0, x dh + y db df — 0, 
