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tuting # by « in y= f(e). Now we must also know in what con- 
nection 2’ is with the other quantities. In general this is done by 
giving : 
Fi(y, 9, 2, =%. 
The problem is therefore reduced to the solution of: 
F(y', y, #, 2} =0 and Fi ty, Hye, 2) = 9 
where we must take into account, that the dependence of y' on «' 
and that of y on « is expressed by the same function. 
From these two equations we may think y' aud #' as being solved: 
e= 71 (2, 9) y = Xe (2, y)- 
Let us now assume an arbitrary value r= 2, and a perfectly 
arbitrary function y= wj(e). Let us now calculate y; = wi (*) and 
v= 71 (*1 4). Then we can represent the quantity y between the 
limits 2; and 2 by the perfectly arbitrary function. If we substitute 
in 7, (vy) and zz (ey) the value y («) for y, we get # and y' both 
expressed in z: 
a! = 7) (2) y= Zr (2) 
By eliminating « from these two equations we get y = Wz (2'). 
If we then determine: 
Yo = Wo(*2) and 23 = Zi (%2 Ys) 
we may represent the value y by we (x), when « ranges between the limits 
za and 73. In exactly the same way we may determine functions 
ws (x), wa(e) ete, which will respectively be a solution for z be- 
tween zz and «4, 2, and «; ete. With the aid of the integrals of 
Fourier we may write the solution as follows: 
ehh 
e= ah js sin (ux) sin (vu) yi (v) dv du + 
Xi 0 
En ic 
+ f ii sin (ux) sin (vu) Wo (w) dv du + etc. 
2 0 
