354 Mr J. Brill, On Solutions of Differential [May 6, 
happens because du/dy and du/dx are rational functions of y and 
respectively ; but in cases where irrational functions turn up, it 
will be found that the work is in general much more difficult. We 
have chosen these two cases because they work out quite comfort- 
ably to the very end, and therefore serve as excellent examples 
of the complete process. We will illustrate the cases in which 
irrational functions turn up with the aid of some new examples, 
which we will not trouble to work beyond the point at which we 
find the expressions for du/dy and 0u/d# along one of the bounding 
curves. 
We will now work out the case in which the boundaries consist 
of the axis of y and the ellipse 
2 2 
= tase 1. 
In the transformed problem the equations of the bounding 
curves will be e+y=0, 
and (2° + y’) (a? — b) — 2ay (a + 0’) + 40°D? = 0; 
and we therefore have the three equations 
“+ y,=9, 
(a7 + y,°) (a — b*) — 2a,y, (a + b’) + 4076? = 0, 
L, + Yy,= 0. 
If we eliminate 2, from the first two of these equations, we obtain 
(y+ Y,) (a — b°) + 2y,y, (a? + 6”) + 40°7b’ =0; 
and differentiating this we have 
ty, (a* — B°) + y,(a° + 6%) } dy, + iy, (@ + 6") + y,(a" — b’)} dy, = 0. 
And if we now write 
es ecm 2. OS Gime 
y,(a" — b*) + y,(a? + 6")? y,(a" + b*) + y,(a* — b*)’ 
this equation becomes dy,/v = dy,/w. 
If we now eliminate y, between the equation giving v and that 
connecting y, with y,, also y, between the equation giving w and 
that connecting y, with y,, we obtain the two equations 
(CLS —y2(@ VY} -20 (yet + +5 ye +0) =0, 
w (a + DF —y,(@ DY} — Qu (ys ta +0) + Bys + B)=0. 
