352 Mr J. Brill, On Solutions of Differential [May 6, 
2 
Therefore 8B -—a=k log - 
and we obtain 
b? 
Up = a+ (8 —a) log 4 log a. 
Thus the solution of the original problem is 
w= at (8a) log 4" [og fe 
6. We will next take as our bounding curves the two circles 
“+ y’—2he + &=0, 
and a+ y? — Ihe + 8 =0. 
In the transformed problem the equations of the bounding 
curves will be 
sy —h(e#+y)+o =0, 
and ay—k(a#@+y)+o=0. 
Thus we shall have the three equations 
2b = h (a, oP y;) +07 = 0, 
CLs k (a, + Yq) + 6° = 0, 
LY,—h (#,+ Y,) +5 = 0. 
Kliminating «, from the first two of these equations we obtain 
(h ae k) (Ws oa os) ca (his ie 6?) (y, By Ys) = 0) 
Differentiating this we have 
dy, (h—k)y,— (hk — 0°)} + dy, (hk) y, + hh — 8} =0; 
and therefore 
dy, [Ys (y, ae Ys) ae (YYs a }°)} ar dy, (Tk =O) Tr Oily = O°} = 0; 
0.@. dy, (Ys et oe) = dy, (yy 7 5’). 
In a similar manner we can obtain from the second and third 
equations dix, (a, — 8) = da, (#7 — 5’). 
Further, from the first equation we have 
te = 
Sian ao, —h D 
and therefore 
LE SI SOHO) My 
i eT oe naa 
