PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 
=f jawem “f(yt2wanz) 
which is the solution of the equation in the form of a definite integral. 
diz Das 
a aS SS ar, 
dx dy 
The first form of the solution is evidently 
ze D+ePa Fy) 
which is reduced to exe” ® o@ f(y + 2acz) 
Hx. 3. 

—e 2 2 
ie dwe * f(y+2acx+2wa,/z) 
ae pe 
as in Example 2. 
dz d ad az 
tls feed eS p Cn, 
Ex. 4. ea ae dt da 
This equation may be written (d—2ad>5+a? 6?—c*)z=0, 
which is of the form of Ex. 3, Class 1, and the solution is 
gaeiee@eP f2aced f(y) 4 p—2a0nd b (y)} 
=e te ah § Fy 4+ 2acz)+h(y—2acz)} 

= f aee* tf(yt2acat+2wa/x)+p(y—2acx+2waJa)} 
dz ad? di dz 
Ex. 5. de ae dy NS ae 
This equation gives (d—2ad}! 6!+.a?0)z=0 
or z=e™™ Fy) tee"? h(y) (Class 1, Ex. 3. Cor. 1.) 
=f(yt+a@ x)+a (y+a?x) 
5 
Ex. 6. Teta g 7 ae 
Fes +a ek . dy! 2+ dy 
This equation may be written (d+a0! d}+6d0—c)z=0 
which coincides with Ex. 3, Class 1; and the solution is 
amer* f(y) +e (y) 
where a?, @? are the roots of the equation in 7 
v+a 0? vt+b0—c=0: 
2 Teayintal) . 
or A pth a OA (G—2) d+ 
VOL. XVI. PART III. 4c 
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