24,2 Mr. Babbage’s essay towards 
but the general solution of such questions is by no means 
easy. 
Problem XXXVII. 
Required the nature of the function -ty such that 
J'doc \J/x = ipa 
the integral being taken between the limits x=o and x=a. 
Assume q> (x, v) such that 
<p (x, v) — <p ( 0 , v) = V 
the form of tp may be ascertained from this equation by means 
already described. Then if we make 
J'dx iJAr = (p (x, -tya) ( 1 ) 
it is evident that between the two limits x — o and x — a, the 
integral will be reduced to -tya, and we have therefore a diffe- 
rential functional equation whose mode of solution has already 
been pointed out. Other more complicated equations may be 
solved in the same way ; these I shall omit. I shall, however, 
make some observations on this method of solution, with a 
view to point out some questions of considerable importance. 
In equation (l) the function indicated by <p is so assumed 
that we may have 
(p (a, ypa) — (p (o, $a) = \pa 
from which, perhaps, it might be imagined, that p (x, \pa) 
must contain only x, \pa and constant quantities, but the con- 
dition would still be fulfilled if it contained y a , -fy 3 a, or i p n a, 
which though not actually variable cannot strictly be regarded 
as constant. To fix our ideas, let us consider the example in 
this Problem ; one value of <p (x, ypa) is evidently p (x, ia) = 
— $a, we have therefore 
Jdx 4/'x=±i/a 
