( 649 ) 
From the initial condition, that ¢ being = 0, P too we will be 
= 0, it follows that A = — B, the formula therefore will stand finally : 
B A (et POT apy oa a ae yee 400) 
II. In the case of our making use of a constant stimulus, the 
quantity of stimulation-substance supplied in each particle of time 
will be the same, consequently : 
f@=Cyt and f@=—C, and f'() =O. 
In this case the third term of (5) becomes: = = = C, the for- 
mula therefore standing as follows: 
P = Ac—*t+ Be-#t + C 
in which, keeping in view the individual significance of both 
exponential functions, we must again take B as a negative value, 
thus obtaining finally: 
| BER Bee nde We gn oe (8) 
If once it has been proved that the expression of an effect 
caused by a stimulus may be represented by a differentia! equation 
of the second degree we cannot allow any restriction about the 
coefficients a + 6 and a, as we may easily imagine circumstances, 
occasioning such alterations in the values of these coefficients, as 
would make impossible the solution of (4) in the form of (5). This 
will be the case if we take for our coefficients the arbitrary values 
2q and #2. Anticipating therefore on this right we alluded to before, 
we will now introduce these new constants. The equation in its 
most general form then will stand thus: 
PP 9 dP 2 p= 9 
tear et aa SP (ho see So) 
In this formula we have also put p(t) directly for the stimulus, 
and will distinguish again between the two cases of g(t) =0 and 
g (t) = constant. 
III. In the case of p(t) = 0, (9) will become: 
dt? dt 
If here k? < q?, the solution (6) will follow from this differential 
equation. 
If on the contrary 4? > q?, we may write the solution in this form : 
Bet A cost + Dan at} oo ee (IE) 
As it follows from the initial condition that t being = 0, P= 0, 
we obtain : 
P= Bee? sinatra a tet Mede hel) 
in which: 
43* 
