( 538 ) 
which y may be represented by a continuous function of z. I shall 
call such solutions function-solutions. 
I shall demonstrate the existence of such a function-solution by 
means of a very simple example. 
Let us take as being given: 
#=Pyy+Qe and y =Poyt+ Qe. 
If we represent the function-solution of y by w(z), we have: 
g=w(e) or ylPv@)+Q2]=Py(e)+ Qe. 
This equation must be fulfilled identically. It is clear that we 
may do so by putting: 
wle =ar. 
Then : 
Pja+ Qja= Pya Qs 
or 
P, a’ + (Q — Po)a — Q = 0 
Py — Q) 1 — 
Nd LN ON ee AS 
ss oF 2P, We Qi)? + 4 P; Q 
S 4. We shall now pass to the discussion of differential equations. 
As example we shall take a single differential equation (and not 
a set of simultaneous differential equations) of the first order ; while 
no differential coefficients occur in the equation, which indicates in 
what way the value of 2’ is found. 
The equation to be solved is then: 
while: 
ran eer 
is also given. 
If we take an arbitrary value 7; and put as a solution which 
only holds for a part: 
y= we), 
we may calculate again : 
