Full text: Advanced calculus

350 
CALCULUS 
Let us study the differential equation from the point of view of 
§ 15. Consider the two differential equations 
It is clear that any solution of (i) in a given interval a ^ x b is a 
solution of (1); and the same remark applies to (if). 
Equation (i) assigns to each 
point (x, y) within the strip 
— 1 S V =i 1 °f the (x, y)-plane a 
positive slope, and to every 
point on the boundary (y = 1, 
Fig. 84 
y — — 1) of the strip the slope 0. If (x 0 , y 0 ) be an arbitrary interior 
point of the strip, there passes through it one and only one solution, 
and to the determination of this solution the analysis above applies: 
v 
A 
B 
Fig. 85 
x 0 = - COS- 1 y 0 + c, 
X = — COS- 1 y + cos-iyo + x 0 , 
(3) 
where each cos -1 means the principal value of the function (Intro 
duction to the Calculus, Chap. VIII, p. 211). 
The solution (3) proceeds forward till it meets the line y = 1 at 
A ; and it runs backward till it touches the line y = — 1 at B. The 
continuation of this solution to the right of A is the function 
or the part of the upper boundary of the strip to the right of A. 
And likewise the solution is carried backward, to the left of B, by 
the function y = — 1, or the part of the lower boundary of the 
strip to the left of B. 
We see, then, that through every interior point of the strip passes 
one and only one integral curve of (i), and this solution is defined 
throughout the whole range of values for x, — oo < x < + oo. 
Not so, however, with a point (ic 0 , y 0 ) on the boundary of the strip. 
Suppose y 0 — — 1. Then that part of any solution passing through 
{ X Q> Vo) which lies to the left of x 0 is uniquely determined; it is 
y = — 1- But to the right we may proceed along this same line for
	        
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