By Ross Honsberger
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Example text
1) in terms of a first. u2 Exercises s" 1. 20. THE SECOND SOLUTION 2. Simplify this in the case where f = sin t/l/F. 3. for large t, and obtain corresponding expansions for u1 and u2 . Are the expansions convergent ? + 4. What is the eventual behavior of the solution of v' v 2 - 1 -f(t) = 0 with v(0) > 1 iff + 0 as t + 03 ? Hint: Consider the behavior of v in the (v, t)-plane. 20. 1) 0 appears to have a solution that approaches -1 as t + GO, provided that f -+ 0 as t -+ 03. We can establish this indirectly as above by using the properties of u p .
Drawing a figure will convince the reader of this. Hence, the points of intersection of u and v are points of inflection. But again, a simple diagram will show that two points of intersection which are points of inflection must be separated by a point of intersection which is not an inflection point-a contradiction. We will provide an analytic proof below to supplement this intuitive geometric proof. 1,et us next consider the possibility of intersection of u with curves of the form v = c. These intersections must again be points of inflection of u and have one of the four forms shown in Fig.
J" f Z d t < e-1, a weaker estimate ? 6) and iterate, we can obtain a still stronger result. Without loss of generality, take c = 0 since the J,,t eZL1gdt, in most cases of interest. 3) with the property that w -+ 0. 1) FIRST- A N D SECOND-ORDER DIFFERENTIAL E Q U A T I O N S 28 upon using the Cauchy-Schwarz inequality. 2) dt,). 0 1 Since u'/u = 1 w , we see that it is the behavior of Sow dt, which we want in order to estimate that of u(t) as t -+00. 3) Integration by parts yields j t w dt, 0 = 1It l t f dt, e-2(t-ti)f(t,) dt, 2 0 2 0 ~ j" Hence U = 0 ( s t j Zdt,).