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Unformatted text preview: 124 Chapter 20. Solutions: Solution of Ordinary Diﬀerential Equations CHALLENGE 20.15.
• Using the notation of the pointer, we let a(t) = 1 > 0, b(t) = 8.125π cot((1 +
t)π/8), c(t) = π 2 > 0, and f (t) = −3π 2 . These are all smooth functions on
[0, 1].
• Since c(t) = π 2 /2 > 0 and
1 [f (t)]2 dt = 0 π4
,
4 the solution exists and is unique.
• Since f (t) < 0, the Maximum Principle tells us that
max u(t) ≤ max(−2.0761, −2.2929, 0) = 0. t∈[0,1] • Letting v (t) = −3, we see that
−v (t) + 8.125π cot((1 + t)π/8)v (t) + π 2 v (t) = −3π 2
and v (0) = v (1) = −3. Therefore the Monotonicity Theorem says that u(t) ≥
v (t) for t ∈ [0, 1].
• Therefore we conclude −3 ≤ u(t) ≤ 0 for t ∈ [0, 1].
Note on how I constructed the problem: The true solution to the problem is u(t) = cos((1 + t)π/8) − 3, which does indeed have the properties we proved
about it. But we can obtain a lot of information about the solution (as illustrated
in this problem) without ever evaluating it! CHALLENGE 20.16. Let y(1) (t) = a(t), y(2) (t) = a (t). Then our system is
y (t) = y(2) (t)
2
y(1) (t) − 5y(2) (t) . function [t,y,z] = solvebvp()
z = fzero(@fvalue,2);
% {\tt now the solution can be obtained by using ode45 with
% {\tt initial conditions [5,z].}}\STATE{\% {\tt For example,
[t,y] = ode45(@yprime,[0:.1:1],[5 z]);
% End of solvebvp ...
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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