2. Consider equation (6) in the text. Find out for which initial conditions s (0), s'(0), s'' (0) there is a solution s(t) such that: (a) s(t) is periodic; (6) 8(4) The roots of the characteristic polynomial +48 (3) + 5s (2) + 4s' + 4s = 0. are X4 + 4x³ + 5x² + 4x + 4 -2, Therefore the general solution is -2, i, -i.
2. Consider equation (6) in the text. Find out for which initial conditions s (0), s'(0), s'' (0) there is a solution s(t) such that: (a) s(t) is periodic; (6) 8(4) The roots of the characteristic polynomial +48 (3) + 5s (2) + 4s' + 4s = 0. are X4 + 4x³ + 5x² + 4x + 4 -2, Therefore the general solution is -2, i, -i.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please use these three theories to solve the problem

Transcribed Image Text:2.
et4 is hyperbolic if and only if for each x 0 either
| e¹4x | → ∞
or
| etªx ||
∞
as
as
t → ∞
t
- 00.

Transcribed Image Text:2. Consider equation (6) in the text. Find out for which initial conditions s(0),
s'(0), s'' (0) there is a solution s(t) such that:
(a) s(t) is periodic;
(6)
The roots of the characteristic polynomial
s(4) + 48 (3) + 5s(2) + 4s' + 48 = 0.
are
λ4 + 4λ³ + 5)² + 4x + 4
-2, -2, i,
Therefore the general solution is
—2, i, -i.
(7)
where A, B, C, D are arbitrary constants.
2. (a) In (7), A = B = 0. Hence s(0) = C, s'(0)
8 (3) (0) = D.
-
s(t) Ae-2t Bte-2t + C cost + D sin t,
=
D, 8(²)(0) -C,
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