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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please use these three theories to solve the problem

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. et4 is hyperbolic if and only if for each x 0 either | e¹4x | → ∞ or | etªx || ∞ as as t → ∞ t - 00.
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,
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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