-1-9 cos² 6t + 12 sin 6t cos 6t -12 sin² 6t+ 9 sin 6t cos 6t y' = that has periodic (therefore bounded) coefficients. Verify that the eigenvalues are X₁ (t) = -1 and A₂ (t) = -10 (constants). Nevertheless, verify that 12 cos² 6t+9 sin 6t cos 6t 6t)) y -(1 + 9 sin² 6t + 12 sin 6t cos 6t), e²t (cos 6t + 2 sin 6t) e2t (2 cos 6t - sin 6t) is a solution. Explain why this implies the steady state y = 0 is unstable. = (2²4) y =
-1-9 cos² 6t + 12 sin 6t cos 6t -12 sin² 6t+ 9 sin 6t cos 6t y' = that has periodic (therefore bounded) coefficients. Verify that the eigenvalues are X₁ (t) = -1 and A₂ (t) = -10 (constants). Nevertheless, verify that 12 cos² 6t+9 sin 6t cos 6t 6t)) y -(1 + 9 sin² 6t + 12 sin 6t cos 6t), e²t (cos 6t + 2 sin 6t) e2t (2 cos 6t - sin 6t) is a solution. Explain why this implies the steady state y = 0 is unstable. = (2²4) y =
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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Transcribed Image Text:-1-9 cos² 6t+ 12 sin 6t cos 6t
-12 sin² 6t+ 9 sin 6t cos 6t
v = (-
y'
12 cos² 6t+9 sin 6t cos 6t
−(1 + 9 sin² 6t + 12 sin 6t cos 6t)
that has periodic (therefore bounded) coefficients. Verify that the eigenvalues
are X₁ (t) = -1 and A₂(t) = −10 (constants). Nevertheless, verify that
e2t (cos 6t + 2 sin 6t)
e2t (2 cos 6t - sin 6t)
is a solution. Explain why this implies the steady state y = 0 is unstable.
y =
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