Z₁' (t) = -6₁2 + 1.12₁ -0.822 Z₂²' (t) = -13.8 +2.12, -1.222 Z₁ (0) =11 Z₂₁₂ (0) = 4

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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Find the equilibrium solution for the linear system of differential equations and sketch the trajectory
(t) = 6,2 + ۱۱ 2 - 022
(4) = -13.6 + 212 - 1222
(0) = \\
Z, (۰) - 4
Transcribed Image Text:(t) = 6,2 + ۱۱ 2 - 022 (4) = -13.6 + 212 - 1222 (0) = \\ Z, (۰) - 4
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