'dr =y (1) {3 = ~ey Wj sin € For e-0, the stable and unstable manifolds of the equilibrium point (TT,0) are respectively 9 2 cos 2 , y = -2cos 2 y = ~2cos 2 , y = +2cos 2 y = 2sin 2 , y = 2sin 2'
Added by Alicia J.
Step 1
First, we need to find the equilibrium point(s) of the system. This is done by setting dr/dt = 0 and solving for r and θ. From equation (1), we have: y(1) = 0 3 = -yWjsinθ Solving for r and θ, we get: r = π θ = ±π/2 So the equilibrium points are (π, π/2) and Show more…
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For e=0, the stable and unstable manifolds of the equilibrium point (π,0) are respectively y = 2 cos x/2, y = -2 cos x/2 y = 2 sin x/2, y = -2 sin x/2 y = -2 cos x/2, y = +2 cos x/2
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