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14. 12 points! The phase portrait (without arrows) of a non-linear syster dz 3-32, 16r dt 2 dy dt a) Mark all critical points

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Answer #1

a) critical points are:

dt dy = 16エー4x3 = z = ±2,0

(0,1), (0,-1), (2,1), (2,-1), (-2,1), (-2,-1)

jacobian matrix is

16-12.2 o

and its eigenvalue is

λ = ±(24g(4-312 ))

so at (0,1) Eigenvalues of the Jacobian matrix is +10,-10, so we get a saddle point, which is an unstable equilibrium point.

at (0,-1) Eigenvalues of the Jacobian matrix is +10i,-10i, so we get critical point that is spiral source, and therefore an unstable equilibrium point.

at (2,1) Eigenvalues of the Jacobian matrix is +14i,-14i, so we get critical point that is spiral source, and therefore an unstable equilibrium point.

at (2,-1) Eigenvalues of the Jacobian matrix is +14,-14, so we get a saddle point, which is an unstable equilibrium point.

at (-2,1) Eigenvalues of the Jacobian matrix is +14i,-14i, so we get critical point that is spiral source, and therefore an unstable equilibrium point.

at (-2,-1) Eigenvalues of the Jacobian matrix is +14,-14, so we get a saddle point, which is an unstable equilibrium point.

b)

dz = 3-3 dy = 161-41 dt 3-3y dx (3 - 3y2)dy

Where, E = total energy

H(x, y)=E=3y - y-82

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PLEASE HELP !!!! thanks 14. 12 points! The phase portrait (without arrows) of a non-linear syster dz 3-32, 16r dt 2 dy dt a) Mark all critical points directly on the 2 diagram, and classify them...
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