Unformatted text preview: 5.] Equilibrium POll'lt Analysis 325
(b) y 3” 27 23 14 10 98 102 73 77 (c) We could compute the eigenvalues and eigenvectors for each of the linear systems. how
ever, we can determine much of the qualitative information about behavior of the solutions
from the ﬁgures in part (b). It is interesting to note the size of the eigenvalues (which are
quite large in absolute value), so solutions move toward or away from the equilibria very
quickly. 10. (a) The equilibrium points in the ﬁrst quadrant are (0. 0), (0, 50) and (100. 0). To classify
these equilibrium points we compute the Jacobian matrix which is 2x  y + 100 —x
—2xy x2 — 3y2 + 2500 '
At (0. 0) the Jacobian matrix is
100 0
0 2500 which has eigenvalues 100 and 2500, so (0. 0) is a source. At (0. 50) the Jacobian matrix is
50 0
O 5000 which has eigenvalues 10 and —5000, so (0. 50) is a saddle. At (100. 0) the Jacobian matrix is
— 100  100
0 900 which has eigenvalues 40 and —7500, so (40. O) is a sink. ...
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 Spring '08
 JU

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