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anonymous
 3 years ago
i still can't understand that:
λ1 = 0 > λ2.
Line of critical points.
The critical points are not isolated –they lie on the line
through 0 with direction v1.
x = c1v1 + c2eλ2tv2
As t → ∞ x → c1v1 along a line parallel to v2.
why do the critical points lie on the line through 0 with direction v1?
anonymous
 3 years ago
i still can't understand that: λ1 = 0 > λ2. Line of critical points. The critical points are not isolated –they lie on the line through 0 with direction v1. x = c1v1 + c2eλ2tv2 As t → ∞ x → c1v1 along a line parallel to v2. why do the critical points lie on the line through 0 with direction v1?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Where is this question in the MIT courseware? Or can you completely reproduce the question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry I didn't make it clear. Suppose a matrix A has real eigenvalues with two independent eigenvectors. Let λ1, λ2 be the eigenvalues and v1 and v2 the corresponding eigenvectors. ⇒ general solution to the differential equation x'=Ax is x = c1e ^(λ1*t)v1 + c2e^(λ2*t)v2. when λ1 = 0 > λ2, The critical points are not isolated –they lie on the line through 0 with direction v1. x = c1v1 + c2e^(λ2*t)v2 As t → ∞ x → c1v1 along a line parallel to v2. why do the critical points lie on the line through 0 with direction v1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here is the graph of the solution to the diffenrential equation dw:1356597736417:dw The question is here: http://ocw.mit.edu/courses/mathematics/1803scdifferentialequationsfall2011/unitivfirstordersystems/qualitativebehaviorphaseportraits/MIT18_03SCF11_s34_6text.pdf
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