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Unformatted text preview: (a) x = c 1 e rt v 1 + c 2 e rt v 2 (b) x = c 1 e rt v + c 2 [ e rt t v + e rt v 2 ] Consider each case in turn. In example 4.9 we Found For two independent eigen-vectors x ( t ) = c 1 e rt y ( t ) = c 2 e rt Hence, x / y = c 1 / c 2 is independent oF t and depends only on the components oF v r and v s and the arbitrary constants c 1 and c 2 . This is a general result and so all solutions lie on straight lines through the origin, as shown in fgure 4.21. In this case the origin is a proper node that is stable. Had the repeated root been positive, then we would have an unstable proper node. It is this situation we gave an example...
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- Fall '11