22BsolnsHW8

# 22BsolnsHW8 - Math 22B Solutions Homework 8 Spring 2008...

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Math 22B Solutions Homework 8 Spring 2008 Section 7.5 2. Solution Let x = ζe rt . Substitute into the ODE to get: 1 - r - 2 3 - 4 - r ζ 1 ζ 2 = 0 0 For a nonzero solution, we must have det ( A - rI ) = r 2 +3 r +2 = 0, which has roots r 1 = - 1 and r 2 = - 2. For r = - 1, we get ζ 1 = ζ 2 . A corresponding eigenvector is v 1 = (1 , 1) T . For r = - 2, we get 3 ζ 1 = 2 ζ 2 . A corresponding eigenvector is v 2 = (2 , 3) T . So x = c 1 (1 , 1) T e - t + c 2 (2 , 3) T e - 2 t . Also see figure 1 for the graph. 24. See figure 2 Section 7.6 13. Solution (a) det ( A - rI ) = ( α - r ) 2 + 1 so r 2 - 2 + α 2 + 1 = 0 gives us r = α ± i . (b) The equilibrium is when α = 0. (c) See figure 3. 28. Solution (a) x 1 = u, x 2 = u 0 x 0 1 = x 2 and x 0 2 = u 00 . So x 0 2 = - k m u = - k m x 1 . (b) det ( A - rI ) = r 2 + k m = 0. So r = ± i q k m . (c) Origin is the center. See figure 4. (d) The general solution is u ( t ) = c 1 cos r k m t + c 2 sin r k m t So the solution to the system is x = c 1 ( r m k sin r k m , cos r k m t ) T + c 2 ( r m k cos r k m , - sin r k m t ) T 1

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The natural frequency is I m ( r 1 , 2 ). Section 7.7 3. Solution
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