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Unformatted text preview: 18.03 PS10 SolutionsSpring 09 1a. Char. eq. ( + 1) 2 = 0 = = 1; A I = A + I = 1 1 2 = 1 1 1 1 . We see by inspection that ( A + I ) 1 1 = , so x 1 = e t 1 1 is a solution. We guess that the other independent solution will have the form x 2 = ( a t + b ) e t for column vectors a = a 1 a 2 , b = b 1 b 2 . Plugging in x 2 = A x 2 , we obtain ( A + I ) a = 0 and ( A + I ) b = a . Take a = 1 1 . Given that choice of a , the condition on b is b 1 + b 2 = 1, so b = 1 works. x 1 = e t 1 1 and x 2 = 1 1 t + 1 e t are two independent solutions. Note that the answer is far from unique. 1b. The 2ndorder equation is x 00 + 2 x + x = 0 . Two solutions are x = e t , y = te t . x x = e t 1 1 = x 1 ; y y = te t 1 1 + e t 1 = x 2 x 1 . 2ac. See Figure 1 for the diagram. Table follows. Region Geom. type Stability Eigenvalues ( T, D ) definition 1 saddle N/A (unstable) 1 < < 2 D < 2 node stable = sink 1 < 2 < T < , < D < T 2 / 4 3 spiral stable = sink r si , r < T < , D > T 2 / 4 4 spiral unstable = source r si , r > T > , D > T 2 / 4 5 node unstable = source < 1 < 2 T > , < D < T 2 / 4 2b. The parabola is given by (i) 1 = 2 , (ii) s = 0, and (iii) D = T 2 / 4. 2d. The equation for the boundary is T = 0 , D > 0. In terms of the eigenvalues, one can say that they must be purely imaginary or that they are complex with r = 0. As we move across the boundary from T < 0 to T > 0, we begin with spirals pointing inward.0, we begin with spirals pointing inward....
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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