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Unformatted text preview: Math 2940 Worksheet: 7.4, 7.5, 8.1, 8.3, 9.1, 9.3 (Solutions) December 3, 2009 1. Let A be an n n matrix and k a scalar. Consider the following two systems: d~x dt = A~x d~ c dt = kA~ c Show that if ~x ( t ) is a solution to the first system, then ~ c ( t ) = ~x ( kt ) is a solution of the second system. Solution : Starting with ~ c ( t ) = ~x ( kt ), we have d~ c dt = k~x ( kt ) = k ( A~x ( kt )) = kA~ c , where the second equality is from the assumption that ~x satisfies the first equation. 2. Consider the interaction of two species of animals in a habitat. We are told that the change of the populations x ( t ) and y ( t ) can be modeled by the equations dx dt = 1 . 4 x 1 . 2 y dy dt = 0 . 8 x 1 . 4 y where time t is measured in years. (a) Sketch a phase portrait for this system. From the nature of the problem, we are interested only in the first quadrant. Solution : For the phase portrait, we must find the eigenvalues and eigenvectors of the matrix associated with this system, A = 1 . 4 1 . 2 . 8 1 . 4 . Eigenvalues f A ( ) = det( A I ) = det 1 . 4  1 . 2 . 8 1 . 4 = 2 1 . 96 + 0 . 96 = 2 1 = (  1)( + 1) = 0 when = 1 , 1 Eigenspaces E 1 = Ker( A I ) = Ker . 4 1 . 2 . 8 2 . 4 = Span 2 1 E 1 = Ker( A + I ) = Ker 2 . 4 1 . 2 . 8 . 4 = Span 1 3 So our system has the solution ~x ( t ) = c 1 e t 2 1 + c 2 e t 1 3 . Using this information, we obtain the phase portrait (b) What will happen in the long term? Does the outcome depend on the initial populations? If so, how? Solution : Yes, the outcome does depend on the initial condition. This is obvious from the phase portrait. If ~x is above E 1 , both populations die out. E 1 is the line 3 x 1 = x 2 , so this region is 0 < 3 x 1 < x 2 where both populations die out. If ~x is between E 1 and E 1 , both populations survive. E 1 is the line x 1 = 2 x 2 . So this region is 3 x 1 > x 2 > 1 2 x 1 . If If ~x is below E 1 , again both populations survive. This is the region 0 < x 2 < 1 2 x 1 . 3. Consider the dynamical system d~x dt = 1 2 2 1 ~x (a) Find all real solutions of this system. Solution : The general solution of the system is given by ~x ( t ) = c 1 e 1 t ~v 1 + c 2 e 2 t ~v 2 , where i s are the eigenvalues and ~v i s the eigenvectors of the matrix above, and c i s are scalars. So first we find the eigenvalues: f A ( ) = det( A I ) = det 1  2 2 1 = 2 + 2 + 5 = 0 Using the quadratic formula to solve for , = 2 p 2 2 4(1)(5) 2 = 1 2 i Since the eigenvalues are complex, the above formula is not what we want it gives complex solutions to the system. All real solutions are in general given by the formula ~x (...
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 Spring '06
 PANTANO
 Multivariable Calculus, Scalar

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