If the two eigenvectors ξ 1 and ξ 2 are linearly

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If the two eigenvectors ξ (1) and ξ (2) are linearly independent, then the general solution is The ratio x 2 / x 1 is independent of t , but depends on the components of ξ (1) and ξ (2) and on c 1 and c 2 . A phase portrait is given on the next slide. rt rt e c e c ) 2 ( 2 ) 1 ( 1 ξ ξ x + =

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Case 3: Star Node (2 of 5) The general solution is Thus every trajectory lies on a line through the origin, as seen in the phase portrait below. Several graphs of x 1 versus t are given below as well, with the case of x 2 versus t similar. The critical point at the origin is called a proper node , or a star point. 0 , ) 2 ( 2 ) 1 ( 1 < + = r e c e c rt rt ξ ξ x
Case 3: Equal Eigenvalues (3 of 5) If the repeated eigenvalue r has only one linearly independent eigenvector ξ , then from Section 7.8 the general solution is For large t , the dominant term is c 2 ξ te rt . Thus every trajectory approaches origin tangent to line through the eigenvector ξ . Similarly, for large negative t the dominant term is again c 2 ξ te rt , and hence every trajectory is asymptotic to a line parallel to the eigenvector ξ . The orientation of the trajectories depends on the relative positions of ξ and η , as we will see. ( 29 rt rt rt e te c e c η ξ ξ x + + = 2 1

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Case 3: Improper Node (4 of 5) We can rewrite the general solution as Note that y determines the direction of x , whereas the scalar quantity e rt affects only the magnitude of x . For fixed values of c 1 and c 2 , the expression for y is a vector equation of line through the point c 1 ξ + c 2 η and parallel to ξ . Using this fact, solution trajectories can be sketched for given coefficients c 1 and c 2 . See phase portrait below. When a double eigenvalue has only one linearly independent eigenvalue, the critical point is called an improper or degenerate node . ( 29 t c c c e e t c c c rt rt ξ η ξ y y ξ η ξ x 2 2 1 2 2 1 , + + = = + + =
Case 3: Phase Portraits (5 of 5) The phase portrait is given in figure (a) along with several graphs of x 1 versus t are given below in figure (b). When the relative orientation of ξ and η are reversed, the phase portrait given in figure (c) is obtained.

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Case 4: Complex Eigenvalues (1 of 5) Suppose the eigenvalues are λ ± i μ , where λ and μ are real, with λ 0 and μ > 0. It is possible to write down the general solution in terms of eigenvalues and eigenvectors, as shown in Section 7.6. However, we proceed in a different way here. Systems having eigenvalues λ ± i μ are typified by We introduce the polar coordinates r , θ given by 2 1 2 2 1 1 x x x x x x λ μ μ λ λ μ μ λ + - = + = - = x x 1 2 2 2 2 1 2 tan , x x x x r = + = θ
Case 4: Polar Equations (2 of 5) Differentiating the polar equations with respect to t , we have or Substituting into these derivative equations, we obtain 2 1 2 2 1 1 , x x x x x x λ μ μ λ + - = + = 1 2 2 2 2 1 2 tan , x x x x r = + = θ ( 29 ( 29 2 1 1 2 2 1 2 2 2 1 1 / / sec , 2 2 2 x dt dx x dt dx x dt d dt dx x dt dx x dt dr r - = + = θ θ ( 29 ( 29 2 1 1 2 2 1 2 2 2 1 1 sec , x x x x x

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