Recall if the Wronskian is nonzero then x 1 t and x 2 t form a fundamental set

Recall if the wronskian is nonzero then x 1 t and x 2

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Recall if the Wronskian is nonzero, then x 1 ( t ) and x 2 ( t ) form a fundamental set of solutions to the system of DEs Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (13/54)
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Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Linear Algebra Result Theorem Let A have real or complex eigenvalues, λ 1 and λ 2 , such that λ 1 6 = λ 2 , and let the corresponding eigenvectors be v 1 = v 11 v 21 and v 2 = v 12 v 22 . If V is the matrix formed from v 1 and v 2 with V = v 11 v 12 v 21 v 22 , then det | V | = v 11 v 12 v 21 v 22 6 = 0 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (14/54)
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Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Real and Different Eigenvalues 1 The two previous slides show that if A has real and different eigenvalues , λ 1 and λ 2 , then the system ˙ x = Ax has a fundamental set of solutions x 1 ( t ) = e λ 1 t v 1 and x 2 ( t ) = e λ 2 t v 2 , where v 1 and v 2 are the corresponding eigenvectors for λ 1 and λ 2 , respectively It follows that the general solution can be written x ( t ) = c 1 e λ 1 t v 1 + c 2 e λ 2 t v 2 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (15/54)
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Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Real and Different Eigenvalues 2 Example 1: Consider the example: ˙ x 1 ˙ x 2 = - 0 . 5 2 0 - 1 x 1 x 2 Find the general solution to this problem and create a phase portrait. From above we need to find the eigenvalues and eigenvectors, so solve det - 0 . 5 - λ 2 0 - 1 - λ = ( λ + 0 . 5)( λ + 1) = 0 , which is the characteristic equation with solutions λ 1 = - 0 . 5 and λ 2 = - 1 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (16/54)
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Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Real and Different Eigenvalues 3 Example 1 (cont): For λ 1 = - 0 . 5 we have: - 0 . 5 - λ 1 2 0 - 1 - λ 1 ξ 1 ξ 2 = 0 2 0 - 0 . 5 ξ 1 ξ 2 = 0 0 This results in the eigenvector ξ (1) = 1 0 . Similarly, for λ 2 = - 1 we have: - 0 . 5 - λ 2 2 0 - 1 - λ 2 ξ 1 ξ 2 = 0 . 5 2 0 0 ξ 1 ξ 2 = 0 0 This results in the eigenvector ξ (2) = 4 - 1 . Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (17/54)
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Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Real and Different Eigenvalues
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