We have � 2 � 1 and ξ 2 ξ 1 Thus x 1 t 1 1 i e t cos2 t i sin2 t u t i w t e t

We have ? 2 ? 1 and ξ 2 ξ 1 thus x 1 t 1 1 i e t

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We have λ 2 = ¯ λ 1 and ξ (2) = ¯ ξ (1) Thus, x 1 ( t ) = 1 1 - i e t (cos(2 t ) + i sin(2 t )) = u ( t ) + i w ( t ) = e t cos(2 t ) e t (cos(2 t ) + sin(2 t )) + i e t sin(2 t ) e t (sin(2 t ) - cos(2 t )) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (35/54)

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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 Complex Eigenvalues 8 Example 5 (cont): From above the general solution is x 1 ( t ) x 2 ( t ) = c 1 e t cos(2 t ) e t (cos(2 t ) + sin(2 t )) + c 2 e t sin(2 t ) e t (sin(2 t ) - cos(2 t )) . This is an unstable spiral . All solutions spiral away from the origin. Solutions with complex eigenvalues with negative real parts spiral toward the origin, creating a stable spiral Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (36/54)
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 Imaginary Eigenvalues 9 Example 6: Consider the example: ˙ x 1 ˙ x 2 = 2 - 5 1 - 2 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 2 - λ - 5 1 - 2 - λ = λ 2 + 1 = 0 , which is the characteristic equation with solutions λ = ± i (purely imaginary eigenvalues) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (37/54)

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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 Imaginary Eigenvalues 10 Example 6 (cont): For λ 1 = i we have: 2 - λ 1 - 5 1 - 2 - λ 1 ξ 1 ξ 2 = 2 - i - 5 1 - 2 - i ξ 1 ξ 2 = 0 0 This results in the eigenvector ξ (1) = 2 + i 1 . We have λ 2 = ¯ λ 1 and ξ (2) = ¯ ξ (1) Thus, x 1 ( t ) = 2 + i 1 (cos( t ) + i sin( t )) = u ( t ) + i w ( t ) = 2 cos( t ) - sin( t ) cos( t ) + i 2 sin( t ) + cos( t ) sin( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (38/54)
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 Imaginary Eigenvalues 11 Example 6 (cont): From above the general solution is x 1 ( t ) x 2 ( t ) = c 1 2 cos( t ) - sin( t ) cos( t ) + c 2 2 sin( t ) + cos( t ) sin( t ) . This is a center . All solutions form ellipses around the origin. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (39/54)

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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 Repeated Eigenvalues 1
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