If we examine the scalar equations then x 1 x 1 x 2 and x 2 x 2 Thus x 2 t c 2

# If we examine the scalar equations then x 1 x 1 x 2

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If we examine the scalar equations, then ˙ x 1 = - x 1 + x 2 and ˙ x 2 = - x 2 Thus, x 2 ( t ) = c 2 e - t , so ˙ x 1 + x 1 = c 2 e - t with μ ( t ) = e t This has the solution x 1 ( t ) = c 2 te - t + c 1 e - t Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (44/54) Subscribe to view the full document.

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 6 Example 8 (cont): Combining the results above we see x ( t ) = x 1 ( t ) x 2 ( t ) = c 1 + c 2 t c 2 e - t = c 1 1 0 e - t + c 2 1 0 t + 0 1 e - t The second solution has the form x 2 ( t ) = v te - t + w e - t Upon differentiation ˙ x 2 ( t ) = v (1 - t ) e - t - w e - t = Ax 2 = A ( v te - t + w e - t ) Since ( A + I ) v = 0 , this reduces to solving for w ( A + I ) w = v or w = 0 1 + k 1 0 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (45/54) 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 Repeated Eigenvalues 7 Example 8 (cont): This DE produces a stable improper node with all solutions moving toward the origin Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (46/54) Subscribe to view the full document.

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 - General Repeated Eigenvalues - Two Dimensional Null Space Suppose the 2 × 2 matrix A has a repeated eigenvalue λ . If the eigenspace spanned by the eigenvectors has dimension 2, v 1 and v 2 , then the solution is simply x ( t ) = c 1 v 1 e λt + c 2 v 2 e λt Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (47/54) 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 Repeated Eigenvalues - General Repeated Eigenvalues - One Dimensional Null Space If the 2 × 2 matrix A has only one eigenvector v associated with λ , then one solution is x 1 ( t ) = v e λt We attempt a second solution of the form x 2 ( t ) = v te λt + w e λt , which upon differentiation gives ˙ x 2 ( t ) = v ( λt + 1) e λt + λ w e λt = Ax 2 = A ( v te λt + w e λt ) Since ( A - λ I ) v = 0 , this reduces to solving for w ( A - λ I ) w = v This gives the second linearly independent solution, x 2 ( t ), above, where w solves this higher order null space problem , which will include a particular solution and any multiple, k v Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (48/54) Subscribe to view the full document.

Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation  • Fall '08
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