u 1 e u 2 e 14 This is readily solved by row reduction row reduced echelon form

U 1 e u 2 e 14 this is readily solved by row

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u 1 e u 2 e = - 14 0 This is readily solved by row reduction ( row reduced echelon form ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (12/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Solve Linear System 1 Solve Linear System: Write [ A : b ], so - 13 8 3 4 . . . - 14 1 4 - 1 4 . . . 0 - 8 13 R 1 -→ 4 R 2 1 - 6 13 . . . 112 13 1 - 1 . . . 0 R 2 - R 1 -→ 1 - 6 13 . . . 112 13 0 - 7 13 . . . - 112 13 - 13 7 R 2 -→ 1 - 6 13 . . . 112 13 0 1 . . . 16 R 1 + 6 13 R 2 -→ 1 0 . . . 16 0 1 . . . 16 or u e = 16 16 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (13/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Solve Linear System 2 Solve Linear System: Linear systems are efficiently solved in MatLab and Maple MatLab - Solving equilibrium Enter matrix, A , and vector, b Use linsolve command or inv(A)*b Augment A with b and use rref Maple - Solving equilibrium Start with(LinearAlgebra) to invoke the Linear Algebra package Enter matrix, A , and vector, b Use LinearSolve( A , b ) command or Multiply( A - 1 , b ) operation Detailed supplemental sheets are provided Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (14/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Solving the System of DEs 1 Model System satisfies ˙ u = Ku + b and has a steady state solution u ( t ) = u e , where Ku e = - b Make a change of variables v ( t ) = u ( t ) - u e , then ˙ v = ˙ u and ˙ v = K ( v + u e ) + b = Kv This change of variables allows considering the simpler system ˙ v = Kv Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (15/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Solving the System of DEs 2 Model System has a Newton’s Law of Cooling , so anticipate an exponential (decaying) solution Try a solution of the form v ( t ) = ξe λt , where ξ = [ v 1 , v 2 ] T is a constant vector, so ˙ v ( t ) = λξe λt The translated Model System ˙ v ( t ) = Kv ( t ) becomes λξe λt = K ξe λt or λξ = K ξ This is the classic eigenvalue problem ( K - λ I ) ξ = 0 , which has eigenvalues , λ , and associated eigenvectors , ξ The solution of the eigenvalue problem gives the solution of the Model System , v ( t ) = ξe λt Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (16/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse Example 1 Example Model: satisfies the DE: ˙ u 1 ˙ u 2 = - 13 8 3 4 1 4 - 1 4 !
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