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Unformatted text preview: Homework 5, Solutions 3. p. 143, problem 5.2.13. a) As a twodimensional linear system, the equations look like this: ˙ x = y , ˙ y = b m y k m x . The matrix is A = 1 m m k b b) The trace of A is τ = b m < , and the determinant is Δ = k m > . This implies that the origin is either a stable node, or a stable spiral. It is a stable spiral if Δ > τ 2 4 , which means k m > b 2 4 m 2 , or b < 2 √ km . It is a stable node if b > 2 √ km . The borderline case b = 2 √ km , will be discussed later. Let us assume b > 2 √ km , so that the origin is a stable node, and compute the eigenvalues and eigenvectors of A . The eigenvalues are solutions of ( λ )( λ + b )+ km = , so λ 2 + b λ + km = , 1 or λ = λ ± = b ± √ b 2 4 km 2 Both of these are negative real numbers; the smaller of the two if λ . The eigenvector associated with an eigenvalue λ is a solution x y of y = λ x , ( k / m ) x ( b / m ) y = λ y ....
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 Spring '09
 Dr.Borgers
 Linear Algebra, Determinant, Equations, Eigenvalue, eigenvector and eigenspace, stable node

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