hw5sol

# hw5sol - Homework 5 Solutions 3 p 143 problem 5.2.13 a As a...

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Unformatted text preview: Homework 5, Solutions 3. p. 143, problem 5.2.13. a) As a two-dimensional 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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hw5sol - Homework 5 Solutions 3 p 143 problem 5.2.13 a As a...

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