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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 =- b m < , and the determinant is = k m > . This implies that the origin is either a stable node, or a stable spiral. It is a stable spiral if > 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 ( - )( + b )+ km = , so 2 + b + km = , 1 or = =- b b 2- 4 km 2 Both of these are negative real numbers; the smaller of the two if - . The eigenvector associated with an eigenvalue is a solution x y of y = x ,- ( k / m ) x- ( b / m ) y = y ....
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