hw6-solutions

# Consequently the eigenvalues of a are given by a11

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Unformatted text preview: 1 = −2 and a2 = −3. Thus, cos(t) − 5 sin(t) . −2 cos(t) − 3 sin(t) y (t) = e−2t Problem 23: (a) The characteristic equation is 0 = det(λI − A) = det λ − a11 −a12 −a21 λ − a22 = λ2 − (a11 + a22 )λ + (a11 a22 − a12 a21 ). Consequently, the eigenvalues of A are given by λ= a11 + a22 ± 2 (a11 + a22 )2 − (a11 a22 − a12 a21 ). 4 Therefore, the eigenvalues of A are purely imaginary if and only a11 + a22 = 0 (a11 + a22 )2 − (a11 a22 − a12 a21 ) &lt; 0. 4 and This is equivalent to saying that a11 + a22 = 0 and a11 a22 − a12 a21 &gt; 0. (b) We now assume that (x(t), y (t)) is a solution of the ODE dx = a11 x + a12 y dt dy = a21 x + a22 y. dt In the next step, we express y (t) as a function of x(t). The resulting function y (x) sat...
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