MATH 107 Fall 2009 Homework 9 Solutions - Section 9.3#42 2...

This preview shows page 1 - 3 out of 8 pages.

Section 9.3 #42If A = 1132, then the determinant of A is (-2)(1)-(-1)(3) = 1 and the trace of A is (-2)+1 = -1. So the determinant is positive and the trace is negative, which means (by the theorem) that the real parts of both eigenvalues are negative. Section 9.3 #44If A = 1210, then the determinant of A is (0)(1)-(-2)(1) = 2 and the trace of A is 0+1 = 1 So the determinant is positive, but the trace is not-negative which means that the real parts of both eigenvalues are not negative. Section 11.1 #14dtdxdtdx21=1412)()(21txtxFirst we want to find the general solution: So we need to find the eigenvalues, and the associated eigenvectors. So find det (A-λI) = det λλ1412=(2-λ)(-1-λ)-(1)(4) =λ2-λ-2 –4 = =λ2-λ-6 =(λ-3)( λ+2) So for λ1= 3 we find the eigenvector: 1412vu= 3 vu==+3vv-4u :And3uv2u :ThusSo –u + v = 0 This leads to 11as an eigenvector. Now for λ2= -2 we find the eigenvector: 1412vu= -2 vu
==+2v-v-4u :And2u-v2u :ThusSo 4u + v = 0 This leads to 41as an eigenvector. Combining this information we get the general solution to be: x(t) = c1e3t11+c2e-2t41Now consider the lines through these vectors: -10-50510-5-3-1135x2x1By examining the vector field pictured with the problem we see that along the blue line (y=x) arrows point away from the origin, and along the magenta line (y=(-1/4)x) the arrows point towards the origin. Solutions on these lines will stay on these lines. Section 11.1 #18dtdxdtdx21=6125)()(21txtxFirst we want to find the general solution: So we need to find the eigenvalues, and the associated eigenvectors. So find det (A-λI) = det λλ6125=(5-λ)(6-λ)-(2)(1) = 30 – 11 λ+ λ2=λ2 – 11 λ+28 = (λ-7)( λSo for λ1= 4 we find the eigenvector: 6125vu= 4 vu=+=+4v6v
–2
-4)

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture