MATH 2214 Midterm: MATH 2214 Virginia Tech 4.7 Exam

• School: Virginia Tech
• Department: Mathematics
• Course: MATH 2214
• Professor: All

Document Summary

Solutions 4. 7: a) the characteristic equation is ( 2 )2 = 0, i. e. we have an algebraically double eigenvalue 2. 0 0(cid:19)(cid:18) x1 x2 (cid:19) = (cid:18) 0. Hence we have only one linearly independent eigenvector, namely (cid:18) 1. 0(cid:19) : the equation for a generalized eigenvector is (cid:18) 0 1. 0 0(cid:19)(cid:18) x1 x2 (cid:19) = (cid:18) 1. We can, for instance, choose the vector (cid:18) 0. We now have the fundamental set y1 = e 2t (cid:18) 1. 0 1(cid:19) . t: the general solution is. For t = 0, we nd y = c1y1 + c2y2. y(0) = (cid:18) c1 c2 (cid:19) = (cid:18) 1. 1: a) the characteristic equation is (5 )(1 ) + 4 = 2. 6 + 9 = ( 3)2 = 0, i. e. we have an algebraically double eigenvalue 3. For an eigenvector, we nd (cid:18) 2 1. 2 (cid:19)(cid:18) x1 x2 (cid:19) = (cid:18) 0.