2 5 and hence the roots of the to nd an eigenvector

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Unformatted text preview: nd an eigenvector for = 0 we must solve 42 u1 0 = 2 1 u2 0 We may choose 1 2 =) 2 u1 u2 = 0, i.e. u2 = 2u1 . as our eigenvector. To find an eigenvector for 5 we must solve 12 v1 0 = 24 v2 0 = . AMATH 350 We may choose Page 4 Assignment #7 Solutions - Fall 2012 v1 + 2v2 = 0, i.e. v1 = 2v2 . =) 2 1 as our eigenvector. Thus, we’ve determined that x h = C1 1 2 + C2 2 1 e 5t . Now, to find a particular solution, we assume that it can be written in the form 1 2 x p = u1 + u2 e 5t . 2 1 As shown in the course notes, this requires us to solve the system u01 (1) + u02 ( 2e 5t 1 t )= (3) 2 t Multiplying (3) by 2, and then subtracting it from (4), we obtain u01 (2) + u02 (e 5e 5t 0 u2 4 Therefore u02 = e5t , so 5 u2 = From (3) we also know that u01 = 5t (4) )=4+ = 4. 4 5t e. 25 1 + 2e t 5t 0 u2 u1 = ln |t| + = 18 + , so t5 8t . 5 Therefore a particular solution is ✓ ◆ 8t 4 1 2 xp = ln |t| + + 2 1 5 25 8 8 ln |t| + 5t 25 = , 16t 4 2 ln |t| + 5 + 25 and the full general solution is 1 2 x = C1 + C2 e 2 1 5t ✓ 8t + ln |t| + 5 ◆ 1 2 4 + 25 2 1 . AMATH 350 Page 5 Assignment #7 Solutions - Fall 2012 4. Commodity with constant supply S = 50,...
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This document was uploaded on 02/09/2014.

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