MTHE 225 Quiz: quiz 6 (10).pdf
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20 Nov 2012
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We are asked to use the eigenvalue method to nd a solution to the following linear system, satisfying the speci ed initial conditions. x . 2 = 6x1 2x2 , x1(0) = 1 , x2(0) = 0. 6 2! has eigenvalues = 3, = 4, as is seen from the characteristic. 5 polynomial: det(a i) = det 9 . To nd the eigenvector v corresponding to = 3, we solve the system. To nd the eigenvector v corresponding to = 4, we solve the system. Thus the two independent solutions are to nd v = 1 x1(t) = e3t 5. 1 and so the general solution is x(t) = c1e3t 5. E4t! c1 c2! x(t) = x(t)c = 5e3t c2! such that. Now we wish to nd c = c1 which can be written as. We nd c1 = 1, c2 = 6, thus or. 1! x1(t) = 5e3t + 6e4t x2(t) = 6e3t.
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False |
False |
False |
False |
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