Amath 350 page 5 assignment 7 solutions fall 2012 4

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Unformatted text preview: demand curve with derivative dD = 2D dt 5p + 10. 7. (a) The model is dD = 2D 5p + 10, dt dp = D S = D 50, dt or y0 = Ay + F where D p y= 2 1 , A= 5 0 (b) Equilibrium solution is given by y(t) = k, t Ak + F = 0. Thus 0 k = A 1 ( F) = . IR where k is a constant 2-vector satisfying 1 1 5 10 50 , F= 10 50 2 5 = 50 22 . (c) First find general solution of associated homogeneous equation: y0 = Ay. 2 5 Characteristic equation is: det = 0 ) 2 2 + 5 = 0. 1 Thus the eigenvalues are: 1 ± 2i. 1 2i 5 v1 0 Eigenvector for = 1 + 2i: = ) v2 = 1 52i v1 . 1 1 2i v2 0 5 5 0 Choose v = = +i = u + iw. The eigenvector for ¯ = 1 2i 1 2i 1 2 ¯ is v = u iw. From Theorem 4.22 in the Course Notes, the general solution of the homogeneous equation is yh (t) = c1 et cos(2t) 5 1 0 2 sin(2t) + c2 et sin(2t) 5 1 + cos(2t) 0 2 . A particular solution is the equilibrium solution found above, i.e. yp (t) = [50, 22]T . Thus the general solution of the DE is et [5c1 cos(2t) + 5c2 sin(2t)] + 50 e [(c1 2c2 ) cos(2t) + (2c1 + c2 ) sin(2t)]...
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This document was uploaded on 02/09/2014.

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