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Unformatted text preview: Solutions to Problem Set 1 ARE 261 Question 1 Begin with the assumption that the solution to the second order differential equation is of the form y = k exp xt , where k and x are constants. Then y = kx exp xt and y = kx 2 exp xt . Substitute these expressions for y, y and y into the differential equation and divide through by k exp xt to get x 2 + ax + b = 0 (1) This gives a quadratic in x which we assume has two distinct root, x 1 and x 2 given by x 1 = a 2 + a 2 4 b 2 (2) and x 2 = a 2 a 2 4 b 2 (3) The solution to the second order differential equation is then given by y = k 1 exp x 1 t + k 2 exp x 2 t (4) where k 1 and k 2 are constants to be determined by boundary or initial conditions. In the next subpart we are given two such initial conditions, y (0) = 0 and y (0) = 1. Substitute these conditions 1 into equation (4) and its first derivative with respect to time to get the following two equations k 1 + k 2 = 0 (5) k 1 x 1 + k 2 x 2 = 0 (6) These can be solved for...
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This note was uploaded on 08/01/2008 for the course ARE 263 taught by Professor Karp during the Fall '06 term at University of California, Berkeley.
 Fall '06
 KARP

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