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sol_set1 - Solutions to Problem Set 1 ARE 261 Question 1...

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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
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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 k 1 and k 2 to get the complete solution for the differential equation. The
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