ECH140FirstExam_2013Soln

# ECH140FirstExam_2013Soln - HANOI UNIVERSITY OF MINING and...

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HANOI UNIVERSITY OF MINING and GEOLOGY Advanced Program in Chemical Engineering ECH140: Mathematical Methods for Chemical Engineering First Exam Solution September 9, 2013 (9:00-11:00 ) (Closed book and notes no cell phones; no questions asked or answered) Prof. Higgins Hanoi, September 2013 Problem 1 (15 pts) (i) Determine the general solution for the following ODEs: a H x L d 2 f dx 2 + b H x L d f dx + c H x L f = 0 H a L a H x L = 1, b H x L = 0, c H x L = -l 2 H b L a H x L = 2, b H x L = - 4 x , c H x L = - l 2 x 2 H c L a H x L = 2, b H x L = - 4, c H x L = - l 2 Solution (a) The equation is d 2 f dx 2 - l 2 f = 0 This is a constant coefficient equation, and thus the general solution is f H x L = c 1 Cosh H l x L + c 2 Sinh H l x L (b) The equation is 2x 2 d 2 f dx 2 - 4x d f dx - l 2 f = 0 This is an equi-dimensional equation and so the general solution is of the form f H x L ~ x p Substituting into the equation gives 2x 2 I p H p - 1 L x p - 2 - 4xpx p - 1 - l 2 x p = 0 9 2 I p 2 - p M - 4p - l 2 = x p = 0 Thus p must satisfy 2p 2 - 6p - l 2 = 0 The solution to this quadratic is

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p = 6 ± 36 + 8 l 2 4 = 3 ± 9 + 2 l 2 2 Thus the general solution is f H x L = c 1 x p 1 + c 2 x p 2 where p 1 = 3 + 9 + 2 l 2 2 , p 2 = 3 - 9 + 2 l 2 2 (c) The equation is 2 d 2 f dx 2 - 4 d f dx - l 2 f = 0 Since this is a constant coefficient equation the general solution is of the form f H x L ~ exp H px L Substituting for f H x L gives I 2p 2 - 4p - l 2 M e px = 0 Thus p must satisfy p = 4 ± 16 + 8 l 2 4 = 1
• Summer '13
• LeThuy

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