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Unformatted text preview: 18.03 Problem Set 5 Fall 2009 Solutions 1. (5 points) Find a particular solution of each of the differential equations d 6 y dx 6 y = e 2 x , d 6 y dx 6 y = e x , d 6 y dx 6 y = e x/ 2 cos( x √ 3 / 2) . We use the method of undetermined coefficients. The derivatives of e 2 x are itself (up to constant multiples) and it is not a solution to the associated homogenous equation y (6) y . So we try y p = Ae 2 x . Then y (6) p = 64 Ae 2 x and so we have 63 Ae 2 x = e 2 x , whereupon A = 1 / 63 and so y p = 1 63 e 2 x is a solution. The derivatives of e x are itself, but it does satisfy the associated homogeneous equation, so we also consider xe x (which does not satisfy it). So we try y p = Bxe x . Then y (6) p = 6 Be x + Bxe x , and so we have 6 Be x = e x , whereupon B = 1 / 6 and so y p = 1 6 xe x is a solution. The expression e x/ 2 cos( x √ 3 / 2) is again a solution to the associated homogenous equation (and x times it is not), and so we consider y p = Cxe x/ 2 cos( x √ 3 / 2) + Dxe x/ 2 sin( x √ 3 / 2) . Then y (6) p = ( Cx + 3 C 3 √ 3 D ) e x/ 2 cos( x √ 3 / 2) + ( Dx + 3 D + 3 √ 3 C ) e x/ 2 sin( x √ 3 / 2) , and so we have 3 C 3 √ 3 D = 1 and 3 √ 3 C + 3 D = 0 , whereupon C = 1 / 12 and D = √ 3 / 12, and so y p = xe x/ 2 1 12 cos( x √ 3 / 2) √ 3 12 sin( x √ 3 / 2) ! is a solution. 2. (10 points) Suppose that the mass in the “spring and dashpot” system described on page 101 of the text is driven by a sinusoidal force: m d 2 x dt 2 + c dx dt + kx = cos( ωt ) . ( Forced Damped Spring ) a) Find a constant solution of (Forced Damped Spring) in case ω = 0 . Explain what this solution means physically, and why you should have been able to guess it without doing any calculation....
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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