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Unformatted text preview: 18.03 Problem Set 5: Solutions 17. [Part I problems 4] (a) [10] Complex replacement ¨ z +4 z = te 2 it . Apply ESL with z = e 2 it u and p ( s ) = s 2 +4: p ( D )( e 2 it u ) = e 2 it p ( D + 2 iI ) u . So z = e 2 it u is a solution provided p ( D + 2 iI ) u = t . p ( s + 2 i ) = ( s + 2 i ) 2 + 4 = s 2 + 4 is , so the equation is ¨ u + 4 i ˙ u = t . To solve this, write v = ˙ u , so ˙ v + 4 iv = t . Solve this by undetermined coefficents: v = at + b , ˙ v = a , so a + 4 i ( at + b ) = t . This forces a = 1 / 2 i = i 4 and a + 4 ib = 0 or b = a 4 i = 1 16 . So v p = it 4 + 1 16 , u p = it 2 8 + t 16 , z p = e 2 it u p , x p = Re z p = t 2 sin(2 t ) 8 + t cos(2 t ) 16 (b) [10] If m, b, k ≥ 0 and m 6 = 0, the roots of the characteristic polynomial have nonpositive real parts. As a result, the homogeneous solutions, which are made up of e rt ’s where r is a root, do not grow without bound as t → ∞ —unless there is a multiple root with zero real part. Since the nonreal roots occur in unequal conjugate pairs, the roots must both be zero, and the characteristic polyomial is p ( s ) = ms 2 . So that is one possibility: b = k = 0. This also shows that if p ( s ) is not of this form and one solution to p ( D ) x = B cos( ωt ) is bounded as t → ∞ , then all solutions are bounded as t → ∞ . The ERF tells us that there is a sinusoidal—and hence bounded—solution to p ( D ) x = B cos( ωt ) unless p ( iω ) = 0, i.e. mω 2 + bωi + k = 0. This forces b = 0 and ω = p k/m . In that case the resonance...
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This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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