ps04sol

# ps04sol - 18.03 Problem Set 4 Fall 2009 Solutions 1. (6...

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Unformatted text preview: 18.03 Problem Set 4 Fall 2009 Solutions 1. (6 points) Find all the real-valued solutions of the differential equation d 6 y dx 6- y = 0 . The characteristic equation is r 6- 1 = 0. As on the previous problem set, the roots of this are ± 1 and ± 1 2 ± √ 3 2 i. We have the following 6 linearly independent (complex) solutions: e x , e- x , e ( 1 2 + √ 3 2 i ) x = e x/ 2 cos √ 3 2 x + i sin √ 3 2 x , e ( 1 2- √ 3 2 i ) x = e x/ 2 cos √ 3 2 x- i sin √ 3 2 x , e (- 1 2 + √ 3 2 i ) x = e- x/ 2 cos √ 3 2 x + i sin √ 3 2 x , e (- 1 2- √ 3 2 i ) x = e- x/ 2 cos √ 3 2 x- i sin √ 3 2 x . A complex linear combination of these will be real-valued precisely when each complex conjugate pair forms a real solution already, and so the first two solutions each need real coefficients, and we may use our typical analysis on the last two pairs. Thus the general real solution is y ( x ) = C 1 e x + C 2 e- x + C 3 e x/ 2 cos √ 3 2 x + C 4 e x/ 2 sin √ 3 2 x + C 5 e- x/ 2 cos √ 3 2 x + C 6 e- x/ 2 sin √ 3 2 x, for arbitrary real constants C 1 ,...,C 6 . 2. (12 points) The “spring and dashpot” system described on page 101 of the text satisfies the differential equation m d 2 x dt 2 + c dx dt + kx = 0 . ( Damped Spring ) a) Calculate the potential energy of the spring at position x (normalizing it to be zero when x = 0 ). As usual for this setup (or by integration), the potential energy (up to to an additive constant) is 1 2 kx 2 , and in fact this is already 0 when x = 0....
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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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ps04sol - 18.03 Problem Set 4 Fall 2009 Solutions 1. (6...

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