250_pdfsam_math 54 differential equation solutions odd

250_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Chapter 4 ⇒ y (t) = e−2t − 22 2 18 cos 6t − sin 6t + √ sin(2t + θ) . 85 255 85 √ √ 22 40 − 8 = , 2π π The resonance frequency for the system is γr = 2π (k/m) − (b2 )/(2m2 ) = 2π where we have used formula (15) on page 223 in the text for γr . 13. The mass attached to the spring is m= 32 lb = 1 slug. 32 ft/sec2 Thus the equation governing the motion, my + by + ky = Fext , with m = 1, b = 2, k = 5, and Fext (t) = 3 cos 4t becomes y + 2y + 5y = 3 cos 4t. This is an underdamped motion because b2 − 4mk = (2)2 − 4(1)(5) = −16 < 0. For the steady-state solution of this equation we use formula (6) on page 221 in the text. Since Fext (t) = 3 cos 4t, we have F0 = 3, and γ = 4. Substituting m, b, k , F0 , and γ into (6), we obtain yp (t) = 3 (1)(4)2 ]2 [5 − + (2)2 (4)2 3 = (8 sin 4t − 11 cos 4t) . 185 page 228 [5 − (1)(4)2] cos 4t + (2)(4) sin 4t REVIEW PROBLEMS: 1. Solving the auxiliary equation, r 2 + 8r − 9 = 0, we find r1 = −9, r2 = 1. Thus a general solution is given by y (t) = c1 er1 t + c2 er2 t = c1 e−9t + c2 et . 3. The auxiliary equation, 4r 2 − 4r + 10 = 0, has roots r1,2 = (1 ± 3i)/2. Therefore a general solution is y (t) = c1 cos 246 3t 2 + c2 sin 3t 2 et/2 . ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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