Unformatted text preview: Exercises 4.9
11. First, we ﬁnd the mass m= 8 lb 1 = slug. 32 ft/sec2 4 Thus the equation (23), describing the motion, with m = 1/4, b = 1, k = 10, and the external force F (t) = 2 cos 2t becomes 1 y + y + 10y = 2 cos 2t, 4 (4.14) with the initial conditions are y (0) = y (0) = 0. A general solution to the corresponding homogeneous equation is given in Section 4.8, formula (16). That is, yh (t) = eαt (C1 cos βt + C2 sin βt) . We compute α=− So, yh (t) = e−2t (C1 cos 6t + C2 sin 6t) . For a particular solution, we use formula (7), page 221 in the text. yp (t) = = F0 sin(γt + θ) (k − mγ 2 )2 + b2 γ 2 2 2 sin(2t + θ) = √ sin(2t + θ), 85 [10 − (1/4)(2)2]2 + (1)2 (2)2 b 1 1 =− = −2 and β = 2m 2(1/4) 2(1/4) 4(1/4)(10) − 12 = 6. where θ = arctan[(k − mγ 2 )/(bγ )] = arctan(9/2) ≈ 1.352 . A general solution to (4.14) is then given by 2 y (t) = e−2t (C1 cos 6t + C2 sin 6t) + √ sin(2t + θ) . 85 From the initial conditions, we ﬁnd √ y (0) = C1 + (2/ 85) sin θ = 0, √ y (0) = −2C1 + 6C2 + (4/ 85) cos θ = 0 √ C1 = −(2/ 85) sin θ = −18/85, ⇒ √ C2 = C1 − (2/ 85) cos θ /3 = −22/255. 245 ...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Homogeneity

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