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Unformatted text preview: Chapter 5
x2 = 2 x3 = 0, x4 = 0, etc. Observe that x2 = 22 In general, xj = 2j Consequently, xn = 0 for n ≥ j . 11. (a) A general solution to equation (6) is given by x(t) = xh (t) + xp (t), where xh (t) = Ae−0.11t sin √ 9879t + φ k 2j (mod 1) = k (mod 1) = 0. 3 22 (mod 1) = 3 (mod 1) = 0. 1 (mod 1) = 1 (mod 1) = 0, 2 is the transient term (a general solution to the corresponding homogeneous equation) and xp (t) = 1 sin t + 0.22 √ 1 sin 2t + ψ , 1 + 2(0.22)2 tan ψ = − 1 √, 0.22 2 is the steadystate term (a particular solution to (6)). (xp (t) can be found, say, by applying formula (7), Section 4.12, and using Superposition Principle of Section 4.7.) Diﬀerentiating x(t) we get 1 v (t) = xh (t) + xp (t) = xh (t) + cos t + 0.22 √ 2 cos √ 2t + ψ . 1 + 2(0.22)2 The steadystate solution does not depend on initial values x0 and v0 ; these values aﬀect only constants A and φ in the transient part. But, as t → ∞, xh (t) and xh (t) tend to zero and so the values of x(t) and v (t) approach the values of xp (t) and xp (t), respectively. Thus the limit set of points (x(t), v (t)) is the same as that of (xp (t), xp (t)) which is independent of initial values. 330 ...
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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