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Unformatted text preview: Exercises 5.5
By solving these equations simultaneously, we ﬁnd c1 = −1, c2 = −1, c3 = −1, and c4 = 0. Therefore, the solution to this springmassdashpot system is x(t) = −e−t − te−t − cos t, 7. In operator notations, D 2 + 5 [x] − 2y = 0, −2x + D 2 + 2 [y ] = 3 sin 2t. Multiplying the ﬁrst equation by (D 2 + 2) and the second equation by 2, and adding the results, we obtain D2 + 2 ⇒ ⇒ D 2 + 5 − 4 [x] = 6 sin 2t D 2 + 6 [x] = 6 sin 2t . y (t) = e−t + te−t − cos t. D 4 + 7D 2 + 6 [x] = 6 sin 2t D2 + 1 (5.35) √ Since the characteristic equation, (r 2 + 1)(r 2 + 6) = 0, has the roots r = ±i and r = ±i 6, a general solution to the corresponding homogeneous equation is given by √ √ xh (t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t . Due to the righthand side in (5.35), a particular solution has the form xp (t) = A cos 2t + B sin 2t . In order to simplify computations, we note that both functions, cos 2t and sin 2t, and so xp (t), satisfy the diﬀerential equation (D 2 + 4)[x] = 0. Thus, D2 + 1 D 2 + 6 [xh ] = (D 2 + 4) − 3 (D 2 + 4) + 2 [xh ] = 2 (D 2 + 4) − 3 [xh ] = −6xh = −6A cos 2t − 6B sin 2t = 6 sin 2t 315 ...
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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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