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Unformatted text preview: Chapter 6
(c) As we have mentioned in (a), the ﬁrst equation in (6.7) implies that y = (x + 5x) /2. Substituting the solution x(t) yields y (t) = 1 2 √ √ c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t √ √ +5 c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t √ √ 1 −c1 cos t − c2 sin t − 6c3 cos 6t − 6c4 sin 6t = 2 √ √ +5 c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t √ √ c3 c4 cos 6t − sin 6t . = 2c1 cos t + 2c2 sin t − 2 2 (d) Initial conditions x(0) = y (0) = 1 and x (0) = y (0) = 0 imply the system of linear equations for c1 , c2 , c3 , and c4 . Namely, x(0) = c1 + c3 = 1, y (0) = 2c1 − (c3 /2) = 1, √ x (0) = c2 + c4 6 = 0, √ y (0) = 2c2 − (c4 6/2) = 0 Thus, the solution to this initial value problem is x(t) = √ 2 3 cos t + cos 6t , 5 5 y (t) = √ 1 6 cos t − cos 6t . 5 5 ⇒ c1 = 3 / 5 , c3 = 2 / 5 , c2 = 0 , c4 = 0 . 35. Solving the characteristic equation yields EIr 4 − k = 0 ⇒ ⇒ r2 = ⇒ or or r4 = k EI k EI k k 4 = ±i . − EI EI k EI k 4 r=± EI r2 = − r=±
4 The ﬁrst two roots are real numbers, the other two are pure imaginary numbers. Therefore, a general solution to the vibrating beam equation is √ √ y (x) = C1 e k/(EI )x + C2 e− k/(EI )x + C3 sin 360
4 k x EI + C4 cos 4 k x. EI ...
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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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