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Unformatted text preview: y (0) = c 1 = 2 , y (0) = 4 c 2 = 0 ⇒ c 1 = 2 , c 2 = 0 ⇒ y ( t ) = 2 cos 4 t . (b) Positive damping b = 100 and stiFness k = 1 imply that the displacement y ( t ) tends to zero, as t → ∞ . To con±rm this prediction, we solve the given initial value problem explicitly. The roots of the associated equation are r = − 100 ± √ 100 2 − 4 2 = − 50 ± √ 2499 . Thus the roots r 1 = − 50 − √ 2499 and r 2 = − 50 + √ 2499 are both negative. A general solution is given by y ( t ) = c 1 e r 1 t + c 2 e r 2 t ⇒ y ( t ) = c 1 r 1 e r 1 t + c 2 r 2 e r 2 t . 183...
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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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