312_pdfsam_math 54 differential equation solutions odd

312_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 For the mass m 2 there are two forces acting on it: the force due to the spring with constant k 2 is k 2 y ; and the force due to the spring with constant k 1 is k 1 ( y x ). So we get m 2 y 0 = k 1 ( x y ) k 2 y. So the system is m 1 x 0 = k 1 ( y x ) , m 2 y 0 = k 1 ( y x ) k 2 y, or, in operator form, ( m 1 D 2 + k 1 ) [ x ] k 1 y =0 , k 1 x + ± m 2 D 2 +( k 1 + k 2 ) ² [ y ]=0 . With m 1 =1 , m 2 =2, k 1 =4 ,and k 2 =10 / 3, we get ( D 2 +4)[ x ] 4 y , 4 x +(2 D 2 +22 / 3) [ y , (5.28) with initial conditions: x (0) = 1 ,x 0 (0) = 0 ,y (0) = 0 0 (0) = 0 . Multiplying the second equation of the system given in (5.28) by 4, applying (2
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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.

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