Explanation Given that, the differential equation of a spring/mass system is x ″ + 16 x = 0 . The boundary conditions are, x ( 0 ) = − 1 and x ′ ( 0 ) = 3 . Here, minus sign indicates that the mass is released from the above the equilibrium position of the spring. The homogeneous equation can also be written as, D 2 x + 16 = 0 (1) Auxiliary equation of the equation (1) can be written using x ′ = m as, m 2 + 16 = 0 m = − 16 m = ± 4 i The solution of differential equation is, x ( t ) = c 1 cos 4 t + c 2 sin 4 t (2) Differentiate the equation (2) with respect to x . x ′ ( t ) = − c 1 4 sin 4 t + c 2 4 cos 4 t Re-arrange the terms to obtain x ′ ( t ) = 4 ( − c 1 sin 4 t + c 2 cos 4 t ) (3) Apply the boundary condition x ( 0 ) = − 1 in equation (2)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3 Solutions