To determine The steady state motion of the mass spring system modeled by the ordinary differential equation ( D 2 + D + 4.25 I ) y = 22.1 cos 4.5 t . Explanation The given ordinary differential equation is ( D 2 + D + 4.25 I ) y = 22.1 cos 4.5 t . Since d 2 y d x = D 2 y , d y d x = D y and I ( y ) = y , the equation ( D 2 + D + 4.25 I ) y = 22.1 cos 4.5 t becomes, ( D 2 + D + 4.25 I ) y = 22.1 cos 4.5 t D 2 y + D y + 4.25 I ( y ) = 22.1 cos 4.5 t y ″ + y ′ + 4.25 y = 22.1 cos 4.5 t The steady state motion of the system modeled by the ordinary differential equation y ″ + y ′ + 4.25 y = 22.1 cos 4.5 t is the particular solution of y ″ + y ′ + 4.25 y = 22.1 cos 4.5 t . Since the right hand side function is of the form K cos w x , the choice for particular solution is K cos w x + M sin w x . Therefore, the particular solution of y ″ + 6 y ′ + 8 y = 42.5 cos 2 t becomes y p = K cos 4.5 t + M sin 4.5 t . Differentiate y p = K cos 4.5 t + M sin 4.5 t with respect to t as shown below. y p ′ = d d t ( K cos 4.5 t + M sin 4.5 t ) = K ( − sin 4.5 t ) ( 4.5 ) + M cos 4.5 t ( 4.5 ) = − 4.5 K sin 4.5 t + 4.5 M cos 4.5 t Again differentiate y p ′ with respect to t and obtain y p ′ ′ as follows. y p ′ ′ = d d t ( − 4.5 K sin 4.5 t + 4.5 M cos 4.5 t ) = − 4.5 K cos 4.5 t ( 4.5 ) + 4.5 M ( − sin 4.5 t ) ( 4.5 ) = − 20.25 K cos 4.5 t − 20.25 M sin 4.5 t Since the particular solution satisfies the given ordinary differential equation, substitute y p , y p ′ , and y p ′ ′ in y ″ + y ′ + 4.25 y = 22.1 cos 4.5 t as shown below. y p ′ ′ + y p ′ + 4.25 y p = 22.1 cos 4.5 t { − 20.25 K cos 4.5 t − 20.25 M sin 4.5 t − 4.5 K sin 4.5 t + 4.5 M cos 4.5 t + 4.25 ( K cos 4.5 t + M sin 4.5 t ) } = 22.1 cos 4

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Chapter 2.8, Problem 5P

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The steady state motion of the mass spring system modeled by the ordinary differential equation (D2+D+4.25I)y=22.1cos4.5t.

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