Explanation Given: The displacement functions x ( t ) and y ( t ) for a couple mass-spring system (similar to the one discussed in Problem 6 ) satisfy the initial value problem x ″ ( t ) + 5 x ( t ) − 2 y ( t ) = 0 ... ( 1 ) y ″ ( t ) + 2 y ( t ) − 2 x ( t ) = 3 sin 2 t … ( 2 ) The initial conditions are given as, x ( 0 ) = 0 x ′ ( 0 ) = 0 . y ( 0 ) = 1 . y ′ ( 0 ) = 0 . Approach: The Hooke’s law states that the force acting on an object due to a spring has magnitude proportional to the displacement of the spring from its natural length and has direction opposite to its displacement. As each mass attached to the spring is free to move, Newton’s law is applied to each object. Newton’s law states that if a body of mass m is moving with an acceleration a , then the force acting on the body is given as, F = m a . Calculation: Express equation ( 1 ) in terms of y as follows, x ″ ( t ) + 5 x ( t ) − 2 y ( t ) = 0 2 y ( t ) = x ″ ( t ) + 5 x ( t ) y ( t ) = x ″ ( t ) + 5 x ( t ) 2 … ( 3 ) Differentiate equation ( 3 ) with respect to t as, y ( t ) = x ″ ( t ) + 5 x ( t ) 2 y ′ ( t ) = x ‴ ( t ) + 5 x ′ ( t ) 2 y ″ ( t ) = x ( 4 ) ( t ) + 5 x ″ ( t ) 2 Substitute the value of y and y ″ in equation ( 2 ) as, y ″ ( t ) + 2 y ( t ) − 2 x ( t ) = 3 sin 2 t x ( 4 ) ( t ) + 5 x ″ ( t ) 2 + 2 x ″ ( t ) + 5 x ( t ) 2 − 2 x ( t ) = 3 sin 2 t x ( 4 ) ( t ) + 5 x ″ ( t ) + 2 x ″ ( t ) + 10 x ( t ) − 4 x ( t ) = 6 sin 2 t x ( 4 ) ( t ) + 7 x ″ ( t ) + 6 x = 6 sin 2 t Therefore, the differential equation is given as, ( D 4 + 7 D 2 + 6 ) x = 6 sin ( 2 t ) ( 4 ) Here, D = d d t . The auxiliary equation is given as, m 4 + 7 m 2 + 6 = 0 ( 5 ) Solve equation ( 5 ) as follows, m 4 + 7 m 2 + 6 = 0 m 2 = − 7 ± 49 − 24 2 m 2 = − 7 ± 5 2 m 2 = − 1 , − 6 m = ± i 6 , ± i Therefore, the general equation for x is given as, x ( t ) = A cos 6 t + B sin 6 t + C cos t + D sin t … ( 6 ) Differentiate equation ( 6 ) with respect to t as, x ′ ( t ) = − 6 A sin 6 t + 6 B cos 6 t − C sin t + D cos t … ( 7 ) x ″ ( t ) = − 6 A cos 6 t − 6 B sin 6 t − C cos t − D sin t … ( 8 ) Substitute the value of x and x ″ in equation ( 3 ) as, y ( t ) = x ″ ( t ) + 5 x ( t ) 2 y ( t ) = 1 2 [ − 6 A cos 6 t − 6 B sin 6 t − C cos t − D sin t + 5 ( A cos 6 t + B sin 6 t + C cos t + D sin t ) ] y ( t ) = 1 2 ( − A cos 6 t − B sin 6 t + 4 C cos t + 4 D sin t ) Therefore, the general equation for y is given as, y ( t ) = 1 2 ( − A cos 6 t − B sin 6 t + 4 C cos t + 4 D sin t ) … ( 9 ) Differentiate equation ( 9 ) with respect to t as, y ′ ( t ) = 1 2 ( 6 A sin 6 t − 6 B cos 6 t − 4 C sin t + 4 D cos t ) … ( 10 ) Substitute the initial conditions y ( 0 ) = 1 in equation ( 9 ) as,

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