Explanation Given: The spring mass system is given by the differential equation as follows. m y ″ + b y ′ + k y = g ( t ) ... ( 1 ) Here, m > 0 is the mass of the spring system, b > 0 is the damping constant, k > 0 is the spring constant, g ( t ) is the force on the system at time t , and y ( t ) is the displacement from the equilibrium of the spring system at time t . The damping constant is b = 2 , the mass is m = 4 , the spring constant is k = 5 , and external force is g ( t ) = 2 sin 3 t + 10 cos 3 t . The boundary condition are given as y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . Approach: The complementary solution for the differential equation if the roots of the auxiliary function are imaginary is given by, y c = c 1 e α t cos β t + c 2 e α t sin β t Here, the roots of the auxiliary equation are α + i β and α − i β and c 1 , c 2 are arbitrary constants. The general solution of the differential equation is given by, y = y c + y p Here, y p is the particular solution and y c is the complementary solution. Calculation: Substitute 1 for m , 5 for k , 2 for b , and ( 2 sin 3 t + 10 cos 3 t ) for g ( t ) in the equation ( 1 ) . ( 1 ) y ″ + ( 2 ) y ′ + ( 5 ) y = ( 2 sin 3 t + 10 cos 3 t ) y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t ... ( 2 ) The auxiliary equation of equation ( 2 ) is as follows. r 2 + 2 r + 5 = 0 Solve the above equation using the solution of quadratic equation. r 2 + 2 r + 5 = 0 r 2 + 2 r + 1 + 4 = 0 ( r + 1 ) 2 + 4 = 0 ( r + 1 ) 2 = − 4 Further solve, ( r + 1 ) = 2 i r = − 1 ± 2 i So the roots are, r 1 = − 1 + 2 i r 2 = − 1 − 2 i The complementary solution is as follows. y c = c 1 e − t cos ( 2 ) t + c 2 e − t sin ( 2 ) t Assume the particular solution of the differential equation as follows. y p = A sin 3 t + B cos 3 t ... ( 3 ) Differentiate the above equation with respect to t . y ′ = d d t ( A sin 3 t + B cos 3 t ) = 3 A cos 3 t − 3 B sin 3 t Again differentiate the above equation with respect to t . y ″ = d d t ( 3 A cos 3 t − 3 B sin 3 t ) = − 9 A sin 3 t − 9 B cos 3 t Substitute − 9 A sin 3 t − 9 B cos 3 t for y ″ , 3 A cos 3 t − 3 B sin 3 t for y ′ , and A sin 3 t + B cos 3 t for y in the equation ( 2 ) . ( − 9 A sin 3 t − 9 B cos 3 t + 2 ( 3 A cos 3 t − 3 B sin 3 t ) + 5 ( A sin 3 t + B cos 3 t ) ) = 2 sin 3 t + 10 cos 3 t ( − 9 A sin 3 t − 9 B cos 3 t + 6 A cos 3 t − 6 B sin 3 t + 5 A sin 3 t + 5 B cos 3 t ) = 2 sin 3 t + 10 cos 3 t ( − 9 A − 6 B + 5 A ) cos 3 t + ( − 9 B + 6 A + 5 B ) sin 3 t = 2 sin 3 t + 10 cos 3 t ( − 4 A − 6 B ) cos 3 t + ( 6 A − 4 B ) sin 3 t = 2 sin 3 t + 10 cos 3 t ... ( 4 ) Compare the coefficients of sin 3 t on both sides in above equation. − 4 A − 6 B = 2 4 A = − 2 − 6 B A = − 2 − 6 B 4 A = − 1 − 3 B 2

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4.5, Problem 44E

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The equation of motion for the spring system.

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Chapter 4.5, Problem 43E

Chapter 4.5, Problem 45E

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Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

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Solution Summary: The author explains how the equation of motion for the spring system is given by the differential equation.

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The equation of motion for the spring system.