Explanation Given: The equation is m 1 x ″ 1 = − k 1 x 1 + k 2 ( x 2 − x 1 ) and m 2 x ″ 1 = − k 2 ( x 2 − x 1 ) , k 1 = 3 , k 2 = 2 , m 1 = 1 , m 2 = 1 and x 1 ( 0 ) = 0 , x ′ 1 ( 0 ) = 1 , x 2 ( 0 ) = 1 , x ′ 2 ( 0 ) = 0 . Concept: The Laplace Transform is defined as follows: L { t n f ( t ) } = ( − 1 ) n d n d s n F ( s ) Calculation: The equation is m 1 x ″ 1 = − k 1 x 1 + k 2 ( x 2 − x 1 ) and m 2 x ″ 1 = − k 2 ( x 2 − x 1 ) . Substitute the values of k 1 = 3 , k 2 = 2 , m 1 = 1 and m 2 = 1 in above equation. Therefore, the resulting equation is x ″ 1 = − 3 x 1 + 2 ( x 2 − x 1 ) and x ″ 2 = − 2 ( x 2 − x 1 ) respectively. Apply Laplace transformation on each solution is as follows: s 2 X 1 ( s ) − s x 1 ( 0 ) − x ′ 1 ( 0 ) = − 5 X 1 ( s ) + 2 X 2 ( s ) s 2 X 2 ( s ) − s x 2 ( 0 ) − x ′ 2 ( 0 ) = − 2 X 2 ( s ) + 2 X 1 ( s ) Substitute the initial condition 0 for x 1 ( 0 ) , 1 for x ′ ( 0 ) , 1 for x 2 ( 0 ) and 0 for x ′ 2 ( 0 ) in above equation. s 2 X 1 ( s ) − 0 − 1 = − 5 X 1 ( s ) + 2 X 2 ( s ) s 2 X 2 ( s ) − s − 0 = − 2 X 2 ( s ) + 2 X 1 ( s ) Further simplify the above equation as follows: ( s 2 + 5 ) X 1 ( s ) − 2 X 2 ( s ) = 1 − 2 X 1 ( s ) + ( s 2 + 2 ) X 2 ( s ) = s Further simplify the equation. L { x 1 } = s 2 + 2 s + 2 s 4 + 7 s 2 + 6 = s 2 + 2 s + 2 ( s 2 + 1 ) ( s 2 + 6 ) Apply partial fraction in above equation. s 2 + 2 s + 2 ( s 2 + 1 ) ( s 2 + 6 ) = A s + B ( s 2 + 1 ) + C s + D ( s 2 + 6 ) s 2 + 2 s + 2 = ( A s + B ) ( s 2 + 6 ) + ( C s + D ) ( s 2 + 1 ) s 2 + 2 s + 2 = A s 3 + 6 A s + B s 2 + 6 B + C s 3 + C s + D s 2 + D s 2 + 2 s + 2 = ( A + C ) s 3 + ( B + D ) s 2 + ( 6 A + C ) s + ( 6 B + D ) Equate coefficient of s , s 2 , s 3 on both sides. A + C = 0 B + D = 1 6 A + C = 2 6 B + D = 2 Subtract the equations A + C = 0 and 6 A + C = 2 as follows: A + C = 0 6 A + C = 2 _ ( Subtract ) − 5 A = − 2 Therefore, from above equation, the value of A is 2 5 . Substitute 2 5 for A in equation A + C = 0 to obtain the value of C . 2 5 + C = 0 C = − 2 5 Therefore, the value of C is − 2 5 . Subtract the equation 6 B + D = 2 and B + D = 1 . 6 B + D = 2 B + D = 1 _ ( Subtract ) 5 B = 1 Therefore, from above equation, the value of B is 1 5 . Substitute 1 5 for B in equation B + D = 1 to obtain the value of D

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 7.6, Problem 13E

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The solution of the system m1x″1=−k1x1+k2(x2−x1) and m2x″1=−k2(x2−x1) if k1=3,k2=2,m1=1,m2=1 and x1(0)=0,x′1(0)=1,x2(0)=1,x′2(0)=0 using Laplace transformation.

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Chapter 7.6, Problem 12E

Chapter 7.6, Problem 14E

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Chapter 7 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - In Problems 118 use Definition 7.1.1 to find f(t)....Ch. 7.1 - Prob. 10E

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The solution of the system m 1 x ″ 1 = − k 1 x 1 + k 2 ( x 2 − x 1 ) and m 2 x ″ 1 = − k 2 ( x 2 − x 1 ) if k 1 = 3 , k 2 = 2 , m 1 = 1 , m 2 = 1 and x 1 ( 0 ) = 0 , x ′ 1 ( 0 ) = 1 , x 2 ( 0 ) = 1 , x ′ 2 ( 0 ) = 0 using Laplace transformation.

Solution Summary: The author explains the Laplace transformation solution of the system m_1x′′

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The solution of the system m1x″1=−k1x1+k2(x2−x1) and m2x″1=−k2(x2−x1) if k1=3,k2=2,m1=1,m2=1 and x1(0)=0,x′1(0)=1,x2(0)=1,x′2(0)=0 using Laplace transformation.

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Chapter 7 Solutions