Explanation Given: A string is stretched along the x axis between ( 0 , 0 ) and ( L , 0 ) . Calculation: The wave equation is a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 . The boundary conditions for the string are obtained as follows: u ( 0 , t ) = 0 and u ( L , t ) = 0 Take Laplace transform on both sides of equation a 2 ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 as follows: a 2 L { ∂ 2 u ∂ x 2 } = L { ∂ 2 u ∂ t 2 } a 2 d 2 U d x 2 = s 2 U ( x , s ) − s u ( x , 0 ) − u t ( x , 0 ) ..... ( 1 ) Take Laplace transform of boundary condition u ( 0 , t ) = 0 as follows: L { u ( 0 , t ) } = L { 0 } U ( 0 , s ) = 0 ..... ( 2 ) Take Laplace transform of boundary condition u ( L , t ) = 0 as follows: L { u ( L , t ) } = L { 0 } U ( L , s ) = 0 ...... ( 3 ) Substitute u ( x , 0 ) = A sin π L x and u t ( x , t ) = 0 in equation (1). a 2 d 2 U d x 2 = s 2 U ( x , s ) − s A sin π L x − ( 0 ) a 2 d 2 U d x 2 = s 2 U ( x , s ) − s A sin π L x d 2 U d x 2 − s 2 a 2 U ( x , s ) = − s a 2 A sin π L x Solve the differential equation by the method undetermined coefficient. The auxiliary equation for the above differential equation as follows: r 2 − s 2 a 2 = 0 Solve the above auxiliary equation as follows: r 2 − s 2 a 2 = 0 r 2 = s 2 a 2 r = ± s a ( Square root ) Therefore, the complementary function of the differential equation d 2 U d x 2 − s 2 a 2 U ( x , s ) = − s a 2 A sin π L x is shown below. C .F = c 1 cosh s a x + c 2 sinh s a x The particular integral of the differential equation d 2 U d x 2 − s 2 a 2 U ( x , s ) = − s a 2 A sin π L x is obtained as follows: ( D 2 − s 2 a 2 ) U ( x , s ) = − s a 2 A sin π L x P .I = 1 D 2 − ( s a ) 2 ( − s a 2 A sin π L x ) ( P .I = U ( x , s ) ) P .I = − s A a 2 ( sin ( π L ) x D 2 − ( s a ) 2 ) P

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 15.2, Problem 1E

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The displacement u(x,t) if the string starts from rest in the initial position Asin(πxL).

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Chapter 15.1, Problem 15E

Chapter 15.2, Problem 2E

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(b) Let I[y] be a functional of y(x) defined by [[y] = √(x²y' + 2xyy' + 2xy + y²) dr, subject to boundary conditions y(0) = 0, y(1) = 1. State the Euler-Lagrange equation for finding extreme values of I [y] for this prob- lem. Explain why the function y(x) = x is an extremal, and for this function, show that I = 2. Without doing further calculations, give the values of I for the functions y(x) = x² and y(x) = x³.

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L sin 2x (1+ cos 3x) dx 59

Chapter 15 Solutions

Advanced Engineering Mathematics

Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E Ch. 15.1 - Prob. 6E Ch. 15.1 - Prob. 8E Ch. 15.1 - Prob. 11E Ch. 15.1 - Prob. 12E Ch. 15.1 - Prob. 13E Ch. 15.1 - Prob. 14E

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The displacement u ( x , t ) if the string starts from rest in the initial position A sin ( π x L ) .

Solution Summary: The author calculates the displacement of a string by taking Laplace transforms on both sides of the equation.

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The displacement u(x,t) if the string starts from rest in the initial position Asin(πxL).

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Chapter 15 Solutions