Explanation Given: The boundary conditions are as follows: u ( 0 , t ) = 0 u ( 1 , t ) = 0 u ( x , 0 ) = 0 ∂ u ∂ t | t = 0 = 2 sin π x + 4 sin 3 π x or u t ( x , 0 ) = 2 sin π x + 4 sin 3 π x Calculation: Take Laplace transform on both sides of equation ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 as follows: L { ∂ 2 u ∂ x 2 } = L { ∂ 2 u ∂ t 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 ( 1 , t ) = 0 as follows: L { u ( 1 , t ) } = L { 0 } U ( 1 , s ) = 0 ...... ( 3 ) Substitute u ( x , 0 ) = 0 and u t ( x , 0 ) = 2 sin π x + 4 sin 3 π x in equation (1). d 2 U d x 2 = s 2 U ( x , s ) − 2 sin π x − 4 sin 3 π x d 2 U d x 2 − s 2 U ( x , s ) = − 2 sin π x − 4 sin 3 π x Solve the differential equation by the method undetermined coefficient. The auxiliary equation for the above differential equation as follows: r 2 − s 2 = 0 Solve above auxiliary equation as follows: r 2 − s 2 = 0 r 2 = s 2 r = ± s ( Square root ) Therefore, the complementary function of the for differential equation d 2 U d x 2 − s 2 U ( x , s ) = − 2 sin π x − 4 sin 3 π x is shown below. C .F = c 1 cosh s x + c 2 sinh s x The particular integral of the differential equation d 2 U d x 2 − s 2 U ( x , s ) = − 2 sin π x − 4 sin 3 π x is obtained as follows: ( D 2 − s 2 ) U ( x , s ) = − 2 sin π x − 4 sin 3 π x P .I = − 2 sin π x D 2 − s 2 − 4 sin 3 π x D 2 − s 2 ( P

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 14.2, Problem 2E

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The solution of boundary value problem ∂2u∂x2=∂2u∂t2.

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Chapter 14 Solutions

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

Ch. 14.1 - (a) Show that erf(t)=10ted. (b) Use part (a), the...Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Use the third and fifth entries in Table 14.1.1 to...

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The solution of boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 .

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The solution of boundary value problem ∂2u∂x2=∂2u∂t2.

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Chapter 14 Solutions