Explanation Given: The given heat equation is k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , the initial temperature is constant u 0 , the rod is insulated at x = 0 , the temperature is kept zero at x = 1 and the given identity is 1 1 + e − 2 s k expand in a geometric series. Calculation: The given heat equation is, k ∂ 2 u ∂ x 2 = ∂ u ∂ t (1) Take Laplace transform of both sides of the above equation, k L { ∂ 2 u ∂ x 2 } = L { ∂ u ∂ t } k d 2 U d x 2 = s U ( x , s ) − u ( x , 0 ) Substitute u ( x , 0 ) = u 0 s in the above equation, k d 2 U d x 2 − s U ( x , s ) = − u 0 s The particular solution of the above equation is, U ( x , s ) = c 1 cosh s k x + c 2 sinh s k x + u 0 s (2) At the given boundary conditions x = 0 and x = L , d U d x | x = 0 = 0 and U ( 1 , s ) = 0 c 1 = − u 0 s cosh s k and c 2 = 0 Substitute the values of c 1 and c 2 in equation (2), U ( x , s ) = u 0 s − u 0 cosh s k x s cosh s k = u 0 s − u 0 e s k x + e − s k x s ( e s k + e s k ) = u 0 s − u 0 e s k ( x − 1 ) + e − s k ( x + 1 ) s ( 1 + e − 2 s k ) = u 0 s − u 0 [ e s k ( x − 1 ) s ( 1 + e − 2 s k )

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 23E

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To solve: The heat equation for the temperature u(x,t).

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Chapter 14.2, Problem 22E

Chapter 14.2, Problem 24E

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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 heat equation for the temperature u ( x , t ) .

Solution Summary: The author explains the solution of the heat equation for the temperature u(x,t).

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To solve: The heat equation for the temperature u(x,t).

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Chapter 14 Solutions