Explanation Given: The boundary conditions are as follows: k ∂ 2 u ∂ x 2 + x e − 2 t = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 0 , t ) = 0 ∂ u ∂ x | x = 1 = − u ( 1 , t ) u ( x , 0 ) = 0 Calculation: The boundary conditions are given as follows: k ∂ 2 u ∂ x 2 + x e − 2 t = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 0 , t ) = 0 ∂ u ∂ x | x = 1 = − u ( 1 , t ) u ( x , 0 ) = 0 Consider the homogenous form of the given differential equation as follows: k ∂ 2 u ∂ x 2 = ∂ u ∂ t … … ( 1 ) Consider the product solution of the equation as follows: u ( x , t ) = X ( x ) T ( t ) … … ( 2 ) Substitute u = X T in equation ( 1 ) . k ∂ 2 ( X T ) ∂ x 2 = ∂ ( X T ) ∂ t k X ″ T = X T ′ X ″ X = T ′ k T Separate the variables with the help of separation constant and take the separation constant λ as follows: X ″ X = T ′ k T = − λ The separated equations are written as follows: X ″ + λ X = 0 … … ( 3 ) T ′ + k λ T = 0 … … ( 4 ) Substitute 0 for x in equation ( 2 ) . u ( 0 , t ) = X ( 0 ) T ( t ) X ( 0 ) T ( t ) = 0 ( u ( 0 , t ) = 0 ) X ( 0 ) = 0 Substitute 0 for t in equation ( 2 ) . u ( x , 0 ) = X ( x ) T ( 0 ) X ( x ) T ( 0 ) = 0 ( u ( x , 0 ) = 0 ) T ( 0 ) = 0 Differentiate equation ( 1 ) with respect to x . u ′ ( x , t ) = X ′ ( x ) T ( t ) Substitute 1 for x in the above equation. u ′ ( 1 , t ) = X ′ ( 1 ) T ( t ) X ′ ( 1 ) T ( t ) = − u ( 1 , t ) ( u ′ ( 1 , t ) = − u ( 1 , t ) ) X ′ ( 1 ) T ( t ) = − X ( 1 ) T ( t ) X ′ ( 1 ) = − X ( 1 ) There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 3 ) . X ″ = 0 The general solution of the above equation is as follows: X ( x ) = c 1 + c 2 x … … ( 5 ) Substitute 0 for x in equation ( 5 ) . X ( 0 ) = c 1 + c 2 ⋅ 0 c 1 = 0 ( X ( 0 ) = 0 ) Substitute 0 for c 1 in equation ( 5 ) . X ( x ) = c 2 x Differentiate the above equation with respect to x . X ′ ( x ) = c 2 Substitute 1 for x in the above equation. X ′ ( 1 ) = c 2 c 2 = − X ( 1 ) ( X ′ ( 1 ) = − X ( 1 ) ) c 2 = − c 2 c 2 = 0 Since, the value of c 1 and c 2 is 0 so, for λ = 0 non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute λ = − α 2 in equation ( 3 ) . X ″ − α 2 X = 0 The general solution of the above equation is as follows: X ( x ) = c 3 cos h ( α x ) + c 4 sin h ( α x ) … … ( 6 ) Substitute 0 for x in equation ( 6 ) . X ( 0 ) = c 3 cos h ( 0 ) + c 4 sin h ( 0 ) c 3 = 0 ( X ( 0 ) = 0 ) Substitute 0 for c 3 in equation ( 6 ) . X ( x ) = c 4 sin h ( α x ) Differentiate the above equation with respect to x . X ′ ( x ) = α c 4 cosh ( α x ) Substitute 1 for x in the above equation. X ′ ( 1 ) = α c 4 cosh ( α ) α c 4 cosh ( α ) = − X ( 1 ) ( X ′ ( 1 ) = − X ( 1 ) ) α c 4 cosh ( α ) = c 4 sin h ( α ) α c 4 cosh ( α ) − c 4 sin h ( α ) = 0 Further solve the above equation as follows: α c 4 cosh ( α ) − c 4 sin h ( α ) = 0 c 4 ( α cosh ( α ) − sin h ( α ) ) = 0 c 4 = 0 Since, the value of c 3 and c 4 so, for λ = − α 2 non-trivial solution does not exist. Third case is λ = α 2 ( λ > 0 ) . Substitute λ = α 2 in equation ( 3 ) . X ″ + α 2 X = 0 The general solution of the above equation is as follows: X ( x ) = c 5 cos ( α x ) + c 6 sin ( α x ) … … ( 7 ) Substitute 0 for x in equation ( 7 ) . X ( 0 ) = c 5 cos ( 0 ) + c 6 sin ( 0 ) c 5 = 0 ( X ( 0 ) = 0 ) Substitute 0 for c 5 in equation ( 7 ) . X ( x ) = c 6 sin ( α x ) … … ( 8 ) Differentiate equation ( 8 ) with respect to x . X ′ ( x ) = α c 6 cos ( α x ) Substitute 1 for x in the above equation

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 12.7, Problem 10E

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The solution u(x,t) of the differential equation k∂2u∂x2+xe−2t=∂u∂t such that u(0,t)=0, ∂u∂x|x=1=−u(1,t) and u(x,0)=0.

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Chapter 12.7, Problem 9E

Chapter 12.8, Problem 1E

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The solution u ( x , t ) of the differential equation k ∂ 2 u ∂ x 2 + x e − 2 t = ∂ u ∂ t such that u ( 0 , t ) = 0 , ∂ u ∂ x | x = 1 = − u ( 1 , t ) and u ( x , 0 ) = 0 .

Solution Summary: The author describes the solution of the differential equation u(x,t), based on the boundary conditions.

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The solution u(x,t) of the differential equation k∂2u∂x2+xe−2t=∂u∂t such that u(0,t)=0, ∂u∂x|x=1=−u(1,t) and u(x,0)=0.

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Chapter 12 Solutions