Explanation Given: The given boundary value problem is ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < π , t > 0 and the given boundary conditions are u ( 0 , t ) = 0 , u ( π , t ) = 0 and u ( x , 0 ) = sin x . Calculation: The given boundary value problem is, ∂ 2 u ∂ x 2 = ∂ u ∂ t . (1) Consider the equation, u ( x , t ) = X ( x ) T ( t ) . (2) Separate the variables of equation (2), X ″ X = − α 2 and T ′ T = − α 2 The equation of the variable X is, X ″ + α 2 X = 0 . (3) The equation of the variable T is, T ′ + α 2 T = 0 . (4) The general solution of the equation (3) is, X ( x ) = c 1 cos α x + c 2 sin α x . (5) The general solution of the equation (4) is, T ( t ) = c 3 e α 2 t + c 4 e − α 2 t . (6) At boundary conditions u ( 0 , t ) = 0 , u ( π , t ) = 0 and u ( x , 0 ) = sin x , c 3 = 0 For non-trivial solution c 4 ≠ 0 , α = n

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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