Explanation The equations is given as, ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + t x , 0 < x < π , t > 0 ... ( 1 ) The boundary conditions for this equation is, u ( 0 , t ) = u ( π , t ) = 0 , t > 0 ... ( 2 ) The initial conditions are, u ( x , 0 ) = sin x , 0 < x < π ... ( 3 ) and ∂ u ∂ t ( x , 0 ) = 5 sin 2 x − 3 sin 5 x , 0 < x < π ... ( 4 ) The solution of a wave equation is given by, u n ( x , t ) = u ( x , t ) + h ( x , t ) ... ( 5 ) where u ( x , t ) can be calculated by D Alembert’s test. u ( x , t ) = 1 2 [ f ( x + c t ) + f ( x − c t ) ] + 1 2 c ∫ x − c t x + c t g ( s ) d s u ( x , t ) = 1 2 [ sin ( x + t ) + sin ( x − t ) ] + 1 2 ∫ x − t x + t ( 5 sin 2 x − 3 sin 5 x ) d x u ( x , t ) = 1 2 ( 2 sin x cos t ) + 1 2 [ − 5 2 cos 2 x + 3 5 cos 5 x ] x + t x − t u ( x , t ) = sin x cos t + 1 2 [ − 5 2 ( cos 2 ( x − t ) − cos 2 ( x + t ) ) − 3 5 ( cos 5 ( x + t ) − cos 5 ( x − t ) ) ] Further solve, u ( x , t ) = sin x cos t + 1 2 [ − 5 2 ( 2 sin 2 x sin 2 t ) − 3 5 ( 2 sin 5 x sin 5 t ) ] u ( x , t ) = sin x cos t + 5 2 sin 2 x sin 2 t − 3 5 sin 5 x sin 5 t Put the value in equation (5), u n ( x , t ) = sin x cos t + 5 2 sin 2 x sin 2 t − 3 5 sin 5 x sin 5 t + h ( x , t ) ... ( 6 ) Now, h ( x , t ) = ∑ n = 1 ∞ h n ( t ) sin ( n π x l ) By variation of parameters, the value becomes, h ( x , t ) = ∑ n = 1 ∞ [ l n π ∫ 0 t h n ( s ) sin n π l ( t − s ) d s ] sin ( n π x l ) ... ( 7 ) where, h n ( s ) = 2 l ∫ 0 l h ( x , t ) sin n π x l d x h n ( s ) = 2 π ∫ 0 π t x sin n π x π d x h n ( s ) = 2 t π ∫ 0 π x sin n x d x h n ( s ) = 2 t π [ − x cos n x n | 0 π − ∫ 0 π − cos n x n d x ] Further solve,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 10.6, Problem 7E

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The formal solution to the given vibrating string problem ∂2u∂t2=∂2u∂x2+tx, 0<x<π, t>0.

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Chapter 10.6, Problem 6E

Chapter 10.6, Problem 8E

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The formal solution to the given vibrating string problem ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + t x , 0 < x < π , t > 0 .

Solution Summary: The author explains the formal solution to the given vibrating string problem.

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The formal solution to the given vibrating string problem ∂2u∂t2=∂2u∂x2+tx, 0<x<π, t>0.

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Chapter 10 Solutions

The formal solution to the given vibrating string problem ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + t x , 0 &lt; x &lt; π , t &gt; 0 .

Solution Summary: The author explains the formal solution to the given vibrating string problem.

Videos

The formal solution to the given vibrating string problem ∂2u∂t2=∂2u∂x2+tx, 0<x<π, t>0.

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Chapter 10 Solutions

The formal solution to the given vibrating string problem ∂ 2 u ∂ t 2 = ∂ 2 u ∂ x 2 + t x , 0 < x < π , t > 0 .