The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .

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Answer 1

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Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Answer 2

Question

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Answer 3

Question

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Answer 4

Question

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Answer 5

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

Answer 6

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

Answer 7

Chapter 13.4, Problem 16E

To determine

To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .

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To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Chapter 13 Solutions

The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 &lt; x &lt; π , t &gt; 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .

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To show: The solution of the boundary value problem ∂2u∂x2=∂2u∂t2+u,0<x<π,t>0 is u(x,t)=4π∑n=1∞(−1)k+1(2k−1)2(sin(2k−1)x)cos((2k−1)2+1t).

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Chapter 13 Solutions

The solution of the boundary value problem ∂ 2 u ∂ x 2 = ∂ 2 u ∂ t 2 + u , 0 < x < π , t > 0 is u ( x , t ) = 4 π ∑ n = 1 ∞ ( − 1 ) k + 1 ( 2 k − 1 ) 2 ( sin ( 2 k − 1 ) x ) cos ( ( 2 k − 1 ) 2 + 1 t ) .