The steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , u ( 0 , t ) = 0 , ∂ u ∂ t | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h > 0 , t > 0 and u ( x , 0 ) = f ( x ) , 0 < x < 1 .

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Answer 1

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Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Answer 2

Question

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Answer 3

Question

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Answer 4

Question

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Answer 5

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

Answer 6

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

Answer 7

Chapter 13.7, Problem 2E

To determine

The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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 steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , u ( 0 , t ) = 0 , ∂ u ∂ t | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h > 0 , t > 0 and u ( x , 0 ) = f ( x ) , 0 < x < 1 .

Solution Summary: The author explains the steady-state temperature u(x,t) for the boundary-value problem.

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The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Chapter 13 Solutions

The steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 &lt; x &lt; 1 , t &gt; 0 , u ( 0 , t ) = 0 , ∂ u ∂ t | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h &gt; 0 , t &gt; 0 and u ( x , 0 ) = f ( x ) , 0 &lt; x &lt; 1 .

Solution Summary: The author explains the steady-state temperature u(x,t) for the boundary-value problem.

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The steady-state temperature u(x,t) for the boundary-value problem k∂2u∂x2=∂u∂t, 0<x<1, t>0, u(0,t)=0, ∂u∂t|x=1=−h(u(1,t)−u0), h>0,t>0 and u(x,0)=f(x), 0<x<1.

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Chapter 13 Solutions

The steady-state temperature u ( x , t ) for the boundary-value problem k ∂ 2 u ∂ x 2 = ∂ u ∂ t , 0 < x < 1 , t > 0 , u ( 0 , t ) = 0 , ∂ u ∂ t | x = 1 = − h ( u ( 1 , t ) − u 0 ) , h > 0 , t > 0 and u ( x , 0 ) = f ( x ) , 0 < x < 1 .