Explanation Given: The boundary value problem is k ∂ 2 u ∂ x 2 = ∂ u ∂ t , − ∞ < x < ∞ , t > 0 , the boundary condition is u ( x , 0 ) = f ( x ) , use given identity F { e − x 2 4 p 2 } = 2 π p e − p 2 α 2 and second identity is when F { f ( x ) } = F ( α ) and F { g ( x ) } = G ( α ) then ∫ − ∞ ∞ f ( τ ) g ( x − τ ) d τ = F − 1 { F ( α ) G ( α ) } . Calculation: The given boundary value problem is, k ∂ 2 u ∂ x 2 = ∂ u ∂ t . (1) Take Fourier transform of both sides of the above equation, k F { ∂ 2 u ∂ x 2 } = F { ∂ u ∂ t } − k α 2 U ( α , t ) + k α u ( 0 , t ) = d U d t − k α 2 U ( α , t ) = d U d t d U d t + k α 2 U ( α , t ) = 0 The particular solution of the above equation, U ( α , t ) = c e − k α 2 t . (2) At the given boundary condition, F { u ( x , 0 ) } = F { f ( x ) } So, the above equation is expressed as, U ( α , 0 ) = F ( α ) . (3) Substitute t = 0 in equation (2), U ( α , 0 ) = c

WEB ASSIGN FOR ZILL'S DIFFERENTIAL EQUAT

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ISBN: 9780357539545

Author: ZILL

Publisher: CENGAGE L

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Chapter 14.4, Problem 23E

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Chapter 14.4, Problem 22E

Chapter 14.4, Problem 24E

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Chapter 14 Solutions

WEB ASSIGN FOR ZILL'S DIFFERENTIAL EQUAT

Ch. 14.1 - (a) Show that erf(t)=10ted. (b) Use part (a), the...Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Use the third and fifth entries in Table 14.1.1 to...

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The given equation is the solution of boundary value problem.

Solution Summary: The author explains that the given equation is the solution of boundary value problem.

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