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Determine u(x,t) para o problema abaixo: a2u 25 ax2 u(0, t) = u(16, t) = 0, t>0 u(x, 0) = 0, du(x, 0) au at2 (0 0 %3D %3D (0 < x < 16) %3! = 2x - 1 Əx O a. 32 25nn u(x, t) = > [1-33(-1)"]sen sen 5n2r (x) 16 16 n=1 O b. -32 3nn u(x, t) = > 3r22 [31(-1)" + 1]sen (x) x sen 16 3n2n2 16 n=1 5nm 32 [1-33(-1)"]sen () 16 u(x, t) = x sen 16 %3D 5n2n2 n=1 O d. 25nn -32 u(x, t) = L 25n²n? [31(-1)" + 1]sen () ser 16 16 n=1 -32 [31(-1)" +1]sen 5nn u(x, t) = 131(-1)" + 1]sen (x) sen ( %3D 5n2n2 n=1 16

Determine u(x,t) para o problema abaixo: a2u 25 ax2 u(0, t) = u(16, t) = 0, t>0 u(x, 0) = 0, du(x, 0) au at2 (0 0 %3D %3D (0 < x < 16) %3! = 2x - 1 Əx O a. 32 25nn u(x, t) = > [1-33(-1)"]sen sen 5n2r (x) 16 16 n=1 O b. -32 3nn u(x, t) = > 3r22 [31(-1)" + 1]sen (x) x sen 16 3n2n2 16 n=1 5nm 32 [1-33(-1)"]sen () 16 u(x, t) = x sen 16 %3D 5n2n2 n=1 O d. 25nn -32 u(x, t) = L 25n²n? [31(-1)" + 1]sen () ser 16 16 n=1 -32 [31(-1)" +1]sen 5nn u(x, t) = 131(-1)" + 1]sen (x) sen ( %3D 5n2n2 n=1 16

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 70EQ
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Determine u(x,t) para o problema abaixo: a2u 25 ax2 u(0, t) = u(16, t) = 0, t>0 u(x, 0) = 0, du(x, 0) au at2 (0 <x < 16,t > 0 %3D %3D (0 < x < 16) %3! = 2x - 1 Əx O a. 32 25nn u(x, t) = > [1-33(-1)"]sen sen 5n2r (x) 16 16 n=1 O b. -32 3nn u(x, t) = > 3r22 [31(-1)" + 1]sen (x) x sen 16 3n2n2 16 n=1 5nm 32 [1-33(-1)"]sen () 16 u(x, t) = x sen 16 %3D 5n2n2 n=1 O d. 25nn -32 u(x, t) = L 25n²n? [31(-1)" + 1]sen () ser 16 16 n=1 -32 [31(-1)" +1]sen 5nn u(x, t) = 131(-1)" + 1]sen (x) sen ( %3D 5n2n2 n=1 16
Transcribed Image Text:Determine u(x,t) para o problema abaixo: a2u 25 ax2 u(0, t) = u(16, t) = 0, t>0 u(x, 0) = 0, du(x, 0) au at2 (0 <x < 16,t > 0 %3D %3D (0 < x < 16) %3! = 2x - 1 Əx O a. 32 25nn u(x, t) = > [1-33(-1)"]sen sen 5n2r (x) 16 16 n=1 O b. -32 3nn u(x, t) = > 3r22 [31(-1)" + 1]sen (x) x sen 16 3n2n2 16 n=1 5nm 32 [1-33(-1)"]sen () 16 u(x, t) = x sen 16 %3D 5n2n2 n=1 O d. 25nn -32 u(x, t) = L 25n²n? [31(-1)" + 1]sen () ser 16 16 n=1 -32 [31(-1)" +1]sen 5nn u(x, t) = 131(-1)" + 1]sen (x) sen ( %3D 5n2n2 n=1 16
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