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Consider the following non-homogeneous boundary value problem (F) : ²u u 0 0 (1) (2) (3) (4) = 4 +x, и (0, t) — 0, и(3, t) — 0, u(x, 0) - u:(r, 0) = sin r + sinar, 00 00 v(x, 0) = -U(r) – +, 00 (1) (2) (3) (4) %3D (1) Based on equations (1) and (2) of (F) and (L), the function U(x) is a solution of the following ODE: a. 4U"(x) +x = 0 and U(0) = 0, U (3) = 0 b. 4U"(x) – x = 0 and U(0) = 0, U(1) = 0. c. 4U'(x) + x = 0 and U(0) = 1, U(3) = 1. d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r) = +* b. U(r) = -+ c. U(2) = -- %3D d. None of the above +00 2nn 2nn (3) The solution of the problem (L) is v(x, t) = >[En cos- -t + Fn sin 3 t] sin æ where 3 n=1 a. F, = 0, E2 = , E3 = and En = 0 Vn e N – {2,3} b. En = 0, F2 = ,F3 = and Fn = 0 V n € N– {2, 3} c. En = 0, F2 = , F3 = and Fn = 0 V n € N – {2, 3} %3D 4 %3D d. None of the above (4) Based on part (3), the solution of the problem (F) is: 4nt, cos(- 3 a. u(x, t) - 1 - cos(2rt) cos(72) 3 Cos(- 3 24 1 4at sin(- 3x b. u(r,t) -) sin(- sin(2at) sin(72) 24 8. с. и(х,t) - 3 4xt - sin( ) sin(- 1 -) + sin(2nt) sin(72) 24 d. None of the above

Consider the following non-homogeneous boundary value problem (F) : ²u u 0 0 (1) (2) (3) (4) = 4 +x, и (0, t) — 0, и(3, t) — 0, u(x, 0) - u:(r, 0) = sin r + sinar, 00 00 v(x, 0) = -U(r) – +, 00 (1) (2) (3) (4) %3D (1) Based on equations (1) and (2) of (F) and (L), the function U(x) is a solution of the following ODE: a. 4U"(x) +x = 0 and U(0) = 0, U (3) = 0 b. 4U"(x) – x = 0 and U(0) = 0, U(1) = 0. c. 4U'(x) + x = 0 and U(0) = 1, U(3) = 1. d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r) = +* b. U(r) = -+ c. U(2) = -- %3D d. None of the above +00 2nn 2nn (3) The solution of the problem (L) is v(x, t) = >[En cos- -t + Fn sin 3 t] sin æ where 3 n=1 a. F, = 0, E2 = , E3 = and En = 0 Vn e N – {2,3} b. En = 0, F2 = ,F3 = and Fn = 0 V n € N– {2, 3} c. En = 0, F2 = , F3 = and Fn = 0 V n € N – {2, 3} %3D 4 %3D d. None of the above (4) Based on part (3), the solution of the problem (F) is: 4nt, cos(- 3 a. u(x, t) - 1 - cos(2rt) cos(72) 3 Cos(- 3 24 1 4at sin(- 3x b. u(r,t) -) sin(- sin(2at) sin(72) 24 8. с. и(х,t) - 3 4xt - sin( ) sin(- 1 -) + sin(2nt) sin(72) 24 d. None of the above

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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part 1 2 3 pde non bondary
Consider the following non-homogeneous boundary value problem (F) : ²u u 0 <x < 3, t>0 (1) (2) (3) (4) = 4 +x, и (0, t) — 0, и(3, t) — 0, u(x, 0) - u:(r, 0) = sin r + sinar, 0<x<3 t>0 0<x<3 27 %3D We consider the change of variable u(x, t) = v(x, t) +U(x) in order to transform the problem (F) into the following homogeneous problem (L): = 4 о(0, е) — 0, о(3,t) — 0, t >0 v(x, 0) = -U(r) – +, 0<r< 3 ve(x,0) = sinr+ sinar, 0< x < 3 0 < x < 3, t>0 (1) (2) (3) (4) %3D (1) Based on equations (1) and (2) of (F) and (L), the function U(x) is a solution of the following ODE: a. 4U"(x) +x = 0 and U(0) = 0, U (3) = 0 b. 4U"(x) – x = 0 and U(0) = 0, U(1) = 0. c. 4U'(x) + x = 0 and U(0) = 1, U(3) = 1. d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r) = +* b. U(r) = -+ c. U(2) = -- %3D d. None of the above +00 2nn 2nn (3) The solution of the problem (L) is v(x, t) = >[En cos- -t + Fn sin 3 t] sin æ where 3 n=1 a. F, = 0, E2 = , E3 = and En = 0 Vn e N – {2,3} b. En = 0, F2 = ,F3 = and Fn = 0 V n € N– {2, 3} c. En = 0, F2 = , F3 = and Fn = 0 V n € N – {2, 3} %3D 4 %3D d. None of the above (4) Based on part (3), the solution of the problem (F) is: 4nt, cos(- 3 a. u(x, t) - 1 - cos(2rt) cos(72) 3 Cos(- 3 24 1 4at sin(- 3x b. u(r,t) -) sin(- sin(2at) sin(72) 24 8. с. и(х,t) - 3 4xt - sin( ) sin(- 1 -) + sin(2nt) sin(72) 24 d. None of the above
Transcribed Image Text:Consider the following non-homogeneous boundary value problem (F) : ²u u 0 <x < 3, t>0 (1) (2) (3) (4) = 4 +x, и (0, t) — 0, и(3, t) — 0, u(x, 0) - u:(r, 0) = sin r + sinar, 0<x<3 t>0 0<x<3 27 %3D We consider the change of variable u(x, t) = v(x, t) +U(x) in order to transform the problem (F) into the following homogeneous problem (L): = 4 о(0, е) — 0, о(3,t) — 0, t >0 v(x, 0) = -U(r) – +, 0<r< 3 ve(x,0) = sinr+ sinar, 0< x < 3 0 < x < 3, t>0 (1) (2) (3) (4) %3D (1) Based on equations (1) and (2) of (F) and (L), the function U(x) is a solution of the following ODE: a. 4U"(x) +x = 0 and U(0) = 0, U (3) = 0 b. 4U"(x) – x = 0 and U(0) = 0, U(1) = 0. c. 4U'(x) + x = 0 and U(0) = 1, U(3) = 1. d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r) = +* b. U(r) = -+ c. U(2) = -- %3D d. None of the above +00 2nn 2nn (3) The solution of the problem (L) is v(x, t) = >[En cos- -t + Fn sin 3 t] sin æ where 3 n=1 a. F, = 0, E2 = , E3 = and En = 0 Vn e N – {2,3} b. En = 0, F2 = ,F3 = and Fn = 0 V n € N– {2, 3} c. En = 0, F2 = , F3 = and Fn = 0 V n € N – {2, 3} %3D 4 %3D d. None of the above (4) Based on part (3), the solution of the problem (F) is: 4nt, cos(- 3 a. u(x, t) - 1 - cos(2rt) cos(72) 3 Cos(- 3 24 1 4at sin(- 3x b. u(r,t) -) sin(- sin(2at) sin(72) 24 8. с. и(х,t) - 3 4xt - sin( ) sin(- 1 -) + sin(2nt) sin(72) 24 d. None of the above
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