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23. For the heat equation u = 2uxx, 0 0, with the boundary and initial conditions u(0,1)=0, u(3,1)= 0, u(x,0) = sin(x), the solution u(x, t) is (a) sin(x) cos(2πt) (b) sin(x) cos(√2) (c) sin(x) e-3 ut щ+ = вихх x t (d) sin(x) ²²1 Ⓒsin(x) e-2n² 0 0 3²=9 - (nz) ² kt nx nt 2² e sk

23. For the heat equation u = 2uxx, 0 0, with the boundary and initial conditions u(0,1)=0, u(3,1)= 0, u(x,0) = sin(x), the solution u(x, t) is (a) sin(x) cos(2πt) (b) sin(x) cos(√2) (c) sin(x) e-3 ut щ+ = вихх x t (d) sin(x) ²²1 Ⓒsin(x) e-2n² 0 0 3²=9 - (nz) ² kt nx nt 2² e sk

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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Please explain in detail where the purple 2/3 comes from and why. Thanks. This is my last question until it resets 

23. For the heat equation ut = 2uxx, 0<x<3, t> 0, with the boundary and initial conditions u(0,1)= 0, u(3,1)= 0, u(x,0) = sin(x), the solution u(x,t) is (a) sin(x) cos(2πt) (b) sin(x) cos(πt√√2) (c) sin(x) e-3 ut = Buxx x t u(0₁ €) = 0 드 کا u(3,t): u(x,0) = sin(xx) (d) sin(x) ²²1 sin(x) e-2n² 0<x<3 L => 0 u(x, t) = NIM A WIN u(x, t) = 3²=9 e (?) K→ 26 L→ 3 ≤₁ An sin (nxx) = sin(xx) a=1 n = 3 n=1 = 1 लात e e 2 - (1²) ² kt nx nt 2² e é An e 2 -2 (9) x² + 2x k ↓ _ 2n²=²+ sin (^_xx) ન x² t -2x² t sin (xx)
Transcribed Image Text:23. For the heat equation ut = 2uxx, 0<x<3, t> 0, with the boundary and initial conditions u(0,1)= 0, u(3,1)= 0, u(x,0) = sin(x), the solution u(x,t) is (a) sin(x) cos(2πt) (b) sin(x) cos(πt√√2) (c) sin(x) e-3 ut = Buxx x t u(0₁ €) = 0 드 کا u(3,t): u(x,0) = sin(xx) (d) sin(x) ²²1 sin(x) e-2n² 0<x<3 L => 0 u(x, t) = NIM A WIN u(x, t) = 3²=9 e (?) K→ 26 L→ 3 ≤₁ An sin (nxx) = sin(xx) a=1 n = 3 n=1 = 1 लात e e 2 - (1²) ² kt nx nt 2² e é An e 2 -2 (9) x² + 2x k ↓ _ 2n²=²+ sin (^_xx) ન x² t -2x² t sin (xx)
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