Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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How come lambda=0 and lambda=a^2 don't satisfy the boundary conditions. Can you please explain it to me. Thank you 

6:29 PM Sat May 1 * 12% AA bartleby.com Bb Sign i Bb What if I add... Sign in Bb Request AA... Sign in Answered: P... = bartleby Q Search for textbooks, step-by-step explanatio... Ask an Expert Math / Advanced Math / Q&A Library | Please find attached the question as you told me to upload.l don't understand w... Please find attached the question as you told me to upload.I don't understand why tan(a... u (0, t) = 0, , = -h (u (1, t) – uo), h> 0,1 > 0 and u (x, 0) = f (x), 0 <x < 1. X GET 10 FREE QUESTIONS Expert Answer Step 1 Given boundary value problem is kPu dx 2 ди 0<x<1, t>0 dt > ди u(0, t) = 0 = uo, ax :-h (u (1, t) – uo), h>0, t>0 and u(x, 0) = f(x), 0<x<1. x=1 We will solve this by variable separation method. Let u(x, t) = X(x)T(t) is the solution of the given boundary value problem, 1 d² x X dx2 d² x 1 dT then kT- d x2 dt kT dt a (positive), we are not able to get any solution which satisfy all the boundary conditions. -a° (negative). For 2 = 0 and 1 So, we can neglect these conditions. Now let 1 %3| 7²kt Then by above equation we get, X А cos (аx) + B sin (аx) and T — Се а'ki. Hence, u (x, t) = [A cos (ax) + B sin (ax)] (Ce-a*ki) ...(1) Step 2 Since, u(0, t) = 0. So, by (1), we get
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Transcribed Image Text:6:29 PM Sat May 1 * 12% AA bartleby.com Bb Sign i Bb What if I add... Sign in Bb Request AA... Sign in Answered: P... = bartleby Q Search for textbooks, step-by-step explanatio... Ask an Expert Math / Advanced Math / Q&A Library | Please find attached the question as you told me to upload.l don't understand w... Please find attached the question as you told me to upload.I don't understand why tan(a... u (0, t) = 0, , = -h (u (1, t) – uo), h> 0,1 > 0 and u (x, 0) = f (x), 0 <x < 1. X GET 10 FREE QUESTIONS Expert Answer Step 1 Given boundary value problem is kPu dx 2 ди 0<x<1, t>0 dt > ди u(0, t) = 0 = uo, ax :-h (u (1, t) – uo), h>0, t>0 and u(x, 0) = f(x), 0<x<1. x=1 We will solve this by variable separation method. Let u(x, t) = X(x)T(t) is the solution of the given boundary value problem, 1 d² x X dx2 d² x 1 dT then kT- d x2 dt kT dt a (positive), we are not able to get any solution which satisfy all the boundary conditions. -a° (negative). For 2 = 0 and 1 So, we can neglect these conditions. Now let 1 %3| 7²kt Then by above equation we get, X А cos (аx) + B sin (аx) and T — Се а'ki. Hence, u (x, t) = [A cos (ax) + B sin (ax)] (Ce-a*ki) ...(1) Step 2 Since, u(0, t) = 0. So, by (1), we get
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