ди dt 00 dx2 ди u(0, г) — 0 %3DИо, %3D -h (u (1,1) — ио), һ>0, t>0 and u(x, 0) %3D f(x), 0
ди dt 00 dx2 ди u(0, г) — 0 %3DИо, %3D -h (u (1,1) — ио), һ>0, t>0 and u(x, 0) %3D f(x), 0
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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Question
How come lambda=0 and lambda=a^2 don't satisfy the boundary conditions. Can you please explain it to me. Thank you
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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.
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F8475c779-c561-4bfc-8a33-c900cbadfa2c%2Fngb9i4p_processed.png&w=3840&q=75)
Transcribed Image Text:6:29 PM Sat May 1
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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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