Determine u(x,t) for the problem below: a²u a²u (0 < x < 16,t > 0 əx² u(0, t) = u(16, t) = 0, t>0 u(x,0) = 0, ди(х, 0) at2 (0 < x < 16) = 2x – 1 əx -32 u(x,t) = > Зпл [31(-1)" + 1]sen (x) sen (t) 3n²n² n=1 O b. 32 [1 – 33(-1)"]se (x) sen 3nn u(x,t) = L5n²n² n=1 32 25пл u(x, t) = [1 – 33(-1)"]sen () se sen 5n2n² 16 n=1 Od. -32 [31(-1)" + 1]sen (x) nn 5nn u(x, t) = 16) 3n²n² \16 n=1 00 '5nn u(x, t) = >: 32 [1– 33(-1)"]sen(G*): x) sen 16 5n²n² 16 n=1
Determine u(x,t) for the problem below: a²u a²u (0 < x < 16,t > 0 əx² u(0, t) = u(16, t) = 0, t>0 u(x,0) = 0, ди(х, 0) at2 (0 < x < 16) = 2x – 1 əx -32 u(x,t) = > Зпл [31(-1)" + 1]sen (x) sen (t) 3n²n² n=1 O b. 32 [1 – 33(-1)"]se (x) sen 3nn u(x,t) = L5n²n² n=1 32 25пл u(x, t) = [1 – 33(-1)"]sen () se sen 5n2n² 16 n=1 Od. -32 [31(-1)" + 1]sen (x) nn 5nn u(x, t) = 16) 3n²n² \16 n=1 00 '5nn u(x, t) = >: 32 [1– 33(-1)"]sen(G*): x) sen 16 5n²n² 16 n=1
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
Determine u(x,t) for the problem below.
*Contains five alternatives as an answer
![Determine u(x,t) for the problem below:
a²u
a²u
(0 < x < 16,t > 0
əx²
u(0, t) = u(16, t) = 0, t>0
u(x,0) = 0,
ди(х, 0)
at2
(0 < x < 16)
= 2x – 1
əx
-32
u(x,t) = >
Зпл
[31(-1)" + 1]sen (x) sen (t)
3n²n²
n=1
O b.
32
[1 – 33(-1)"]se (x) sen
3nn
u(x,t) = L5n²n²
n=1
32
25пл
u(x, t) =
[1 – 33(-1)"]sen () se
sen
5n2n²
16
n=1
Od.
-32
[31(-1)" + 1]sen (x)
nn
5nn
u(x, t) =
16)
3n²n²
\16
n=1
Oe.
00
32
[1 – 33(-1)"]sen (x):
'5nn
u(x, t) = >:
:と5nene
16)
5n²n²
16 sen
n=1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf913378-ade0-4628-ac57-fdb243e5fba4%2Ff78c1403-657d-4a23-b13e-7b897f476c8b%2Fdydrd_processed.png&w=3840&q=75)
Transcribed Image Text:Determine u(x,t) for the problem below:
a²u
a²u
(0 < x < 16,t > 0
əx²
u(0, t) = u(16, t) = 0, t>0
u(x,0) = 0,
ди(х, 0)
at2
(0 < x < 16)
= 2x – 1
əx
-32
u(x,t) = >
Зпл
[31(-1)" + 1]sen (x) sen (t)
3n²n²
n=1
O b.
32
[1 – 33(-1)"]se (x) sen
3nn
u(x,t) = L5n²n²
n=1
32
25пл
u(x, t) =
[1 – 33(-1)"]sen () se
sen
5n2n²
16
n=1
Od.
-32
[31(-1)" + 1]sen (x)
nn
5nn
u(x, t) =
16)
3n²n²
\16
n=1
Oe.
00
32
[1 – 33(-1)"]sen (x):
'5nn
u(x, t) = >:
:と5nene
16)
5n²n²
16 sen
n=1
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