a2u a2u 25 ax2 (0 0 %3D at? u(0, t) = u(16, t) = 0, t> 0 u(x, 0) = 0, %3D %3D (0 < x < 16) %3D du(x, 0) = 2x – 1 ax O a. 32 25nt u(x, t) = 2 5n272 (1 – 33(-1)"]sen x) sen 16 n=1 00 -32 3nn u(x, t) = 3n2n? n=1 [31(-1)" + 1]sen (,*)s x sen 16 00 C. -32 5nt u(x,t) = L 5n²n² [31(-1)" +1]sen (-*) s %3D 16 16 n=1 d. 32 25nt1-33(-1)"lsen ) sen () G u(x, t) = 16 n=1 25пл u(x, t) = L 25n²n² -32 [31(-1)" + 1]sen *) sen 16 16 %3D n=1 b.
a2u a2u 25 ax2 (0 0 %3D at? u(0, t) = u(16, t) = 0, t> 0 u(x, 0) = 0, %3D %3D (0 < x < 16) %3D du(x, 0) = 2x – 1 ax O a. 32 25nt u(x, t) = 2 5n272 (1 – 33(-1)"]sen x) sen 16 n=1 00 -32 3nn u(x, t) = 3n2n? n=1 [31(-1)" + 1]sen (,*)s x sen 16 00 C. -32 5nt u(x,t) = L 5n²n² [31(-1)" +1]sen (-*) s %3D 16 16 n=1 d. 32 25nt1-33(-1)"lsen ) sen () G u(x, t) = 16 n=1 25пл u(x, t) = L 25n²n² -32 [31(-1)" + 1]sen *) sen 16 16 %3D n=1 b.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Determine u(x,t) for the problem below.
![a2u a2u
25
ax2
(0 <x < 16,t > 0
%3D
at?
u(0, t) = u(16, t) = 0, t> 0
u(x, 0) = 0,
%3D
%3D
(0 < x < 16)
%3D
du(x, 0)
= 2x – 1
ax
O a.
32
25nt
u(x, t) = 2 5n272 (1 – 33(-1)"]sen
x) sen
16
n=1
00
-32
3nn
u(x, t) =
3n2n?
n=1
[31(-1)" + 1]sen (,*)s
x sen
16
00
C.
-32
5nt
u(x,t) = L 5n²n²
[31(-1)" +1]sen (-*) s
%3D
16
16
n=1
d.
32
25nt1-33(-1)"lsen ) sen ()
G
u(x, t) =
16
n=1
25пл
u(x, t) = L 25n²n²
-32
[31(-1)" + 1]sen *)
sen
16
16
%3D
n=1
b.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf913378-ade0-4628-ac57-fdb243e5fba4%2F3f7a1cd3-f361-4f5e-aa3b-5b1e4784d98a%2F2g8kx3a_processed.png&w=3840&q=75)
Transcribed Image Text:a2u a2u
25
ax2
(0 <x < 16,t > 0
%3D
at?
u(0, t) = u(16, t) = 0, t> 0
u(x, 0) = 0,
%3D
%3D
(0 < x < 16)
%3D
du(x, 0)
= 2x – 1
ax
O a.
32
25nt
u(x, t) = 2 5n272 (1 – 33(-1)"]sen
x) sen
16
n=1
00
-32
3nn
u(x, t) =
3n2n?
n=1
[31(-1)" + 1]sen (,*)s
x sen
16
00
C.
-32
5nt
u(x,t) = L 5n²n²
[31(-1)" +1]sen (-*) s
%3D
16
16
n=1
d.
32
25nt1-33(-1)"lsen ) sen ()
G
u(x, t) =
16
n=1
25пл
u(x, t) = L 25n²n²
-32
[31(-1)" + 1]sen *)
sen
16
16
%3D
n=1
b.
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