The Fourier sine series of the function f(x) = x³ on the interval [0,T] is 32(-1)+1 (nπ)²-6 -sin(nx). The solution of the equation m-1 is equal to? (a) u(x,t) = e-toz³ θα J²u = მე2 It u(x, 0) = 1³ u(0,t) = 0, u(π,t) = 0 (b) u(x,t) = 2(−1)÷1 (n)²–6 on²t sin(nx) (c) u(x,t)=2(−1)n−1 (nx)²–6¸ e sin(nx) ○ (d) u(x,t) = 2(-1)+1 (na)²=6e-on²t cos(nx) ○ (e) It's impossible to solve this equation
The Fourier sine series of the function f(x) = x³ on the interval [0,T] is 32(-1)+1 (nπ)²-6 -sin(nx). The solution of the equation m-1 is equal to? (a) u(x,t) = e-toz³ θα J²u = მე2 It u(x, 0) = 1³ u(0,t) = 0, u(π,t) = 0 (b) u(x,t) = 2(−1)÷1 (n)²–6 on²t sin(nx) (c) u(x,t)=2(−1)n−1 (nx)²–6¸ e sin(nx) ○ (d) u(x,t) = 2(-1)+1 (na)²=6e-on²t cos(nx) ○ (e) It's impossible to solve this equation
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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![The Fourier sine series of the function f(x) = x³ on the interval [0,T] is
32(-1)+1
(nπ)²-6
-sin(nx).
The solution of the equation
m-1
is equal to?
(a) u(x,t) = e-toz³
θα
J²u
=
მე2
It
u(x, 0) = 1³
u(0,t) = 0, u(π,t) = 0
(b) u(x,t) = 2(−1)÷1 (n)²–6
on²t sin(nx)
(c) u(x,t)=2(−1)n−1 (nx)²–6¸
e
sin(nx)
○ (d) u(x,t) = 2(-1)+1 (na)²=6e-on²t cos(nx)
○ (e) It's impossible to solve this equation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F912ee020-db0c-48be-8b51-55ea98f76c9b%2Fd223f539-f05b-4e65-8d9d-5247aed1e330%2Fytrli6sj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The Fourier sine series of the function f(x) = x³ on the interval [0,T] is
32(-1)+1
(nπ)²-6
-sin(nx).
The solution of the equation
m-1
is equal to?
(a) u(x,t) = e-toz³
θα
J²u
=
მე2
It
u(x, 0) = 1³
u(0,t) = 0, u(π,t) = 0
(b) u(x,t) = 2(−1)÷1 (n)²–6
on²t sin(nx)
(c) u(x,t)=2(−1)n−1 (nx)²–6¸
e
sin(nx)
○ (d) u(x,t) = 2(-1)+1 (na)²=6e-on²t cos(nx)
○ (e) It's impossible to solve this equation
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