The Fourier sine series of the function f(x) = x³ on the interval [0, π] is The solution of the equation 3 x³ = 2(-1)+1 n=1 (Nπ) 2 - 6 n3 -sin(nx). is equal to? ди a²u Ət მ2 u(x, 0) = x³ - u(0,t) = 0, u(T,t)=0 O (a) u(x,t) = e -tox³ ○ (b) u(x,t) = 2Σx²-1 (−1) n+1 (nπ)² - -6 -on²t e n3 sin(nx) n=1 ○ (c) u(x,t) = 2 (−1) n+1 (nm) 2-6 sin(nx) Σx-1 n3 ○ (d) u(x,t) = 2x-1 (−1) n+1 (nm) 2-6 e-on²+ cos(nx) n=1 n³ (e) It's impossible to solve this equation
The Fourier sine series of the function f(x) = x³ on the interval [0, π] is The solution of the equation 3 x³ = 2(-1)+1 n=1 (Nπ) 2 - 6 n3 -sin(nx). is equal to? ди a²u Ət მ2 u(x, 0) = x³ - u(0,t) = 0, u(T,t)=0 O (a) u(x,t) = e -tox³ ○ (b) u(x,t) = 2Σx²-1 (−1) n+1 (nπ)² - -6 -on²t e n3 sin(nx) n=1 ○ (c) u(x,t) = 2 (−1) n+1 (nm) 2-6 sin(nx) Σx-1 n3 ○ (d) u(x,t) = 2x-1 (−1) n+1 (nm) 2-6 e-on²+ cos(nx) n=1 n³ (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
Related questions
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![The Fourier sine series of the function f(x) = x³ on the interval [0, π] is
The solution of the equation
3
x³ = 2(-1)+1
n=1
(Nπ) 2 - 6
n3
-sin(nx).
is equal to?
ди
a²u
Ət
მ2
u(x, 0) = x³
- u(0,t) = 0, u(T,t)=0
O (a) u(x,t) = e
-tox³
○
(b) u(x,t) = 2Σx²-1 (−1) n+1 (nπ)² -
-6
-on²t
e
n3
sin(nx)
n=1
○ (c) u(x,t) = 2 (−1) n+1 (nm) 2-6 sin(nx)
Σx-1
n3
○ (d) u(x,t) = 2x-1 (−1) n+1 (nm) 2-6 e-on²+ cos(nx)
n=1
n³
(e) It's impossible to solve this equation](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fdcb783-5702-49ba-a873-fdccebd2669a%2F2ff648bd-d3cb-4bb3-abff-504c3876481b%2F40mdwv4_processed.png&w=3840&q=75)
Transcribed Image Text:The Fourier sine series of the function f(x) = x³ on the interval [0, π] is
The solution of the equation
3
x³ = 2(-1)+1
n=1
(Nπ) 2 - 6
n3
-sin(nx).
is equal to?
ди
a²u
Ət
მ2
u(x, 0) = x³
- u(0,t) = 0, u(T,t)=0
O (a) u(x,t) = e
-tox³
○
(b) u(x,t) = 2Σx²-1 (−1) n+1 (nπ)² -
-6
-on²t
e
n3
sin(nx)
n=1
○ (c) u(x,t) = 2 (−1) n+1 (nm) 2-6 sin(nx)
Σx-1
n3
○ (d) u(x,t) = 2x-1 (−1) n+1 (nm) 2-6 e-on²+ cos(nx)
n=1
n³
(e) It's impossible to solve this equation
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