2. Suppose that the odd, 2-periodic function f(x) has the Fourier series expansion 1 1+n³ (A) g(x) = Σ n=1 If g(x) is a 2-periodic solution of the differential equation y" = f(x) and the average value of g(x) on the interval [-1, 1] is 2 (i.e. f, g(x) dx = 2), then g(x) has the Fourier series expansion: -2 n²π² (1+n³) 00 (D) g(x) = Σ n=1 n=1 → (C) g(x) ~2-Σ 1 (B) g(x)=1-Σ n²π² (1+n³) 1 n²π² (1+n³) 8 f(x) = Σ n=1 n=1 8 -1 n²π² (1+n³) (E) g(x)=1-Σ n=1 sin(nxx) cos(nπx) 1 n²π² (1+³) sin(nxx) cos(nTX) sin(nлx), -∞0 < x <∞0. Solution. yp(x) = − En=1 n²x² (1+n³) sin(nxx) and ye(a) = A + Bx. Thus, 8 g(x) = A +Bx - Σ 1 n²π² (1+n³) n=1 Since g(x) is 2-periodic, B ₁9(x) dx = 2, A = 2. Thus, 8 g(x) = 2 - Σ n=1 (sin(nx) +n cos(nπx)) sin(ntx). = 0 and, since 1 n²² (1+n³) sin(nxx)
2. Suppose that the odd, 2-periodic function f(x) has the Fourier series expansion 1 1+n³ (A) g(x) = Σ n=1 If g(x) is a 2-periodic solution of the differential equation y" = f(x) and the average value of g(x) on the interval [-1, 1] is 2 (i.e. f, g(x) dx = 2), then g(x) has the Fourier series expansion: -2 n²π² (1+n³) 00 (D) g(x) = Σ n=1 n=1 → (C) g(x) ~2-Σ 1 (B) g(x)=1-Σ n²π² (1+n³) 1 n²π² (1+n³) 8 f(x) = Σ n=1 n=1 8 -1 n²π² (1+n³) (E) g(x)=1-Σ n=1 sin(nxx) cos(nπx) 1 n²π² (1+³) sin(nxx) cos(nTX) sin(nлx), -∞0 < x <∞0. Solution. yp(x) = − En=1 n²x² (1+n³) sin(nxx) and ye(a) = A + Bx. Thus, 8 g(x) = A +Bx - Σ 1 n²π² (1+n³) n=1 Since g(x) is 2-periodic, B ₁9(x) dx = 2, A = 2. Thus, 8 g(x) = 2 - Σ n=1 (sin(nx) +n cos(nπx)) sin(ntx). = 0 and, since 1 n²² (1+n³) sin(nxx)
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
![2.
Suppose that the odd, 2-periodic function f(x) has the Fourier series expansion
1
1+n³
(A) 9(2) = Σ
n=1
If g(x) is a 2-periodic solution of the differential equation y" = f(r) and the average value of g(x) on the
interval [1,1] is 2 (i.e. ₁ g(x) dx = 2), then g(x) has the Fourier series expansion:
(D) g(x) ~
(B) g(x)=1-Σ
n=1
→ (C) g(x) ~2-Σ
n=1
(E) g(x) ~1-
-2
n²π² (1+n³)
f(x) = Σ
n=1
n=1
1
n²π² (1+n³)
n=1
sin(nxx)
-1
n²π² (1+n³)
1
n²π² (1+n³)
cos(nxx)
1
n²π² (1+n³)
sin(nxx)
cos(nxx)
sin(nπx), -∞ < x < 00.
Solution. Yp(x) = − Σn=1_n²x² (1+n³) sin(nπx)
and ye(x) = A + Br. Thus,
1
n²π² (1+n³)
n=1
g(x) = A + Bx Σ
Since g(x) is 2-periodic, B
₁ g(x) dx = 2, A = 2. Thus,
g(x) = 2 - Σ
n=1
(sin(nx) +n cos(nπx))
1
n²π² (1+³)
sin(nπx).
0 and, since
sin(n+x)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca6b5de9-d666-4be4-bec5-372f49facd74%2F247f3467-d0d8-487a-8672-ab76ded224d9%2Fba7o2fr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2.
Suppose that the odd, 2-periodic function f(x) has the Fourier series expansion
1
1+n³
(A) 9(2) = Σ
n=1
If g(x) is a 2-periodic solution of the differential equation y" = f(r) and the average value of g(x) on the
interval [1,1] is 2 (i.e. ₁ g(x) dx = 2), then g(x) has the Fourier series expansion:
(D) g(x) ~
(B) g(x)=1-Σ
n=1
→ (C) g(x) ~2-Σ
n=1
(E) g(x) ~1-
-2
n²π² (1+n³)
f(x) = Σ
n=1
n=1
1
n²π² (1+n³)
n=1
sin(nxx)
-1
n²π² (1+n³)
1
n²π² (1+n³)
cos(nxx)
1
n²π² (1+n³)
sin(nxx)
cos(nxx)
sin(nπx), -∞ < x < 00.
Solution. Yp(x) = − Σn=1_n²x² (1+n³) sin(nπx)
and ye(x) = A + Br. Thus,
1
n²π² (1+n³)
n=1
g(x) = A + Bx Σ
Since g(x) is 2-periodic, B
₁ g(x) dx = 2, A = 2. Thus,
g(x) = 2 - Σ
n=1
(sin(nx) +n cos(nπx))
1
n²π² (1+³)
sin(nπx).
0 and, since
sin(n+x)
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