Mathematic Methods HW Solutions 59

# Mathematic Methods HW Solutions 59 - 2.14 For f(x = x − 5...

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Unformatted text preview: Chapter 13 59 2.14 For f (x) = x − 5: T = − 40 π2 ∞ 1 odd n nπx −nπy/10 1 cos e 2 n 10 For f (x) = x: Add 5 to the answer just given. 2.15 For f (x) = 100, T = 100 − 10y/3 40 1 For f (x) = x, T = (30 − y)− 2 6 π 3.2 3.3 3.4 3.5 3.6 3.7 u= 400 π ∞ 1 odd n u = 100 − u= 40 π ∞ 1 odd n −2 ∞ ∞ 2 even n 2 n=2+4k 1 −(nπα/l)2 t nπx e sin n l nπx 1 −(nπα/10)2 t sin e n 10 0, even n 2 ∞ 1 2 nπx where bn = n2 π 2 − nπ , n = 1 + 4k u = 100 + 400 bn e−(nπα/2) t sin 2 −2 1 1 − , n = 3 + 4k n2 π 2 nπ Add to (3.15) : uf = 20 + 30x/l. Note: Any linear function added to both u0 and uf leaves the Fourier series unchanged. u= l 4l − 2 π2 ∞ nπx −(nπα/l)2 t 1 cos e n2 l 1 odd n ∞ 200 π u = 50x + 3.9 400 u = 100− π 1 ∞ 0 nπx (−1)n −(nπα/2)2 t e sin n 2 (−1)n −[(2n+1)πα/4]2 t e cos 2n + 1 n2 ¯ 2 h 4 3.11 En = , Ψ(x, t) = 2m π o dd n 8 n2 π 2 ¯ 2 h , Ψ(x, t) = 3.12 En = 2m π 4.3 1 odd n 1 nπ nπx sinh (30 − y ) cos n2 sinh 3nπ 10 10 1 −(nπα/10)2 t nπx e sin n 10 100x 400 − l π 3.8 4.2 ∞ 2n + 1 πx 4 sin nx −iEn t/h ¯ e n o dd n sin nπx −iEn t/h ¯ e n(4 − n2 ) ∞ √ nπx 8h nπvt 1 Bn sin cos , where B1 = 2 − 1, B2 = , 2 π l l 2 n=1 1√ B3 = ( 2 + 1), B4 = 0, · · · , Bn = (2 sin nπ/4 − sin nπ/2)/n2 9 ∞ nπvt nπ nπ 16h nπx cos where Bn = 2 sin − sin y= 2 Bn sin / n2 π l l 8 4 1 y= 8h π2 4.4 y= 4.5 8hl y= 3 πv ∞ 1 1 nπ 2nπx 2nπvt sin sin cos 2 n 2 l l ∞ 1 odd n nπ nπx nπvt 1 sin sin sin 3 n 2 l l ...
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