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Matakuliah Tahun : K0124 / Matematika Teknik II : 2006/2007 PERTEMUAN 17 & 18 VIBRATING STRING EQUATION 1 To solve this boundy-value problem , let y = XT as usual. Then XT " a 2 X "T or T " a 2T X " X Calling the separation constant 2 , we have T " 2 a 2T 0, X " 2 X 0 2 and T A1 sin at B1 cos at, A2 sin ax B2 cos ax A1 s A solution is thus given by y( x, t ) XT A2 sin x B2 cos x A1 sin t B1 cos t From y(0, t ) 0, A2 0 , 3 then y( x, t ) B2 sin x A1 sin t B1 cos t sin Asin t B cos t From y ( L, t ) 0 , we have sin L Asin t B cos t 0 4 So that sin L 0, L m or m L since the second factor must not be equal to zero. Now yt ( x, t ) sin x Aa cos t Ba sin at And yt ( x,0) sin x a 0 from which A0 . Thus m x m at y( x, t ) B sin cos L L 5 PERTEMUAN 18 VIBRATING STRING EQUATION (LANJUTAN) 6 To solve this boundy-value problem , let y = XT as usual. Then XT " a 2 X "T or T " a 2T X " X To satisfy the condition y ( x,0) f ( x) , it will be necessary to superimpose solutions. 7 This yields y ( x, t ) Bm sin m 1 m x m at cos L L Then m x y ( x,0) f ( x) Bm sin L m 1 8 And from the theory of Fourier series, m x 2 L Bm f ( x) sin dx L 0 L The result is m x m x m at 2 L y ( x, t ) f ( x) sin dx sin cos 0 L L L m 1 L which can be verified as the solution. 9 TERIMA KASIH 10