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# 1 πx 2 the c n s must satisfy 5 cos 9 πx 2 n 1 c n

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1) πx 2 . The c n ’s must satisfy 5 cos 9 πx 2 = n =1 c n cos (2 n 1) πx 2 . Thus c 5 = 5, c n = 0 if n ̸ = 5. The solution is u ( x, t ) = 5 e 81 π 2 t/ 4 cos 9 πx 2 .

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3 (c) (20 points) u tt = 4 u xx , 0 < x < 1 , t > 0 , u (0 , t ) = 0 , u (1 , t ) = 0 , t > 0 , u ( x, 0) = sin πx 2 sin 5 πx, u t ( x, 0) = sin 7 πx, 0 < x < 1 . Solution. The solution one obtains by separation of variables of the PDE and the boundary conditions is: u ( x, t ) = n =1 ( a n cos 2 nπt + b n sin 2 nπt ) sin nπx. The a n ’s, b n ’s are found by sin πx 2 sin 5 πx = n =1 a n sin nπx, sin 7 πx = n =1 2 nπb n sin nπx. We get a 1 = 1, a 5 = 2, a n = 0 if n ̸ = 1 , 5, b 7 = 1 / 14, b n = 0 if n ̸ = 7. The solution is u ( x, t ) = cos 2 πt sin πx 2 cos 10 πt sin 5 πx + 1 14 sin 14 πt sin 7 πx.
• Spring '13
• Schonbek
• Sin, Boundary value problem, Partial differential equation, Sturm–Liouville theory

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