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The c n ’s must satisfy 3 sin 5 x − 4 sin 7 x =

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Unformatted text preview: The c n ’s must satisfy 3 sin 5 x − 4 sin 7 x = ∞ ∑ n =1 c n sin nx. Thus c 5 = 3 , c 7 = − 4, c n = 0 if n ̸ = 5 , 7. The solution is u ( x, t ) = 3 e − 25 t sin 5 x − 4 e − 49 t sin 7 x. (b) (20 points) u t = 4 u xx , < x < 1 , t > , u x (0 , t ) = , u (1 , t ) = 0 , t > u ( x, 0) = 5 cos 9 πx 2 , < x < 1 . Solution. The solution of the PDE solving the boundary conditions obtained by separation of vari- ables is u ( x, t ) = ∞ ∑ n =1 c n e − (2 n − 1) 2 π 2 t/ 4 cos (2 n − 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 . 3 (c) (20 points) u tt = 4 u xx , < x < 1 , t > , u (0 , t ) = , u (1 , t ) = 0 , t > , u ( x, 0) = sin πx − 2 sin 5 πx, u t ( x, 0) = sin 7 πx, < 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....
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The c n ’s must satisfy 3 sin 5 x − 4 sin 7 x = ∞ ∑...

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