Differential Equations Lecture Work Solutions 140

Differential Equations Lecture Work Solutions 140 - 2. ut =...

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2. u t = u xx + e t 0 <x<π, t> 0 u ( x, 0) = cos 2 x 0 <x<π u x (0 ,t )= u x ( π, t )=0 Since the boundary conditions are homogeneous we can immediately expand u ( x, t ), the right hand side and the initial temperature distribution in terms of the eigenfunctions. These eigenfunctions are φ n =co s nx λ = n 2 n =0 , 1 , 2 , ... u ( x, t )= 1 2 u 0 ( t )+ X n =1 u n ( t )cos nx u ( x, 0) = 1 2 u 0 (0) + X n =1 u n (0) cos nx =co s2 x . . . by initial condition u n (0) = 0 n 6 =2 u 2 (0) = 1 e t = X n =1 s n ( t )cos nx + 1 2 s 0 ( t ) s n ( t )= R π 0 e t cos nx dx R π 0 cos 2 nx dx = e t R π 0 cos
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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