Differential Equations Lecture Work Solutions 86

Differential Equations Lecture Work Solutions 86 - The...

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The eigenfunctions and eigenvalues are sin n +1 / 2 L πx n / 2 L π ! 2 ,n =1 , 2 ,... Thus, we can expand v ( x, t )= X n =1 v n ( t )sin n / 2 L At t =0wehave x 1= v ( x, 0) = X n =1 v n (0) sin n / 2 L so v n (0) = 2 L Z L 0 ( x +1)sin n / 2 L πxdx Substitute the expansion is the PDE and equate coefficients ˙ v n ( t )+ k n / 2 L π ! 2 v n ( t )=0 v n (0) = 2 L Z L 0 ( x n / 2 L πxdx The solution is then v n ( t v n (0) e k ( n +1 / 2 L π ) 2 t and v ( x, t X n =1
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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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