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