Differential Equations Lecture Work Solutions 155

Differential Equations Lecture Work Solutions 155 - vn vn(0...

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˙ v n (0) = R L 0 x + x 2 2 L cos L xdx R L 0 cos 2 L xdx n =0 , 1 , 2 , ··· ¨ v n + L 2 c 2 v n = s n ( t ) ¨ v 0 = s 0 ( t ) The solution of the ODE for n = 0 is obtained by integration twice and using the initial conditions v 0 ( t )= Z t 0 Z ξ 0 s 0 ( τ ) ! + v 0 (0) + t ˙ v 0 (0) v n = C n cos c L t + D n sin c L t + Z t 0 s n ( τ ) sin c L ( t τ ) c L v n (0) = C n ˙ v n (0) = D n c L v n ( t )= v n (0) cos cnπ L t + L ˙
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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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