Differential Equations Lecture Work Solutions 158

Differential Equations Lecture Work Solutions 158 - t s n (...

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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) 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) = 0 = D n c L D n =0 v n ( t )= v n (0) cos cnπ L t + Z
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Unformatted text preview: t s n ( ) sin cn L ( t ) cn L d Now that we have all the coecients in the expansion of v , recall that u = v + w . 158...
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