Unformatted text preview: „x H f L = ï 2 C 2 è!!! l Sin I è!!! l p M = Hence the only way we can satisfy both requirements (55) and (6) and not have a trivial solution is (7) l = n 2 , n = 1, 2, 3, … Thus the solution is (8) f H x L = f H x p L = C 1 Sin @ n H x p LD + C 2 Cos @ n H x p LD = C 1 H1 L n Sin H n x L + C 2 H1 L 2 Cos H n x L which can be expressed as (9) f H x L = C 1 ' Sin H n x L + C 2 ' Cos H n x L...
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 Fall '09
 Stroeve
 Trigraph, C1 sin, Sin Hn xL, Cos Hn xL

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