Unformatted text preview: λL = 0 If we take B 1 = 0, we get a trivial solution, but we have another choice, namely √ λ cos √ λ L + sin √ λL = 0 If cos √ λL = 0 then we are left with sin √ λL = 0 which is not possible (the cosine and sine functions do not vanish at the same points). Thus cos √ λL 6 = 0 and upon dividing by it we get − √ λ = tan √ λL This can be solved graphically or numerically (see ±gure). The points of intersection are values of √ λ n . The solution is then depending on n , and obtained by substituting λ n X n ( x ) = sin ± λ n x 24...
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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.
- Fall '08