Differential Equations Lecture Work Solutions 15

# Differential Equations Lecture Work Solutions 15 - and the...

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and the second condition X ( π )= =0 This implies that A = 0 and therefore we again have a trivial solution. Case 3: λ> 0 In this case the two roots are imaginary r = ± i λ Thus the solution is a combination of sine and cosine X = A 1 cos λx + B 1 sin λx Substitute the boundary condition at zero X (0) = A 1 Thus A 1 = 0 and the solution is X = B 1 sin λx Now use the condition at π X ( π )= B 1 sin λπ =0 If we take B 1 = 0, we get a trivial solution, but we have another choice, namely sin λπ =0 This implies that the argument of the sine function is a multiple of π ± λ n π =
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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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