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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