Differential Equations Lecture Work Solutions 19

Differential Equations Lecture Work Solutions 19 - Case 2:...

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Case 2: λ =0 In this case we have a double root r =0andasweknowfromODEstheso lut ionis X = Ax + B The boundary condition at zero yields X (0) = B =0 and the second condition X 0 ( L )= A =0 This implies that B = A = 0 and therefore we have again 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 DiFerentiate X 0 = λA 1 sin λx + λB 1 cos λx Substitute the boundary condition at zero X (0) = A 1 Thus A
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