Differential Equations Lecture Work Solutions 22

Differential Equations Lecture Work Solutions 22 - 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 (0) = A =0 and the second condition X ( L )= B =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 (0) = λB 1 Thus B 1 = 0 and the solution is X = A 1 cos λx Now use the condition at L X ( L )= A 1 cos λL =0 If we take A
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