Case 2: λ =0 In this case we have a double root r =0andasweknowfromODEstheso lut ionis X = Ax + BThe boundary condition at zero yields X 0 (0) = A =0and the second condition X 0 ( L )= A =0 This implies that A = 0 and therefore we have no restriction on B . Thusinth iscasethe solution is a constant and we take X ( x)=1 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
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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.