1a. y0 − λ 2 y =00 <x<a y0 (0) = y0 ( a )=0 case 1: λ =0 y0 =0 implies y = Ax + B. Using the boundary conditions, we get y0 ( x )= A = 0. Thus the solution in this case is y ( x )= B case 2: λ 6 =0 The solution is y = Ae λx + Be − λx or y = C cosh λx + D sinh λx To use the boundary conditions we need y0 ( x )= Cλ sinh λx + Dλ cosh λx The condition y0 (0) = 0 implies λD = 0 and since in this case λ 6 =0 ,wehave D =0
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Boundary value problem, Boundary conditions, sinh nπi, cosh λx, Dλ cosh λx