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Differential Equations Lecture Work Solutions 168

Differential Equations Lecture Work Solutions 168 - 1a y 2...

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1a. y 0 λ 2 y =0 0 <x<a y 0 (0) = y 0 ( a )=0 case 1: λ =0 y 0 =0 implies y = Ax + B. Using the boundary conditions, we get y 0 ( 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 y 0 ( x )= sinh λx + cosh λx The condition y 0 (0) = 0 implies λD = 0 and since in this case λ 6 =0 ,wehave D =0
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