Differential Equations Lecture Work Solutions 169

Differential Equations Lecture Work Solutions 169 - 1b. y 2...

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1b. y 0 λ 2 y =0 0 <x<a y (0) = 0 ,y ( a )=1 case 1: λ =0 y 0 =0 implies y = Ax + B. Using the boundary conditions, we get y (0) = B =0 ,and y ( a )= Aa + B =1 ,or A = 1 a . Thus the solution in this case is y ( x )= 1 a x case 2: λ 6 =0 The solution is y = Ae λx + Be λx or y = C cosh λx + D sinh λx The frst boundary condition gives C = 0. Now apply the other boundary condition D sinh λa =1 This implies D = 1 sinh λa . The problem is when
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