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Differential Equations Lecture Work Solutions 170

Differential Equations Lecture Work Solutions 170 - 1d y 2...

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1d. y + λ 2 y = 0 0 < x < a y (0) = 1 , y ( a ) = 0 case 1: λ = 0 y = 0 implies y = Ax + B. Using the boundary conditions, we get y (0) = B = 1, and y ( a ) = A = 0. Thus the solution in this case is y ( x ) = B case 2: λ = 0 The solution is y = C cos λx + D sin λx The first boundary condition gives C = 1. Now apply the other boundary condition λ sin λa + λD cos λa = 0 Suppose sin λa = 0, then this implies D = tan λa (since
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