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Differential Equations Lecture Work Solutions 169

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

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1b. y λ 2 y = 0 0 < x < a y (0) = 0 , y ( a ) = 1 case 1: λ = 0 y = 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 = 0 The solution is y = Ae λx + Be λx or y = C cosh λx + D sinh λx The first 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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