Differential Equations Lecture Work Solutions 170

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

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1d. y 0 + λ 2 y =0 0 <x<a y (0) = 1 ,y 0 ( a )=0 case 1: λ =0 y 0 =0 implies y = Ax + B. Using the boundary conditions, we get y (0) = B =1 ,and y 0 ( a )= A =0 . Thu sthe solution in this case is y ( x )= B case 2: λ 6 =0 The solution is y = C cos λx + D sin λx The frst boundary condition gives C = 1. Now apply the other boundary condition λ sin λa + λD cos λa =0 Suppose sin λa 6 = 0, then this implies D =tan λa (since λ
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