Differential Equations Lecture Work Solutions 174

Differential Equations Lecture Work Solutions 174 - 4c

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4c. y 0 + λy =0 π<x<π y 0 ( π )=0 y 0 ( π )=0 case 1: λ = 0, the solution is y = Ax + B The boundary conditions give A =0 The solution is then y = B for λ =0. case 2: λ< 0, the solution is trivial case 3: λ> 0, the solution is y = A cos λx + B sin λx DiFerentiate and use the boundary conditions, we get λ ( A sin λπ + B cos λπ )=0 λ ( A sin λπ + B cos λπ )=0 To solve the homogeneous system (we can drop the factor λ since it is not zero), we must have the determinat equals zero sin λπ cos λπ sin λπ cos λπ =0 2sin λπ cos λπ
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