Unformatted text preview: μx ) + c 2 sin( μx ). Then, y =μc 1 sin( μx ) + μc 2 cos( μx ). So, y (0) = μc 2 = 0 i.e. c 2 = 0. Also, y ( π ) = μc 1 sin( μπ ) = 0. So, μπ = nπ . i.e. μ = n . Thus, eigenvelues are λ n = n 2 with corresponding eigenfunctions y n = cos( nx ) for n = 1 , 2 , 3 ,.... Case3:: λ =μ 2 < ,r = ± μ : y = c 1 cosh( μx )+ c 2 sinh( μx ). Then, y = μc 1 sinh( μx )+ μc 2 cosh( μx ). So, y (0) = μc 2 = 0 i.e. c 2 = 0. Also, y ( π ) = μc 1 sinh( μπ ) = 0. This implies that sinh( μπ ) = 0.i.e. μ = 0. Not possible. Thus c 1 = 0. Then the solution is y = 0 . There is no nontrivial solution....
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 Spring '07
 COSTIN
 Math, Differential Equations, Equations, Partial Differential Equations, Sin, Boundary value problem, Eigenvalue, eigenvector and eigenspace, Eigenfunction, c2 sin

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