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Chem Differential Eq HW Solutions Fall 2011 97

# Chem Differential Eq HW Solutions Fall 2011 97 - Section...

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Section 6.2 Sturm-Liouville Theory 97 Since α = 0, it follows that sinh α = 0 and this implies that 1 - α 2 = 0 or α = ± 1. We take α = 1, because the value - 1 does not yield any new eigenfunctions. For α = 1, the corresponding solution is X = c 1 cosh x + c 2 sinh x = - c 2 cosh x + c 2 sinh x, because c 1 = - αc 2 = - c 2 . So in this case we have one negative eigenvalue λ = - α 2 = - 1 with corresponding eigenfunction X = cosh x - sinh x . Case III If λ = α 2 > 0, then the general solution of the differential equation is X = c 1 cos αx + c 2 sin αx. We have X = - c 1 α sin αx + c 2 α cos αx . In order to have nonzero solutions, one of the coefficients c 1 or c 2 must be = 0. Using the boundary conditions, we obtain c 1 + αc 2 = 0 c 1 (cos α - α sin α ) + c 2 (sin α + α cos α ) = 0 The first equation implies that c 1 = - αc 2 and so both c 1 and c 2 are neq 0. From the second equation, we obtain - αc 2 (cos α - α sin α ) + c 2 (sin α + α cos α ) = 0 - α (cos α - α sin α ) + (sin α + α cos α ) = 0 sin α ( α 2 + 1) = 0 Since α 2 + 1 = 0, then sin α = 0, and so α = , where n = 1 , 2 ,... . Thus the
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• Fall '11
• StuartChalk
• Sin, Boundary value problem, Eigenvalue, eigenvector and eigenspace, Eigenfunction, Sturm–Liouville theory, c2 sin

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