By Mejlbro L.

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Its derivative is cL > 1 for u = 0, and it decreases towards 0 for u increasing. Hence this curve has precisely one intersection with the curve z = u, which again means that the curve given by (4) has precisely one solutionu = αL. We have now proved that if cL > 1, then there is just one negative eigenvalue λ = −α 2 , where α = u/L, and where u is the unique positive solution of (4). A corresponding generating eigenfunction is yα (x) = sinh(αx). Now put cL = 6, so (4) is written u = 6 tanh u, u > 0.

2) Prove that λ = 0 is an eigenvalue and ﬁnd a corresponding eigenfunction. 3) Prove that the remaining eigenfunctions are given by yn (x) = sin αn x, where αn is the n-th positive root of the equation tan z = z. Sketch the roots. 1) Put λ = −k 2 < 0, where k > 0. • The complete solution. The characteristic equation R2 + λ = R2 − k 2 = (R − k)(R + k) = 0 has the solutions R = ±k, and the diﬀerential equation is homogeneous, so the complete solution is y(x) = c1 cosh(kx) + c2 sinh(kx) where y (x) = c1 k sinh(kx) + c2 k cosh(kx).

2) Prove that every λ ∈ R is an eigenvalue for the eigenvalue problem under consideration, and that y = ex , x ∈ [0, 1], is a corresponding eigenfunction. 1) We get by insertion of λ = −2 that d2 y dy −2 + y = 0, dx2 dx x ∈ [0, 1], with the characteristic polynomial R2 − 2R + 1 = (R − 1)2 . Since the root R = 1 has multiplicity 2, the complete solution is y = c1 ex + c2 xex where y = (c1 + c2 )ex + c2 xex . wanted: ambitious people Please click the advert At NNE Pharmaplan we need ambitious people to help us achieve the challenging goals which have been laid down for the company.