The boundary conditions require that 0 c1 and using

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Unformatted text preview: 1 e⌥x + C2 xe⌥x . The boundary conditions require that 0 = C1 and (using the fact that C1 = 0) 0 = C 2 e ⌥4 . This gives C2 = 0, so again, the only solution is the trivial solution y = 0. Case III: |k | < 4 p Under this assumption m = k ± 16 k 2 i, h i p p so y = e kx C1 cos 16 k 2 x + C2 sin 16 k 2 x . Enforcing the boundary conditions, we find that 0 = C1 p and 0 = C2 e k sin 16 k 2 . We conclude that either C2 = 0 (which gives us the trivial solution again) p or sin 16 k 2 = 0. This occurs if p 16 k 2 = n⇡ , p where n is any integer. Solving gives k = ± p 16 n2 ⇡ 2 , and we can see that there are only two eigenvalues: k = ± 16 ⇡ 2 (choosing n = 0 violates our assumption that |k | < 4, so we can have only n = 1). The corresponding eigenfunctions are y = Ce⌥ p 16 ⇡ 2 x sin(⇡ x). AMATH 350 Assignment #6 Solutions - Solutions AMATH 350 Assignment 5 Fall 2012 Page Fall 2010 3 c S.A. Campbell 2010 Consider 2.1.Consider y 00 + a1 (x)y 0 + a2 (x)y = f (x) z 00 + b1 (x)z 0 + b2 (x)z =...
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